JEE Statistics Questions

Given ten observations $$x_1, x_2, \ldots, x_{10}$$ with:

$$\sum_{i=1}^{10} (x_i - 2) = 30, \quad \sum_{i=1}^{10} (x_i - \beta)^2 = 98, \quad \beta > 2,$$

and variance $$\frac{4}{5}$$. We need to find $$\frac{\beta \mu}{\sigma^2}$$, where $$\mu$$ and $$\sigma^2$$ are the mean and variance of the new data set $$y_i = 2(x_i - 1) + 4\beta$$.

First, recall that variance is given by $$\frac{1}{n} \sum (x_i - \bar{x})^2$$. With $$n = 10$$ and variance $$\frac{4}{5}$$:

$$\frac{1}{10} \sum_{i=1}^{10} (x_i - \bar{x})^2 = \frac{4}{5}$$

Multiplying both sides by 10:

$$\sum_{i=1}^{10} (x_i - \bar{x})^2 = 8 \quad \text{(Equation 1)}$$

From $$\sum_{i=1}^{10} (x_i - 2) = 30$$:

$$\sum_{i=1}^{10} (x_i - 2) = \sum x_i - 20 = 30$$

So:

$$\sum x_i = 50 \quad \text{(Equation 2)}$$

The mean $$\bar{x}$$ is:

$$\bar{x} = \frac{50}{10} = 5 \quad \text{(Equation 3)}$$

Using Equation 1 and $$\bar{x} = 5$$:

$$\sum_{i=1}^{10} (x_i - 5)^2 = \sum (x_i^2 - 10x_i + 25) = \sum x_i^2 - 10 \sum x_i + 250 = 8$$

Substituting $$\sum x_i = 50$$:

$$\sum x_i^2 - 10 \times 50 + 250 = 8 \implies \sum x_i^2 - 500 + 250 = 8 \implies \sum x_i^2 = 258 \quad \text{(Equation 4)}$$

Given $$\sum_{i=1}^{10} (x_i - \beta)^2 = 98$$:

$$\sum (x_i^2 - 2\beta x_i + \beta^2) = \sum x_i^2 - 2\beta \sum x_i + 10\beta^2 = 98$$

Substituting Equations 2 and 4:

$$258 - 2\beta \times 50 + 10\beta^2 = 98 \implies 258 - 100\beta + 10\beta^2 = 98$$

Rearranging:

$$10\beta^2 - 100\beta + 160 = 0 \implies \beta^2 - 10\beta + 16 = 0$$

Solving the quadratic equation:

Discriminant $$D = (-10)^2 - 4 \times 1 \times 16 = 100 - 64 = 36$$

$$\beta = \frac{10 \pm \sqrt{36}}{2} = \frac{10 \pm 6}{2}$$

So $$\beta = 8$$ or $$\beta = 2$$. Given $$\beta > 2$$, we have $$\beta = 8$$.

Now, define the new data set:

$$y_i = 2(x_i - 1) + 4\beta = 2x_i - 2 + 4\beta$$

Substituting $$\beta = 8$$:

$$y_i = 2x_i - 2 + 4 \times 8 = 2x_i + 30$$

So $$y_i = 2x_i + 30$$.

For a linear transformation $$y_i = a x_i + b$$, the new mean $$\mu = a \bar{x} + b$$ and new variance $$\sigma^2 = a^2 \times (\text{original variance})$$.

Here, $$a = 2$$, $$b = 30$$, $$\bar{x} = 5$$, original variance $$= \frac{4}{5}$$.

Thus:

$$\mu = 2 \times 5 + 30 = 10 + 30 = 40$$

$$\sigma^2 = (2)^2 \times \frac{4}{5} = 4 \times \frac{4}{5} = \frac{16}{5}$$

Now compute:

$$\frac{\beta \mu}{\sigma^2} = \frac{8 \times 40}{\frac{16}{5}} = \frac{320}{\frac{16}{5}} = 320 \times \frac{5}{16} = \frac{1600}{16} = 100$$

Therefore, $$\frac{\beta \mu}{\sigma^2} = 100$$.

Mean = 5.5 for 10 values, $$\sum x_i^2 = 371$$. Two values 4,5 should be 6,8. Find corrected variance.

The original sum of the observations is $$\sum x_i = 10 \times 5.5 = 55$$, and after replacing 4 and 5 with 6 and 8 the corrected sum becomes $$55 - 4 - 5 + 6 + 8 = 60$$ so that the corrected mean is $$60/10 = 6$$.

The original sum of squares is 371, and after removing 16 and 25 and adding 36 and 64 the corrected sum of squares becomes $$371 - 16 - 25 + 36 + 64 = 430$$.

The corrected variance is then given by $$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{430}{10} - 36 = 43 - 36 = 7$$.

The correct answer is Option 3: 7.

The eight observations are $$6,\,4,\,a,\,8,\,b,\,12,\,10,\,13$$. Number of observations $$n = 8$$ and the given mean (arithmetic average) is $$\mu = 9$$.

Formula for the mean: $$\mu = \dfrac{\Sigma x}{n}$$. Hence $$\Sigma x = n\mu = 8 \times 9 = 72$$.

Adding the six known numbers: $$6 + 4 + 8 + 12 + 10 + 13 = 53$$. Therefore $$a + b = 72 - 53 = 19$$ $$-(1)$$

The variance (population form) is given to be $$\sigma^{2} = 9.25$$. For a population, $$\sigma^{2} = \dfrac{\Sigma(x - \mu)^2}{n} = \dfrac{\Sigma x^{2}}{n} - \mu^{2}$$.

Substituting the known values:
$$9.25 = \dfrac{\Sigma x^{2}}{8} - 9^{2}$$

Re-arranging:
$$\dfrac{\Sigma x^{2}}{8} = 9.25 + 81 = 90.25$$
$$\Sigma x^{2} = 90.25 \times 8 = 722$$

Squares of the six known numbers:
$$6^{2} = 36,\;4^{2} = 16,\;8^{2} = 64,\;12^{2} = 144,\;10^{2} = 100,\;13^{2} = 169$$
Sum $$= 36 + 16 + 64 + 144 + 100 + 169 = 529$$.

Hence
$$a^{2} + b^{2} = 722 - 529 = 193$$ $$-(2)$$

Square of $$a + b$$ from $$(1)$$:
$$(a + b)^{2} = 19^{2} = 361$$
But $$(a + b)^{2} = a^{2} + b^{2} + 2ab$$.
Substituting $$(2)$$:
$$361 = 193 + 2ab$$
$$2ab = 361 - 193 = 168$$
$$ab = 84$$.

Required expression:
$$a + b + ab = 19 + 84 = 103$$.

Therefore, the correct value is $$103$$   ⟹   Option B.

The given information can be arranged in a two-row table.

Values (x): 4 5 8 9 6 12 11
Frequencies (f): 5 $$f_1$$ $$f_2$$ 2 1 1 3

Adding all the frequencies must give 19:

$$5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 19 \;\Longrightarrow\; f_1 + f_2 = 7 \qquad -(1)$$

The median is stated to be 6. Since the total number of observations is 19 (an odd number), the median is the 10-th item in the ordered data.

Arranging the values in ascending order:
4 (5), 5 ($$f_1$$), 6 (1), 8 ($$f_2$$), 9 (2), 11 (3), 12 (1)

Cumulative frequency just before 6 is $$5 + f_1$$ and including the single 6 it becomes $$6 + f_1$$. To make the 10-th observation land on the value 6 we require

$$5 + f_1 \lt 10 \le 6 + f_1 \;\Longrightarrow\; f_1 = 4$$

Using $$f_1 + f_2 = 7$$ from $$(1)$$ gives $$f_2 = 3$$.

The completed frequency table is

4(5), 5(4), 6(1), 8(3), 9(2), 11(3), 12(1)   with   $$\sum f = 19$$.

1. Mean $$\mu$$
$$\sum fx = 4\cdot5 + 5\cdot4 + 6\cdot1 + 8\cdot3 + 9\cdot2 + 11\cdot3 + 12\cdot1 = 133$$
$$\mu = \frac{133}{19} = 7$$

2. Mean deviation about the mean $$\alpha$$

$$\sum f|x-\mu| = 15 + 8 + 1 + 3 + 4 + 12 + 5 = 48$$
$$\alpha = \frac{48}{19}\;\Longrightarrow\; 19\alpha = 48$$

3. Mean deviation about the median $$\beta$$ (median = 6)

$$\sum f|x-6| = 10 + 4 + 0 + 6 + 6 + 15 + 6 = 47$$
$$\beta = \frac{47}{19}\;\Longrightarrow\; 19\beta = 47$$

4. Variance $$\sigma^{2}$$

$$\sum f(x-\mu)^2 = 45 + 16 + 1 + 3 + 8 + 48 + 25 = 146$$
$$\sigma^{2} = \frac{146}{19}\;\Longrightarrow\; 19\sigma^{2} = 146$$

5. Expression $$7f_1 + 9f_2$$
$$7f_1 + 9f_2 = 7\cdot4 + 9\cdot3 = 28 + 27 = 55$$

Matching with List-II
(P) $$7f_1 + 9f_2 = 55$$ → (5)
(Q) $$19\alpha = 48$$ → (3)
(R) $$19\beta = 47$$ → (2)
(S) $$19\sigma^{2} = 146$$ → (1)

Thus the correct matching is
(P)→(5), (Q)→(3), (R)→(2), (S)→(1)

Option C which is: (P)→(5), (Q)→(3), (R)→(2), (S)→(1)

We need to find the variance of $$x$$, where $$x$$ is the number of defective oranges when 2 oranges are drawn from a lot of 3 defective and 7 good oranges.

For a hypergeometric distribution (sampling without replacement), the variance is given by $$\text{Var}(x) = E(x^2) - [E(x)]^2$$. Alternatively, we can compute it directly from the probability distribution of $$x$$.

First, we find the probability distribution of $$x$$. The total number of ways to choose 2 from 10 is $$\binom{10}{2} = 45$$.

For $$x = 0$$, both oranges are good, so $$P(x=0) = \frac{\binom{7}{2}}{\binom{10}{2}} = \frac{21}{45} = \frac{7}{15}$$.

For $$x = 1$$, one defective and one good, so $$P(x=1) = \frac{\binom{3}{1}\binom{7}{1}}{\binom{10}{2}} = \frac{3 \times 7}{45} = \frac{21}{45} = \frac{7}{15}$$.

For $$x = 2$$, both defective, so $$P(x=2) = \frac{\binom{3}{2}}{\binom{10}{2}} = \frac{3}{45} = \frac{1}{15}$$.

Next, the expected value of $$x$$ is calculated as $$E(x) = 0 \times \frac{7}{15} + 1 \times \frac{7}{15} + 2 \times \frac{1}{15} = 0 + \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5}$$.

Then, we compute $$E(x^2) = 0^2 \times \frac{7}{15} + 1^2 \times \frac{7}{15} + 2^2 \times \frac{1}{15} = 0 + \frac{7}{15} + \frac{4}{15} = \frac{11}{15}$$.

Finally, substituting these values into the variance formula gives $$\text{Var}(x) = E(x^2) - [E(x)]^2 = \frac{11}{15} - \left(\frac{3}{5}\right)^2 = \frac{11}{15} - \frac{9}{25} = \frac{11 \times 5}{75} - \frac{9 \times 3}{75} = \frac{55}{75} - \frac{27}{75} = \frac{28}{75}$$.

The correct answer is Option (1): 28/75.

Since the coin is fair, each toss has two equally likely outcomes: head (H) or tail (T). With three tosses, the sample space has $$2^3 = 8$$ outcomes, each with probability $$\frac{1}{8}$$. We list all outcomes and determine $$X$$ for each by counting the number of times a tail immediately follows a head in consecutive tosses.

The outcomes and corresponding $$X$$ values are:

• HHH: Pairs (H,H) and (H,H) → No tail follows head → $$X = 0$$
• HHT: Pair (H,H) → No; Pair (H,T) → Yes → $$X = 1$$
• HTH: Pair (H,T) → Yes; Pair (T,H) → No → $$X = 1$$
• HTT: Pair (H,T) → Yes; Pair (T,T) → No → $$X = 1$$
• THH: Pair (T,H) → No; Pair (H,H) → No → $$X = 0$$
• THT: Pair (T,H) → No; Pair (H,T) → Yes → $$X = 1$$
• TTH: Pair (T,T) → No; Pair (T,H) → No → $$X = 0$$
• TTT: Pairs (T,T) and (T,T) → No → $$X = 0$$

Tabulating the values:

• $$X = 0$$ for HHH, THH, TTH, TTT → 4 outcomes
• $$X = 1$$ for HHT, HTH, HTT, THT → 4 outcomes

Note that $$X$$ can only be 0 or 1 because it is impossible to have two occurrences (as that would require overlapping pairs like (H,T) and (T,H), which conflict on the middle toss).

The probability mass function is:

• $$P(X=0) = \frac{4}{8} = \frac{1}{2}$$
• $$P(X=1) = \frac{4}{8} = \frac{1}{2}$$

The mean $$\mu$$ is the expected value of $$X$$. The formula for expected value is $$\mu = E(X) = \sum x_i \cdot P(X = x_i)$$.

So,

$$\mu = 0 \cdot P(X=0) + 1 \cdot P(X=1) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$$

The variance $$\sigma^2$$ is given by $$\sigma^2 = E(X^2) - [E(X)]^2$$. First, compute $$E(X^2)$$:

$$E(X^2) = 0^2 \cdot P(X=0) + 1^2 \cdot P(X=1) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$$

Then,

$$\sigma^2 = E(X^2) - [E(X)]^2 = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$

Now, compute $$\mu + \sigma^2$$:

$$\mu + \sigma^2 = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$$

Then, $$64(\mu + \sigma^2) = 64 \cdot \frac{3}{4} = 64 \cdot 0.75 = 48$$.

Thus, the value is 48, which corresponds to option D.

Let the eight observations be $$2,\,3,\,3,\,4,\,5,\,7,\,a,\,b$$.

Given mean $$\bar{x}=4$$.
For $$n=8$$ observations, the sum satisfies $$2+3+3+4+5+7+a+b = 8\times 4 = 32$$ $$24 + a + b = 32 \;\;\Longrightarrow\;\; a + b = 8$$ $$-(1)$$

The standard deviation is $$\sqrt{2}$$, so the variance is $$2$$.
Variance formula: for $$n$$ observations, $$\sigma^2=\dfrac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^2$$ $$-(2)$$

Compute the squared deviations of the known numbers from $$\bar{x}=4$$:
$$(2-4)^2=4,\;(3-4)^2=1,\;(3-4)^2=1,\;(4-4)^2=0,\;(5-4)^2=1,\;(7-4)^2=9$$ Sum of these six values $$=4+1+1+0+1+9=16$$.

Include the unknowns: Total squared deviation $$=16 + (a-4)^2 + (b-4)^2$$.

Using $$\sigma^2=2$$ and $$n=8$$ in $$(2)$$: $$\dfrac{16 + (a-4)^2 + (b-4)^2}{8}=2$$ $$16 + (a-4)^2 + (b-4)^2 =16$$ $$(a-4)^2 + (b-4)^2 =0$$.

Since a square is non-negative, both must be zero: $$a-4=0,\;b-4=0\;\;\Longrightarrow\;\; a=4,\;b=4$$.

The complete data set is $$2,\,3,\,3,\,4,\,4,\,4,\,5,\,7$$.

Mode: the value that occurs most often is $$4$$ (three times).

Mean deviation about the mode (M.D.) is $$\text{M.D.}=\dfrac{1}{n}\sum_{i=1}^{n}\lvert x_i - \text{mode}\rvert.$$

Compute the absolute deviations from $$4$$: $$|2-4|=2,\;|3-4|=1,\;|3-4|=1,\;|4-4|=0,\;|4-4|=0,\;|4-4|=0,\;|5-4|=1,\;|7-4|=3$$ Sum $$=2+1+1+0+0+0+1+3=8$$.

Therefore $$\text{M.D.}=\dfrac{8}{8}=1$$.

Hence, the mean deviation about the mode is $$1$$ (Option A).

We need to find the variance of the arithmetic progression 8, 21, 34, 47, ..., 320.

The first term is a = 8 and the common difference is d = 13. Since the last term is 320, we set $$a + (n-1)d = 320$$ which leads to $$8 + (n-1)(13) = 320$$ and hence $$(n-1) = \frac{312}{13} = 24$$ so that $$n = 25$$.

For an AP with n terms and common difference d, the variance is given by $$\sigma^2 = \frac{d^2(n^2 - 1)}{12}$$. Substituting d = 13 and n = 25 yields

$$\sigma^2 = \frac{13^2(25^2 - 1)}{12} = \frac{169 \times 624}{12} = 169 \times 52 = 8788$$.

The answer is 8788.

The five given observations are $$x_1 = 1,\; x_2 = 3,\; x_3 = a,\; x_4 = 7,\; x_5 = b$$ with $$a \gt b$$.

Their mean is given to be $$5$$, so

$$\frac{1 + 3 + a + 7 + b}{5} = 5 \; \Longrightarrow \; 11 + a + b = 25 \; \Longrightarrow \; a + b = 14 \; -(1)$$

The variance is given to be $$10$$. For a population of five values,

$$\text{Variance} = \frac{1}{5}\sum_{i=1}^{5}(x_i - \mu)^2 = 10,\quad \mu = 5.$$

Compute the three known squared deviations:

$$(1-5)^2 = 16,\; (3-5)^2 = 4,\; (7-5)^2 = 4.$$

Let $$S = (a-5)^2 + (b-5)^2$$. Then

$$\frac{16 + 4 + 4 + S}{5} = 10 \; \Longrightarrow \; 24 + S = 50 \; \Longrightarrow \; S = 26 \; -(2)$$

Expand $$S$$ using $$a + b = 14$$:

$$S = (a-5)^2 + (b-5)^2 = a^2 - 10a + 25 + b^2 - 10b + 25 \\ = (a^2 + b^2) - 10(a + b) + 50.$$

Since $$a^2 + b^2 = (a + b)^2 - 2ab = 196 - 2ab$$,

$$S = 196 - 2ab - 140 + 50 = 106 - 2ab.$$

Using $$S = 26$$ from $$(2)$$,

$$106 - 2ab = 26 \; \Longrightarrow \; 2ab = 80 \; \Longrightarrow \; ab = 40 \; -(3)$$

Equations $$(1)$$ and $$(3)$$ give the quadratic

$$t^2 - 14t + 40 = 0 \quad (\text{with roots } a,\,b).$$

Discriminant $$D = 196 - 160 = 36 \;\;\Longrightarrow\;\; t = \frac{14 \pm 6}{2} = 10 \text{ or } 4.$$

Since $$a \gt b$$, we have $$a = 10,\; b = 4.$$

The original data are therefore $$1,\,3,\,10,\,7,\,4.$$

Now form the new observations $$n + x_n$$ for $$n = 1,2,\ldots,5$$:

$$\begin{aligned} n=1 &: 1 + 1 = 2\\ n=2 &: 2 + 3 = 5\\ n=3 &: 3 + 10 = 13\\ n=4 &: 4 + 7 = 11\\ n=5 &: 5 + 4 = 9 \end{aligned}$$

Thus the new set is $$2,\,5,\,13,\,11,\,9.$$

Mean of the new set:

$$\bar{y} = \frac{2 + 5 + 13 + 11 + 9}{5} = \frac{40}{5} = 8.$$

Compute squared deviations from $$8$$:

$$ (2-8)^2 = 36,\; (5-8)^2 = 9,\; (13-8)^2 = 25,\; (11-8)^2 = 9,\; (9-8)^2 = 1.$$

Sum of squared deviations $$= 36 + 9 + 25 + 9 + 1 = 80.$$

Variance (population) $$= \frac{80}{5} = 16.$$

Therefore, the variance of the observations $$n + x_n$$ is $$16$$.

Option D is correct.

We are given a grouped frequency distribution with median = 14, median class 12-18, median class frequency $$f = 12$$, and cumulative frequency below 12 is $$F = 18$$.

$$\text{Median} = L + \left(\frac{N/2 - F}{f}\right) \times h$$

where $$L = 12$$ (lower boundary), $$F = 18$$ (cumulative frequency before median class), $$f = 12$$ (median class frequency), $$h = 6$$ (class width).

$$14 = 12 + \left(\frac{N/2 - 18}{12}\right) \times 6$$

$$2 = \frac{N/2 - 18}{2}$$

$$4 = N/2 - 18$$

$$N/2 = 22$$

$$N = 44$$

The correct answer is Option 3: 44.

The random variable $$X$$ can take four values, so we write its probability distribution:

$$P(X = 0) = p,\; P(X = 1) = p,\; P(X = 2) = q,\; P(X = 3) = q.$$

Step 1: Use the fact that total probability equals $$1$$.
$$2p + 2q = 1 \;$$ $$-(1)$$

Step 2: Compute the expected value $$E(X)$$.
Formula: $$E(X) = \sum x_i P(X = x_i).$$
$$E(X) = 0\cdot p + 1\cdot p + 2\cdot q + 3\cdot q = p + 5q.$$ $$-(2)$$

Step 3: Compute the second moment $$E(X^2).$$
Formula: $$E(X^2) = \sum x_i^2 P(X = x_i).$$
$$E(X^2) = 0^2\cdot p + 1^2\cdot p + 2^2\cdot q + 3^2\cdot q = p + 13q.$$ $$-(3)$$

Step 4: Apply the given condition $$E(X^2) = 2E(X).$$
Using $$(2)$$ and $$(3)$$:
$$p + 13q = 2(p + 5q).$$
Simplify:
$$p + 13q = 2p + 10q$$
$$0 = 2p + 10q - p - 13q = p - 3q$$
$$\Rightarrow\; p = 3q.$$ $$-(4)$$

Step 5: Solve for $$p$$ and $$q$$ with $$(1)$$ and $$(4).$$
Substitute $$p = 3q$$ into $$(1):$$
$$2(3q) + 2q = 1$$
$$6q + 2q = 1$$
$$8q = 1 \;\Rightarrow\; q = \frac{1}{8}.$$
Then from $$(4):$$
$$p = 3q = 3 \times \frac{1}{8} = \frac{3}{8}.$$

Step 6: Evaluate the required expression $$8p - 1.$$/> $$8p - 1 = 8 \times \frac{3}{8} - 1 = 3 - 1 = 2.$$

Hence $$8p - 1 = 2$$, which corresponds to Option B.

We are given five observations with mean $$\frac{24}{5}$$ and variance $$\frac{194}{25}$$, and the mean of the first four observations is $$\frac{7}{2}$$. We need to find the variance of the first four observations.

Since the mean of all five observations is $$\frac{24}{5}$$, their sum is $$\sum_{i=1}^{5}x_i = 5 \times \frac{24}{5} = 24$$, and because the mean of the first four observations is $$\frac{7}{2}$$, we have $$\sum_{i=1}^{4}x_i = 4 \times \frac{7}{2} = 14$$. It follows that the fifth observation is $$x_5 = 24 - 14 = 10$$.

Using the variance formula $$\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$$ for the five observations, we write $$\frac{194}{25} = \frac{\sum_{i=1}^{5}x_i^2}{5} - \left(\frac{24}{5}\right)^2$$. Since $$\left(\frac{24}{5}\right)^2 = \frac{576}{25}$$, this gives $$\frac{\sum_{i=1}^{5}x_i^2}{5} = \frac{194 + 576}{25} = \frac{770}{25}$$ and hence $$\sum_{i=1}^{5}x_i^2 = 5 \times \frac{770}{25} = 154$$.

For the first four observations we then have $$\sum_{i=1}^{4}x_i^2 = 154 - x_5^2 = 154 - 100 = 54$$. Their variance is therefore $$\sigma_4^2 = \frac{\sum_{i=1}^{4}x_i^2}{4} - \left(\frac{\sum_{i=1}^{4}x_i}{4}\right)^2 = \frac{54}{4} - \left(\frac{14}{4}\right)^2 = \frac{216}{16} - \frac{196}{16} = \frac{20}{16} = \frac{5}{4}$$.

Answer: Option 3 — $$\frac{5}{4}$$

Mean = 2: (-3+4+7-6+α+β)/6 = 2 → α+β = 10

Variance = 23: (1/6)∑(xi-2)² = 23

∑(xi-2)² = 25+4+25+64+(α-2)²+(β-2)² = 138

118 + (α-2)² + (β-2)² = 138

(α-2)² + (β-2)² = 20

With α+β=10: let α=2+a, β=2+b, then a+b=6 and a²+b²=20

2ab = 36-20 = 16, ab = 8. So a,b are roots of t²-6t+8=0 → t=2,4

α=4,β=6 or α=6,β=4

Mean deviation = (1/6)(|{-3}-2|+|4-2|+|7-2|+|{-6}-2|+|4-2|+|6-2|)

= (1/6)(5+2+5+8+2+4) = 26/6 = 13/3

The correct answer is Option 1: 13/3.

Total students = 40

Given frequencies:
5 + 8 + 5 + 12 + x + y = 40
$$\Rightarrow30+x+y=40$$
$$\Rightarrow x+y=10\quad...(1)$$
Cumulative frequencies:

• 15 → 5
• 16 → 13
• 17 → 18
• 18 → 30

So 20th and 21st observations lie in age 18 ⇒ median = 18

$$\text{MD}=\frac{\sum_{ }^{ }f|x-18|}{40}=1.25$$

$$\sum_{ }^{ }f|x-18|=1.25\times40=50$$

Now compute:
$$=5|15-18|+8|16-18|+5|17-18|+12|18-18|+x|19-18|+y|20-18|$$

= 5(3) + 8(2) + 5(1) + 12(0) + x(1) + y(2)
= 15 + 16 + 5 + x + 2y
= 36 + x + 2y
36 + x + 2y = 50
$$\Rightarrow x+2y=14\quad...(2)$$

Solve (1) & (2)
x + y = 10
x + 2y = 14

Subtract:
$$y=4\Rightarrow x=6$$

4x + 5y = 4(6) + 5(4) = 24 + 20 = 44

Given 10 observations with mean $$\bar{x} = \frac{6}{5}$$ and variance $$\sigma^2 = \frac{84}{25}$$. Also, $$\sum_{i=1}^{10} (x_i - \alpha) = 2$$ and $$\sum_{i=1}^{10} (x_i - \beta)^2 = 40$$, where $$\alpha$$ and $$\beta$$ are positive integers.

First, use the sum of deviations from $$\alpha$$ to find $$\alpha$$. The sum of deviations from a constant $$A$$ is given by $$\sum (x_i - A) = n(\bar{x} - A)$$. Here, $$n = 10$$, so:

$$\sum (x_i - \alpha) = 10 \left( \frac{6}{5} - \alpha \right) = 2$$

Solve for $$\alpha$$:

$$\frac{6}{5} - \alpha = \frac{2}{10} = \frac{1}{5}$$

$$\alpha = \frac{6}{5} - \frac{1}{5} = \frac{5}{5} = 1$$

Thus, $$\alpha = 1$$.

Next, use the variance to find $$\sum (x_i - \bar{x})^2$$. The variance formula is $$\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$$, so:

$$\sum (x_i - \bar{x})^2 = n \sigma^2 = 10 \times \frac{84}{25} = \frac{840}{25} = \frac{168}{5}$$

Now, use the given $$\sum (x_i - \beta)^2 = 40$$. The formula relating sum of squared deviations from $$\beta$$ to the sum of squared deviations from the mean is:

$$\sum (x_i - \beta)^2 = \sum (x_i - \bar{x})^2 + n (\bar{x} - \beta)^2$$

Substitute the known values:

$$40 = \frac{168}{5} + 10 \left( \frac{6}{5} - \beta \right)^2$$

Solve for $$\beta$$:

$$40 - \frac{168}{5} = 10 \left( \frac{6}{5} - \beta \right)^2$$

$$\frac{200}{5} - \frac{168}{5} = \frac{32}{5} = 10 \left( \frac{6}{5} - \beta \right)^2$$

$$\frac{32}{50} = \frac{16}{25} = \left( \frac{6}{5} - \beta \right)^2$$

Take the square root:

$$\left| \frac{6}{5} - \beta \right| = \frac{4}{5}$$

This gives two cases:

Case 1: $$\frac{6}{5} - \beta = \frac{4}{5}$$

$$\beta = \frac{6}{5} - \frac{4}{5} = \frac{2}{5}$$

But $$\beta$$ must be a positive integer, so $$\frac{2}{5}$$ is invalid.

Case 2: $$\frac{6}{5} - \beta = -\frac{4}{5}$$

$$\beta = \frac{6}{5} + \frac{4}{5} = \frac{10}{5} = 2$$

This is a positive integer, so $$\beta = 2$$.

Thus, $$\alpha = 1$$ and $$\beta = 2$$, so:

$$\frac{\beta}{\alpha} = \frac{2}{1} = 2$$

The ratio $$\frac{\beta}{\alpha}$$ is 2, which corresponds to option A.

We have a discrete frequency distribution:

• (x = c, 2c, 3c, 4c, 5c, 6c)
• (f = 2, 1, 1, 1, 1, 1)

Total frequency:
N = 2 + 1 + 1 + 1 + 1 + 1 = 7

$$\overline{x}=\frac{\sum_{ }^{ }fx}{N}$$

$$\sum_{ }^{ }fx=2(c)+1(2c)+1(3c)+1(4c)+1(5c)+1(6c)$$

= 2c + 2c + 3c + 4c + 5c + 6c = 22c

$$\overline{x}=\frac{22c}{7}$$

$$\text{Var}=\frac{\sum_{ }^{ }fx^2}{N}-\overline{x}^2$$

Compute:
$$\sum_{ }^{ }fx^2=2c^2+(2c)^2+(3c)^2+(4c)^2+(5c)^2+(6c)^2$$

$$=2c^2+4c^2+9c^2+16c^2+25c^2+36c^2$$

$$=92c^2$$

$$\text{Var}=\frac{92c^2}{7}-\left(\frac{22c}{7}\right)^2$$

$$=\frac{92c^2}{7}-\frac{484c^2}{49}$$

$$=\frac{644c^2-484c^2}{49}$$
$$=\frac{160c^2}{49}$$

$$\frac{160c^2}{49}=160$$

$$c^2=49\Rightarrow c=7$$

Step 1: Calculate the Cumulative Frequency ($$CF$$)

To find the median, we first construct a cumulative frequency table:

• Total Frequency ($$N$$): $$36$$

Step 2: Identify the Median Class

The median position is given by:

$$\frac{N}{2} = \frac{36}{2} = 18$$

Look for the first cumulative frequency that is greater than or equal to $$18$$. This value is $$22$$, which corresponds to the class interval $$8 - 12$$.

• Median Class: $$8 - 12$$

Step 3: Apply the Median Formula

The formula for the median of a grouped frequency distribution is:

$$M = L + \left( \frac{\frac{N}{2} - CF_{\text{pre}}}{f} \right) \times h$$

Where:

• $$L$$ (Lower limit of the median class) = $$8$$
• $$\frac{N}{2}$$ = $$18$$
• $$CF_{\text{pre}}$$ (Cumulative frequency of the preceding class) = $$12$$
• $$f$$ (Frequency of the median class) = $$10$$
• $$h$$ (Class width) = $$4$$

Substituting these values into the formula:

$$M = 8 + \left( \frac{18 - 12}{10} \right) \times 4$$  $$M = 8 + \left( \frac{6}{10} \right) \times 4$$  $$M = 8 + 2.4 = 10.4$$

Step 4: Calculate $$20M$$

Now, substitute the value of $$M$$ to find the final answer:

$$20M = 20 \times 10.4 = 208$$

Answer:

208

The median of the 7 observations is given as 170. Since there are 7 observations (an odd number), the median is the 4th observation when arranged in ascending order. The known observations are 125, 170, 190, 210, and 230. To have the median as 170, the 4th position must be 170, which requires at least three observations ≤ 170 and at least three observations ≥ 170.

Currently, there is one observation < 170 (125) and one observation = 170. There are three observations > 170 (190, 210, 230). Thus, there are two observations ≤ 170 (125 and 170) and four observations ≥ 170 (170, 190, 210, 230). To satisfy the median condition, both unknown observations $$a$$ and $$b$$ must be ≤ 170, ensuring exactly three observations ≤ 170 and three ≥ 170 (with the 4th being 170).

The mean deviation about the median is given as $$\frac{205}{7}$$. The formula for mean deviation about the median is:

$$\text{MD}(\text{median}) = \frac{1}{n} \sum_{i=1}^{n} |x_i - \text{median}|$$

Here, $$n = 7$$, median = 170, and $$\text{MD}(\text{median}) = \frac{205}{7}$$. Thus,

$$\sum_{i=1}^{7} |x_i - 170| = 205$$

The observations are 125, $$a$$, $$b$$, 170, 190, 210, 230. Since $$a \leq 170$$ and $$b \leq 170$$, $$|a - 170| = 170 - a$$ and $$|b - 170| = 170 - b$$. The absolute deviations are:

• $$|125 - 170| = 45$$
• $$|a - 170| = 170 - a$$
• $$|b - 170| = 170 - b$$
• $$|170 - 170| = 0$$
• $$|190 - 170| = 20$$
• $$|210 - 170| = 40$$
• $$|230 - 170| = 60$$

Summing these:

$$45 + (170 - a) + (170 - b) + 0 + 20 + 40 + 60 = 205$$

Simplifying the constants:

$$45 + 170 + 170 + 20 + 40 + 60 = 505$$

So,

$$505 - a - b = 205$$

$$-a - b = -300$$

$$a + b = 300$$

Next, find the mean deviation about the mean. First, compute the mean. The sum of the observations is:

$$125 + a + b + 170 + 190 + 210 + 230$$

Substituting $$a + b = 300$$:

$$125 + 300 + 170 + 190 + 210 + 230 = 1225$$

The mean $$\bar{x}$$ is:

$$\bar{x} = \frac{1225}{7} = 175$$

The mean deviation about the mean is:

$$\text{MD}(\text{mean}) = \frac{1}{7} \sum_{i=1}^{7} |x_i - 175|$$

The absolute deviations from 175 are:

• $$|125 - 175| = 50$$
• $$|a - 175| = 175 - a$$ (since $$a \leq 170 < 175$$)
• $$|b - 175| = 175 - b$$ (since $$b \leq 170 < 175$$)
• $$|170 - 175| = 5$$
• $$|190 - 175| = 15$$
• $$|210 - 175| = 35$$
• $$|230 - 175| = 55$$

Summing these:

$$50 + (175 - a) + (175 - b) + 5 + 15 + 35 + 55$$

Simplifying the constants:

$$50 + 175 + 175 + 5 + 15 + 35 + 55 = 510$$

So,

$$510 - a - b = 510 - 300 = 210$$

Thus, the mean deviation about the mean is:

$$\frac{210}{7} = 30$$

Therefore, the mean deviation about the mean is 30.

n = 20, mean = 10, σ = 2, so σ² = 4.

Σxᵢ = 200. Σxᵢ² = n(σ² + mean²) = 20(4 + 100) = 2080.

Corrected: Σxᵢ(new) = 200 - 8 + 12 = 204. Mean(new) = 204/20 = 10.2.

Σxᵢ²(new) = 2080 - 64 + 144 = 2160.

σ²(new) = 2160/20 - (10.2)² = 108 - 104.04 = 3.96.

σ(new) = √3.96.

The correct answer is Option (3): √3.96.

Given: $$\sum_{k=1}^{10} a_k = 50$$ and $$\sum_{k < j} a_k a_j = 1100$$.

We know that $$\left(\sum_{k=1}^{10} a_k\right)^2 = \sum_{k=1}^{10} a_k^2 + 2\sum_{k < j} a_k a_j$$

$$50^2 = \sum a_k^2 + 2(1100)$$

$$2500 = \sum a_k^2 + 2200$$

$$\sum a_k^2 = 300$$

Variance: $$\sigma^2 = \frac{\sum a_k^2}{n} - \left(\frac{\sum a_k}{n}\right)^2 = \frac{300}{10} - \left(\frac{50}{10}\right)^2 = 30 - 25 = 5$$

Standard deviation: $$\sigma = \sqrt{5}$$

The answer is $$\sqrt{5}$$, which corresponds to Option (2).

Mean=55

(a+b+68+44+48+60)/6=55. 

a+b=330-220=110.

Variance=194: Σ(xᵢ-55)²/6=194. Σ(xᵢ-55)²=1164.

(a-55)²+(b-55)²+(68-55)²+(44-55)²+(48-55)²+(60-55)²=1164.

(a-55)²+(b-55)²+169+121+49+25=1164. (a-55)²+(b-55)²=800.

Let a-55=p, b-55=q. p+q=110-110=0, so q=-p. p²+p²=800, p²=400, p=±20.

a>b: a=75, b=35. a+3b=75+105=180.

Original: $$n = 15$$, $$\bar{x} = 12$$, $$\sigma = 3$$, $$\sigma^2 = 9$$.

$$\sum x_i = 15 \times 12 = 180$$. $$\sum x_i^2 = n(\sigma^2 + \bar{x}^2) = 15(9 + 144) = 15 \times 153 = 2295$$.

Corrected: Replace 10 with 12.

$$\sum x_i' = 180 - 10 + 12 = 182$$. $$\mu = 182/15$$.

$$\sum x_i'^2 = 2295 - 100 + 144 = 2339$$.

$$\sigma^2 = \frac{2339}{15} - \left(\frac{182}{15}\right)^2 = \frac{2339}{15} - \frac{33124}{225} = \frac{35085 - 33124}{225} = \frac{1961}{225}$$.

$$15(\mu + \mu^2 + \sigma^2) = 15\mu + 15\mu^2 + 15\sigma^2$$

$$= 182 + 15\left(\frac{182}{15}\right)^2 + 15 \times \frac{1961}{225}$$

$$= 182 + \frac{182^2}{15} + \frac{1961}{15}$$

$$= 182 + \frac{33124 + 1961}{15} = 182 + \frac{35085}{15} = 182 + 2339 = 2521$$

The answer is $$\boxed{2521}$$.

Hypergeometric distribution: $$X$$ = number of defective items in sample of 5 from lot of 10 (3 defective).

Mean: $$E(X) = n \cdot ?\frac{K}{N} = 5 \cdot ?\frac{3}{10} = ?\frac{3}{2}$$

Variance: $$\sigma^2 = n \cdot ?\frac{K}{N} \cdot ?\frac{N-K}{N} \cdot ?\frac{N-n}{N-1} = 5 \cdot ?\frac{3}{10} \cdot ?\frac{7}{10} \cdot ?\frac{5}{9} = ?\frac{525}{900} = ?\frac{7}{12}$$

$$96\sigma^2 = 96 \times ?\frac{7}{12} = 56$$.

The answer is 56.

Sum $$= 65+68+58+44+48+45+60+\alpha+\beta+60 = 448+\alpha+\beta$$.

Mean=56: $$448+\alpha+\beta=560$$, $$\alpha+\beta=112$$.

Σx²=65²+68²+...+60²+α²+β²=sum of known squares+α²+β². 

Known: $$4225+4624+3364+1936+2304+2025+3600+3600=25678$$.

Variance=66.2: Σx²/10 - 56² = 66.2. Σx²=10(3136+66.2)=32022.

α²+β²=32022-25678=6344.

Calculate Mean ($$\mu$$) and Variance ($$\sigma^2$$)

• $$\mu = \frac{\sum f_i x_i}{\sum f_i} = \frac{176}{22} = 8$$

• $$\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2 = \frac{2048}{22} - 8^2 = \frac{1024}{11} - 64$$

• $$k = 93.09 - 64 = 29.09$$

• The value of $$[k] = \mathbf{29}$$

We have five observations $$9, 25, a, b, c$$ with $$a < b < c$$ and $$a, b, c \in \mathbb{N}$$. Since the mean is 18,

$$\frac{9 + 25 + a + b + c}{5} = 18 \implies a + b + c = 56 \quad (i)$$

The mean deviation about the mean is 4, so

$$\frac{|9-18| + |25-18| + |a-18| + |b-18| + |c-18|}{5} = 4.$$

Evaluating the known terms gives

$$9 + 7 + |a-18| + |b-18| + |c-18| = 20 \implies |a-18| + |b-18| + |c-18| = 4 \quad (ii)$$

The variance is $$\tfrac{136}{5}$$, hence

$$\frac{(9-18)^2 + (25-18)^2 + (a-18)^2 + (b-18)^2 + (c-18)^2}{5} = \frac{136}{5}.$$

Since $$(9-18)^2 + (25-18)^2 = 81 + 49 = 130$$, it follows that

$$130 + (a-18)^2 + (b-18)^2 + (c-18)^2 = 136 \implies (a-18)^2 + (b-18)^2 + (c-18)^2 = 6 \quad (iii)$$

Let $$p = a-18,\; q = b-18,\; r = c-18$$. Then from (i), (ii), (iii) we get

$$p+q+r = 56 - 54 = 2,\quad |p|+|q|+|r| = 4,\quad p^2+q^2+r^2 = 6\,. $$

Because $$p+q+r=2$$ and $$|p|+|q|+|r|=4$$, the total negative contribution is $$\tfrac{4-2}{2}=1$$, so exactly one of $$p,q,r$$ equals $$-1$$ and the other two are positive summing to 3. Substituting one variable as $$-1$$ in the sum of squares gives

$$1 + q^2 + r^2 = 6 \implies q^2 + r^2 = 5,\quad q + r = 3\,. $$

Writing $$r = 3 - q$$ yields

$$q^2 + (3-q)^2 = 5 \implies 2q^2 - 6q + 9 = 5 \implies q^2 - 3q + 2 = 0 \implies q = 1 \text{ or } 2\,. $$

Thus $$(p,q,r) = (-1,1,2)$$ or $$(-1,2,1)$$. The ordering $a

$$a = 17,\quad b = 19,\quad c = 20\,. $$

Finally,

$$2a + b - c = 2(17) + 19 - 20 = 33\,. $$

The answer is $$\boxed{33}$$.

Total probability = 1: $$\frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1 \implies K = \frac{1}{4}$$.

 Mean ($$\mu$$): $$\sum X \cdot P(X) = \frac{\alpha}{3} - \frac{1}{2}$$.

 $$E(X^2)$$: $$\sum X^2 \cdot P(X) = \frac{\alpha^2}{3} + \frac{5}{2}$$

 Using $$\sigma = \mu + 2 \implies \sigma^2 = \mu^2 + 4\mu + 4$$, and substituting $$\sigma^2 = E(X^2) - \mu^2$$ yields $$\alpha = 6$$.

 $$\mu = \frac{3}{2}$$ and $$\sigma = \frac{7}{2}$$.

 $$\sigma + \mu = \frac{7}{2} + \frac{3}{2} = 5$$

The six distinct observations are $$x_i=3,\,4,\,5,\,8,\,10,\,11$$ with corresponding frequencies $$f_i=5,\,4,\,4,\,2,\,2,\,3$$.

Total number of observations
$$N=\sum f_i = 5+4+4+2+2+3 = 20$$

Case P : Mean
The mean is $$\bar x = \dfrac{\sum f_i x_i}{N}$$.
$$\sum f_i x_i = 3\!(5)+4\!(4)+5\!(4)+8\!(2)+10\!(2)+11\!(3)=15+16+20+16+20+33=120$$
$$\bar x=\dfrac{120}{20}=6$$
So $$\bar x=6$$, which is entry (3).

Case Q : Median
The cumulative frequencies are $$5,\,9,\,13,\,15,\,17,\,20$$.
For $$N=20$$ (even), the median is the average of the $$10^{\text{th}}$$ and $$11^{\text{th}}$$ items.
Both the $$10^{\text{th}}$$ and $$11^{\text{th}}$$ items lie in the class $$x=5$$ (cumulative frequency just exceeds 9).
Hence $$\text{Median}=5$$, which is entry (2).

Case R : Mean deviation about the mean
Mean deviation about the mean is $$\dfrac{\sum f_i \lvert x_i-\bar x\rvert}{N}$$.
$$\begin{aligned} \sum f_i \lvert x_i-\bar x\rvert &= 5\lvert3-6\rvert +4\lvert4-6\rvert +4\lvert5-6\rvert +2\lvert8-6\rvert \\ &\quad +2\lvert10-6\rvert +3\lvert11-6\rvert \\ &=5(3)+4(2)+4(1)+2(2)+2(4)+3(5)\\ &=15+8+4+4+8+15=54 \end{aligned}$$
$$\text{M.D. about mean}= \dfrac{54}{20}=2.7$$, which is entry (4).

Case S : Mean deviation about the median
Mean deviation about the median is $$\dfrac{\sum f_i \lvert x_i-\text{Median}\rvert}{N}$$.
$$\begin{aligned} \sum f_i \lvert x_i-5\rvert &= 5\lvert3-5\rvert +4\lvert4-5\rvert +4\lvert5-5\rvert +2\lvert8-5\rvert \\ &\quad +2\lvert10-5\rvert +3\lvert11-5\rvert \\ &=5(2)+4(1)+4(0)+2(3)+2(5)+3(6)\\ &=10+4+0+6+10+18=48 \end{aligned}$$
$$\text{M.D. about median}= \dfrac{48}{20}=2.4$$, which is entry (5).

Thus the correct matching is
$$P\rightarrow(3),\;Q\rightarrow(2),\;R\rightarrow(4),\;S\rightarrow(5).$$
Therefore, the correct option is Option A.

We are given $$9 = x_1 < x_2 < \ldots < x_7$$ in an AP with common difference $$d$$, standard deviation = 4, and mean = $$\bar{x}$$, and we need to find $$\bar{x} + x_6$$.

First, express the terms of the arithmetic progression by noting that for $$k = 1,2,\ldots,7$$ we have $$x_k = 9 + (k-1)d$$.

Next, recall that for an AP with an odd number of terms, the mean equals the middle term, so $$\bar{x} = x_4 = 9 + 3d$$.

To find the variance, observe that the terms differ by multiples of $$d$$, so the variance of the set $$\{x_1,x_2,\ldots,x_7\}$$ is the same as the variance of $$\{0,d,2d,3d,4d,5d,6d\}$$, namely $$\sigma^2 = d^2 \cdot \text{Var}(\{0,1,2,3,4,5,6\})$$. Since $$\text{Var}(\{0,1,2,3,4,5,6\}) = \frac{0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{7} - \left(\frac{0+1+2+3+4+5+6}{7}\right)^2 = \frac{91}{7} - \left(\frac{21}{7}\right)^2 = 13 - 9 = 4\,,$$ it follows that $$\sigma^2 = 4d^2$$.

Given that $$\sigma = 4$$ implies $$\sigma^2 = 16$$, we set $$4d^2 = 16$$ which leads to $$d^2 = 4$$ and hence $$d = 2$$, taking the positive value since the sequence is increasing.

Now substituting back gives $$\bar{x} = 9 + 3(2) = 15$$ and $$x_6 = 9 + 5(2) = 19$$.

Therefore, $$\bar{x} + x_6 = 15 + 19 = 34$$.

The correct answer is Option B: 34.

Given: Mean = 10, Variance = 4 for $$n$$ students. One student's marks changed from 8 to 12. New mean = 10.2.

Finding n:

Old sum = $$10n$$. New sum = $$10n + 4$$.

New mean: $$\frac{10n + 4}{n} = 10.2 \implies n = 20$$

Finding old $$\Sigma x^2$$:

Variance = $$\frac{\Sigma x^2}{n} - \bar{x}^2 = 4$$

$$\frac{\Sigma x^2}{20} = 104 \implies \Sigma x^2 = 2080$$

Computing new variance:

New $$\Sigma x^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160$$

New variance = $$\frac{2160}{20} - (10.2)^2 = 108 - 104.04 = 3.96$$

The answer is Option C: $$3.96$$.

Given: 5 observations with mean = 5 and variance = 8. Three of the observations are 1, 3, 5.

Let the remaining two observations be $$a$$ and $$b$$.

Using the mean:

$$ \frac{1 + 3 + 5 + a + b}{5} = 5 $$

$$ 9 + a + b = 25 $$

$$ a + b = 16 \quad \ldots (1) $$

Using the variance:

$$ \text{Variance} = \frac{\sum x_i^2}{n} - \bar{x}^2 = 8 $$

$$ \frac{\sum x_i^2}{5} = 8 + 25 = 33 $$

$$ \sum x_i^2 = 165 $$

$$ 1^2 + 3^2 + 5^2 + a^2 + b^2 = 165 $$

$$ 1 + 9 + 25 + a^2 + b^2 = 165 $$

$$ a^2 + b^2 = 130 \quad \ldots (2) $$

From (1): $$(a+b)^2 = a^2 + 2ab + b^2$$

$$ 256 = 130 + 2ab $$

$$ ab = 63 \quad \ldots (3) $$

Now, the sum of cubes:

$$ a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)[(a^2+b^2) - ab] $$

$$ = 16 \times (130 - 63) = 16 \times 67 = 1072 $$

Therefore, the sum of cubes of the remaining two observations is 1072.

Given: Set 1: $$n_1 = 15$$, $$\bar{x}_1 = 12$$, $$\sigma_1^2 = 14$$. Set 2: $$n_2 = 15$$, $$\bar{x}_2 = 14$$, $$\sigma_2^2 = \sigma^2$$. Combined variance = 13.

The combined mean is:

$$ \bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{15(12) + 15(14)}{30} = \frac{180 + 210}{30} = 13 $$

The deviations of each set's mean from the combined mean:

$$ d_1 = \bar{x}_1 - \bar{x} = 12 - 13 = -1, \quad d_2 = \bar{x}_2 - \bar{x} = 14 - 13 = 1 $$

The combined variance formula is:

$$ \sigma_c^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2} $$

Substituting:

$$ 13 = \frac{15(14 + 1) + 15(\sigma^2 + 1)}{30} $$

$$ 390 = 15(15) + 15(\sigma^2 + 1) = 225 + 15\sigma^2 + 15 $$

$$ 390 = 240 + 15\sigma^2 $$

$$ 15\sigma^2 = 150 \implies \sigma^2 = 10 $$

The correct answer is $$\sigma^2 = \mathbf{10}$$.

We are given 6 observations: 1, 2, 4, 5, $$x$$, $$y$$ with mean = 5 and variance = 10.

$$ \frac{1 + 2 + 4 + 5 + x + y}{6} = 5 $$

$$ 12 + x + y = 30 \Rightarrow x + y = 18 $$

$$ \text{Variance} = \frac{\sum x_i^2}{n} - \bar{x}^2 = 10 $$

$$ \frac{1 + 4 + 16 + 25 + x^2 + y^2}{6} = 10 + 25 = 35 $$

$$ 46 + x^2 + y^2 = 210 \Rightarrow x^2 + y^2 = 164 $$

From $$x + y = 18$$ and $$x^2 + y^2 = 164$$:

$$ (x + y)^2 = x^2 + 2xy + y^2 = 324 $$

$$ 2xy = 324 - 164 = 160 \Rightarrow xy = 80 $$

So $$x, y$$ are roots of $$t^2 - 18t + 80 = 0$$:

$$ t = \frac{18 \pm \sqrt{324 - 320}}{2} = \frac{18 \pm 2}{2} $$

$$ x = 10, \; y = 8 \text{ (or vice versa)} $$

$$ \text{M.D.} = \frac{1}{6}\sum|x_i - \bar{x}| $$

$$ = \frac{|1-5| + |2-5| + |4-5| + |5-5| + |10-5| + |8-5|}{6} $$

$$ = \frac{4 + 3 + 1 + 0 + 5 + 3}{6} = \frac{16}{6} = \frac{8}{3} $$

The mean deviation about the mean is Option C: $$\frac{8}{3}$$.

Given: Mean $$\bar{x} = 20$$, standard deviation $$\sigma = 8$$, $$n = 10$$.

$$\sum x_i = n\bar{x} = 200$$

$$\sigma^2 = 64$$, so $$\frac{\sum x_i^2}{n} - \bar{x}^2 = 64$$

$$\frac{\sum x_i^2}{10} = 64 + 400 = 464$$

$$\sum x_i^2 = 4640$$

Correcting: Replace 50 with 40.

New $$\sum x_i = 200 - 50 + 40 = 190$$

New $$\bar{x} = \frac{190}{10} = 19$$

New $$\sum x_i^2 = 4640 - 50^2 + 40^2 = 4640 - 2500 + 1600 = 3740$$

New variance:

$$\sigma^2 = \frac{3740}{10} - 19^2 = 374 - 361 = 13$$

The correct variance is $$\mathbf{13}$$.

$$\sum f_i = (k+2) + 2k + (k^2-1) + (k^2-1) + (k^2+1) + (k-3) = 3k^2 + 4k - 2 = 62$$

$$3k^2 + 4k - 64 = 0$$. Using quadratic formula: $$k = \frac{-4 + \sqrt{16+768}}{6} = \frac{-4+28}{6} = 4$$.

Frequencies: 6, 8, 15, 15, 17, 1. Total = 62 ✓

$$\mu = \frac{0(6)+1(8)+2(15)+3(15)+4(17)+5(1)}{62} = \frac{0+8+30+45+68+5}{62} = \frac{156}{62} \approx 2.516$$

$$\sum f_ix_i^2 = 0+8+60+135+272+25 = 500$$. $$\sigma^2 = 500/62 - (156/62)^2 = 8.065 - 6.33 = 1.735$$.

$$\mu^2 + \sigma^2 = 6.33 + 1.735 = 8.065$$. $$[\mu^2 + \sigma^2] = 8$$.

The correct answer is Option 2: 8.

Let the six numbers be $$a, a+d, a+2d, a+3d, a+4d, a+5d$$.

$$a + (a + 2d) = 10 \implies 2a + 2d = 10 \implies a + d = 5 \implies a = 5 - d$$

$$\mu = \frac{a + (a+d) + (a+2d) + (a+3d) + (a+4d) + (a+5d)}{6} = \frac{6a + 15d}{6} = a + 2.5d$$

$$\frac{19}{2} = (5 - d) + 2.5d$$

$$9.5 = 5 + 1.5d \implies 4.5 = 1.5d \implies d = 3$$

$$a = 5 - 3 = 2$$

The six numbers are $$2, 5, 8, 11, 14, 17$$.

Using the formula for the variance of $$n$$ terms in an A.P.:

$$\sigma^2 = \frac{(n^2 - 1)d^2}{12}$$

$$\sigma^2 = \frac{(6^2 - 1) \times 3^2}{12} = \frac{35 \times 9}{12} = \frac{35 \times 3}{4} = \frac{105}{4}$$

$$8\sigma^2 = 8 \times \left( \frac{105}{4} \right) = 2 \times 105 = 210$$

We are given 12 observations with mean $$\frac{9}{2}$$ and variance 4. Two observations 9 and 10 were incorrectly recorded instead of 7 and 14. We need to find the correct variance $$\frac{m}{n}$$ (where $$m, n$$ are coprime) and compute $$m + n$$.

Find the original sum and sum of squares.

Mean = $$\frac{9}{2}$$, so $$\sum x_i = 12 \times \frac{9}{2} = 54$$.

Variance = $$\frac{\sum x_i^2}{n} - \bar{x}^2$$, so:

$$4 = \frac{\sum x_i^2}{12} - \left(\frac{9}{2}\right)^2 = \frac{\sum x_i^2}{12} - \frac{81}{4}$$

$$\frac{\sum x_i^2}{12} = 4 + \frac{81}{4} = \frac{97}{4}$$

$$\sum x_i^2 = 12 \times \frac{97}{4} = 291$$

Compute the corrected sum and sum of squares.

Corrected sum: $$54 - 9 - 10 + 7 + 14 = 56$$

Corrected mean: $$\frac{56}{12} = \frac{14}{3}$$

Corrected $$\sum x_i^2 = 291 - 9^2 - 10^2 + 7^2 + 14^2 = 291 - 81 - 100 + 49 + 196 = 355$$

Compute the corrected variance.

$$\text{Variance} = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{355}{12} - \left(\frac{14}{3}\right)^2 = \frac{355}{12} - \frac{196}{9}$$

Finding a common denominator (LCM of 12 and 9 is 36):

$$= \frac{355 \times 3}{36} - \frac{196 \times 4}{36} = \frac{1065 - 784}{36} = \frac{281}{36}$$

Check if the fraction is in lowest terms.

We need $$\gcd(281, 36)$$. Since 281 is not divisible by 2 or 3 (the prime factors of 36), and checking: 281 is prime (not divisible by any prime up to $$\sqrt{281} \approx 16.8$$, i.e., not by 2, 3, 5, 7, 11, or 13).

So $$m = 281$$, $$n = 36$$, and they are coprime.

Compute the answer.

$$m + n = 281 + 36 = 317$$

The answer is Option D: 317.

Total apples ($$N$$ ) = 10
Rotten apples ($$M$$ ) = 3
Good apples ($$N-M$$ ) = 7
Apples drawn ($$n$$ ) = 4 

Using the Distribution formula $$P(X=k) = \frac{\binom{M}{k} \binom{N-M}{n-k}}{\binom{N}{n}}$$ :The total ways to draw 4 apples is $$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$ .

• $$P(X=0) = \frac{\binom{3}{0} \binom{7}{4}}{210} = \frac{1 \times 35}{210} = \frac{35}{210}$$
• $$P(X=1) = \frac{\binom{3}{1} \binom{7}{3}}{210} = \frac{3 \times 35}{210} = \frac{105}{210}$$
• $$P(X=2) = \frac{\binom{3}{2} \binom{7}{2}}{210} = \frac{3 \times 21}{210} = \frac{63}{210}$$
• $$P(X=3) = \frac{\binom{3}{3} \binom{7}{1}}{210} = \frac{1 \times 7}{210} = \frac{7}{210}$$

For the any distribution, we know that $$\sigma^2 = E[X^2] - \mu^2$$, so $$10(\mu^2 + \sigma^2) = 10E[X^2]$$ 

Now we calculate $$E[X^2]$$ :

$$E[X^2] = \sum x_i^2 P(x_i)$$ $$E[X^2] = 0^2\left(\frac{35}{210}\right) + 1^2\left(\frac{105}{210}\right) + 2^2\left(\frac{63}{210}\right) + 3^2\left(\frac{7}{210}\right)$$

$$E[X^2] = \frac{105 + 252 + 63}{210} = \frac{420}{210} = 2.0 $$

$$10(\mu^2 + \sigma^2) = 10 \times E[X^2] = 10 \times 2.0 = 20$$ 

The final value is 20.

Let the five terms of the G.P. be

$$a,\ ar,\ ar^2,\ ar^3,\ ar^4$$

Given that their mean is $$\frac{31}{10}$$,

$$\frac{a+ar+ar^2+ar^3+ar^4}{5}=\frac{31}{10}$$

$$a(1+r+r^2+r^3+r^4)=\frac{31}{2}\tag{1}$$

The mean of their reciprocals is $$\frac{31}{40}$$,

$$\frac{\frac1a+\frac1{ar}+\frac1{ar^2}+\frac1{ar^3}+\frac1{ar^4}}{5}=\frac{31}{40}$$

$$\frac{1+r+r^2+r^3+r^4}{ar^4}=\frac{31}{8}\tag{2}$$

Dividing (1) by (2),

$$a^2r^4=4$$

Since all terms are positive,

$$ar^2=2$$

Also,

$$ar^2+ar^3+ar^4=14$$

$$ar^2(1+r+r^2)=14$$

$$2(1+r+r^2)=14$$

$$1+r+r^2=7$$

$$r^2+r-6=0$$

$$(r+3)(r-2)=0$$

Since $$r>0$$,

$$r=2$$

Using $$ar^2=2$$,

$$a(2)^2=2$$

$$a=\frac12$$

Hence the five terms are

$$\frac12,\ 1,\ 2,\ 4,\ 8$$

Now,

$$\sum a_i^2=\left(\frac12\right)^2+1^2+2^2+4^2+8^2$$

$$=\frac14+1+4+16+64$$

$$=\frac{341}{4}$$

Variance,

$$\sigma^2=\frac{\sum a_i^2}{5}-(\text{Mean})^2$$

$$\sigma^2=\frac{\frac{341}{4}}{5}-\left(\frac{31}{10}\right)^2$$

$$=\frac{341}{20}-\frac{961}{100}$$

$$=\frac{1705-961}{100}$$

$$=\frac{744}{100}$$

$$=\frac{186}{25}$$

Thus,

$$m=186,\quad n=25$$

$$m+n=186+25=211$$

$$\boxed{211}$$

To solve for $$x$$ and $$y$$, we use the definitions of mean and variance for the 8 numbers: $$x, y, 10, 12, 6, 12, 4, 8$$.

1. Use the Mean

The mean $$\bar{x} = 9$$.

$$\frac{x + y + 10 + 12 + 6 + 12 + 4 + 8}{8} = 9$$

$$x + y + 52 = 72 \implies \mathbf{x + y = 20} \quad \dots \text{(Eq. 1)}$$

2. Use the Variance

The variance $$\sigma^2 = 9.25$$. The formula is $$\frac{\sum x_i^2}{n} - (\bar{x})^2 = \sigma^2$$.

$$\frac{x^2 + y^2 + 10^2 + 12^2 + 6^2 + 12^2 + 4^2 + 8^2}{8} - 9^2 = 9.25$$

$$\frac{x^2 + y^2 + 100 + 144 + 36 + 144 + 16 + 64}{8} - 81 = 9.25$$

$$\frac{x^2 + y^2 + 504}{8} = 90.25$$

$$x^2 + y^2 + 504 = 722 \implies \mathbf{x^2 + y^2 = 218} \quad \dots \text{(Eq. 2)}$$

3. Solve for $$x$$ and $$y$$

Using $$(x+y)^2 = x^2 + y^2 + 2xy$$:

$$20^2 = 218 + 2xy \implies 400 - 218 = 2xy \implies 2xy = 182 \implies \mathbf{xy = 91}$$

We need two numbers that add to $$20$$ and multiply to $$91$$. These are 13 and 7.

Since the problem states $$x > y$$, we have:

• $$x = 13$$
• $$y = 7$$

4. Final Calculation

Find the value of $$3x - 2y$$:

$$3(13) - 2(7) = 39 - 14 = \mathbf{25}$$

We have 7 observations with mean 8 and variance 16, and one observation, 14, is removed.

The sum of the 7 observations is $$7 \times 8 = 56$$. After removing 14, the sum of the remaining 6 observations becomes $$56 - 14 = 42$$, so their mean is $$a = \frac{42}{6} = 7$$.

Recall that variance is given by $$\frac{\sum x_i^2}{n} - \bar{x}^2$$. For the original 7 observations, $$16 = \frac{\sum x_i^2}{7} - 64$$ implies $$\sum x_i^2 = 7 \times 80 = 560$$. After removing 14, the sum of squares is $$560 - 14^2 = 560 - 196 = 364$$. Therefore, the variance of the remaining observations is $$b = \frac{364}{6} - 49 = \frac{364 - 294}{6} = \frac{70}{6} = \frac{35}{3}$$.

Finally, we compute $$a + 3b - 5 = 7 + 3 \times \frac{35}{3} - 5 = 7 + 35 - 5 = 37$$, so the answer is $$\boxed{37}$$.

The frequency distribution below is known to have mean = 9 and variance = 15.08.

Using the condition for the mean, the total number of observations is $$N = 40 + \alpha + \beta$$ and the sum of the products is

$$\sum f_i x_i = 8 + 16 + 6\alpha + 120 + 80 + 12\beta + 56 + 80 = 360 + 6\alpha + 12\beta.$$

Since the mean equals 9,

$$\frac{360 + 6\alpha + 12\beta}{40 + \alpha + \beta} = 9$$ which leads to $$360 + 6\alpha + 12\beta = 360 + 9\alpha + 9\beta$$ and hence $$3\beta = 3\alpha \implies \alpha = \beta.$$

Next, applying the variance condition, we compute

$$\sum f_i x_i^2 = 16 + 64 + 36\alpha + 960 + 800 + 144\beta + 784 + 1280 = 3904 + 36\alpha + 144\beta.$$

With $$\alpha = \beta$$, this becomes $$\sum f_i x_i^2 = 3904 + 180\alpha$$ and $$N = 40 + 2\alpha.$$

Setting the variance equal to 15.08 gives

$$\frac{3904 + 180\alpha}{40 + 2\alpha} - 81 = 15.08$$ so $$\frac{3904 + 180\alpha}{40 + 2\alpha} = 96.08$$ and therefore $$3904 + 180\alpha = 96.08(40 + 2\alpha) = 3843.2 + 192.16\alpha,$$ leading to $$60.8 = 12.16\alpha \implies \alpha = 5.$$

It follows that $$\alpha = \beta = 5$$, and substituting these values gives

$$\alpha^2 + \beta^2 - \alpha\beta = 25 + 25 - 25 = 25.$$

The answer is $$25$$.

To find the value of $$\alpha$$, we first determine the mean $$\bar{x}$$ of the given frequency distribution.

The total frequency $$N$$ is:

$$N = \sum f_i = 3 + 6 + 16 + \alpha + 9 + 5 + 6 = 45 + \alpha$$

The sum of the products of observations and their frequencies $$\sum f_i x_i$$ is:

$$\sum f_i x_i = (2 \times 3) + (3 \times 6) + (4 \times 16) + (5 \times \alpha) + (6 \times 9) + (7 \times 5) + (8 \times 6)$$

$$\sum f_i x_i = 6 + 18 + 64 + 5\alpha + 54 + 35 + 48 = 225 + 5\alpha$$

---

Step 1: Calculate the mean $$\bar{x}$$

$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{225 + 5\alpha}{45 + \alpha} = \frac{5(45 + \alpha)}{45 + \alpha} = 5$$

Since the mean is independent of $$\alpha$$, we find $$\bar{x} = 5$$.

---

Step 2: Set up the variance formula

The variance $$\sigma^2$$ is given by:

$$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} = 3$$

Now, we calculate the term $$\sum f_i (x_i - 5)^2$$ for all observations:

For $$x_i = 2$$: $$3 \times (2 - 5)^2 = 3 \times 9 = 27$$

For $$x_i = 3$$: $$6 \times (3 - 5)^2 = 6 \times 4 = 24$$

For $$x_i = 4$$: $$16 \times (4 - 5)^2 = 16 \times 1 = 16$$

For $$x_i = 5$$: $$\alpha \times (5 - 5)^2 = \alpha \times 0 = 0$$

For $$x_i = 6$$: $$9 \times (6 - 5)^2 = 9 \times 1 = 9$$

For $$x_i = 7$$: $$5 \times (7 - 5)^2 = 5 \times 4 = 20$$

For $$x_i = 8$$: $$6 \times (8 - 5)^2 = 6 \times 9 = 54$$

Summing these deviations multiplied by their frequencies:

$$\sum f_i (x_i - 5)^2 = 27 + 24 + 16 + 0 + 9 + 20 + 54 = 150$$

---

Step 3: Solve for $$\alpha$$

Substituting this value into the variance equation:

$$\frac{150}{45 + \alpha} = 3$$

$$150 = 3(45 + \alpha)$$

$$50 = 45 + \alpha \implies \alpha = 5$$

Therefore, the value of $$\alpha$$ is equal to 5.

Mean = 28. Frequency distribution:

$$\sum f = 14 + x$$

$$\sum fx = 10 + 45 + 25x + 175 + 180 = 410 + 25x$$

Mean: $$\frac{410 + 25x}{14 + x} = 28$$

$$410 + 25x = 392 + 28x$$

$$18 = 3x$$, $$x = 6$$

$$\sum f = 20$$, $$\sum fx = 560$$

$$\sum fx^2 = 2(25) + 3(225) + 6(625) + 5(1225) + 4(2025) = 50 + 675 + 3750 + 6125 + 8100 = 18700$$

Variance = $$\frac{\sum fx^2}{\sum f} - \left(\frac{\sum fx}{\sum f}\right)^2 = \frac{18700}{20} - 784 = 935 - 784 = 151$$

The variance is 151.

Let $$X = \{11,12,13,\ldots,40,41\}$$ and $$Y = \{61,62,63,\ldots,90,91\}$$. Both sets contain $$31$$ elements, so the total number of observations in $$X \cup Y$$ is $$n = 62$$.

The mean of $$X$$ is $$\bar{x} = \frac{11 + 41}{2} = 26$$ and the mean of $$Y$$ is $$\bar{y} = \frac{61 + 91}{2} = 76$$. Hence the combined mean of $$X \cup Y$$ is $$\mu = \frac{31 \times 26 + 31 \times 76}{62} = \frac{26 + 76}{2} = 51$$.

To find the variance $$\sigma^2$$, we compute $$\frac{1}{n}\sum (x_i - \mu)^2$$ over all observations.

For elements of $$X$$, as $$k$$ ranges from $$11$$ to $$41$$, $$(k - 51)$$ ranges from $$-40$$ to $$-10$$, so $$\sum_{k=11}^{41}(k-51)^2 = \sum_{j=10}^{40} j^2 = \frac{40 \times 41 \times 81}{6} - \frac{9 \times 10 \times 19}{6} = 22140 - 285 = 21855$$.

Similarly, for elements of $$Y$$, as $$k$$ ranges from $$61$$ to $$91$$, $$(k - 51)$$ ranges from $$10$$ to $$40$$, and one similarly finds $$\sum_{k=61}^{91}(k-51)^2 = \sum_{j=10}^{40} j^2 = 21855$$.

It follows that $$\sigma^2 = \frac{21855 + 21855}{62} = \frac{43710}{62} = 705$$.

Hence $$|\bar{x} + \bar{y} - \sigma^2| = |26 + 76 - 705| = |102 - 705| = \boxed{603}$$.

We are given 50 observations with mean 15 and standard deviation 2. One observation was wrongly recorded, and the sum of the correct and incorrect observations is 70. The corrected mean is 16. We need to find the corrected variance.

First, let the wrong observation be $$w$$ and the correct observation be $$c$$. Since $$w + c = 70$$, we can express both values in terms of their sum. Next, the original sum of observations is $$\sum x_i = 50 \times 15 = 750$$, and the corrected sum becomes $$750 - w + c = 50 \times 16 = 800$$, which implies $$c - w = 50$$. Solving these equations gives $$c = 60$$ and $$w = 10$$.

Now we find the original sum of squares. Using the relation $$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$$ and substituting $$4 = \frac{\sum x_i^2}{50} - 225$$ yields $$\sum x_i^2 = 50 \times 229 = 11450$$.

Substituting the wrong and correct values into the sum of squares gives the corrected sum of squares as $$\sum x_i^2 \text{(corrected)} = 11450 - w^2 + c^2 = 11450 - 100 + 3600 = 14950$$.

Finally, the corrected variance is calculated as $$\text{Variance} = \frac{\sum x_i^2 \text{(corrected)}}{50} - \bar{x}_{new}^2 = \frac{14950}{50} - 16^2 = 299 - 256 = 43$$. Therefore, the answer is Option C: $$43$$.

We are given data: $$4, 5, 6, 6, 7, 8, x, y$$ where $$x < y$$, with mean = 6 and variance = $$\frac{9}{4}$$.

Use the mean condition

Mean $$= \frac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6$$

$$36 + x + y = 48$$

$$x + y = 12 \quad \cdots(1)$$

Use the variance condition

Variance $$= \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{9}{4}$$

$$\frac{\sum x_i^2}{8} - 36 = \frac{9}{4}$$

$$\frac{\sum x_i^2}{8} = 36 + \frac{9}{4} = \frac{153}{4}$$

$$\sum x_i^2 = \frac{153 \times 8}{4} = 306$$

Calculate $$\sum x_i^2$$

$$4^2 + 5^2 + 6^2 + 6^2 + 7^2 + 8^2 + x^2 + y^2 = 306$$

$$16 + 25 + 36 + 36 + 49 + 64 + x^2 + y^2 = 306$$

$$226 + x^2 + y^2 = 306$$

$$x^2 + y^2 = 80 \quad \cdots(2)$$

Solve the system

From (1): $$y = 12 - x$$

Substituting in (2): $$x^2 + (12-x)^2 = 80$$

$$x^2 + 144 - 24x + x^2 = 80$$

$$2x^2 - 24x + 64 = 0$$

$$x^2 - 12x + 32 = 0$$

$$(x-4)(x-8) = 0$$

$$x = 4$$ or $$x = 8$$

Since $$x < y$$: $$x = 4, y = 8$$

Calculate $$x^4 + y^2$$

$$x^4 + y^2 = 4^4 + 8^2 = 256 + 64 = 320$$

Therefore, $$x^4 + y^2 = 320$$.

The correct answer is Option A: $$320$$.

We need to find the number of values of $$a \in \mathbb{N}$$ such that the variance of 3, 7, 12, $$a$$, $$43 - a$$ is a natural number. The mean is $$\bar{x} = \frac{3 + 7 + 12 + a + (43 - a)}{5} = \frac{65}{5} = 13$$, and hence the variance is $$\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{5} = \frac{(3-13)^2 + (7-13)^2 + (12-13)^2 + (a-13)^2 + (43-a-13)^2}{5} = \frac{100 + 36 + 1 + (a-13)^2 + (30-a)^2}{5} = \frac{137 + (a-13)^2 + (30-a)^2}{5}.$$

Now $$(a-13)^2 + (30-a)^2 = a^2 - 26a + 169 + 900 - 60a + a^2 = 2a^2 - 86a + 1069$$ so $$\sigma^2 = \frac{137 + 2a^2 - 86a + 1069}{5} = \frac{2a^2 - 86a + 1206}{5}.$$

For the variance to be a natural number, we need $$\frac{2a^2 - 86a + 1206}{5} \in \mathbb{N}$$, which requires $$2a^2 - 86a + 1206 \equiv 0 \pmod{5}.$$ Reducing modulo 5, since $$86 = 17 \times 5 + 1$$ and $$1206 = 241 \times 5 + 1$$, we get $$2a^2 - 86a + 1206 \equiv 2a^2 - a + 1 \pmod{5}.$$ Checking each residue class mod 5: for $$a \equiv 0$$: $$0 - 0 + 1 = 1 \not\equiv 0$$; for $$a \equiv 1$$: $$2 - 1 + 1 = 2 \not\equiv 0$$; for $$a \equiv 2$$: $$8 - 2 + 1 = 7 \equiv 2 \not\equiv 0$$; for $$a \equiv 3$$: $$18 - 3 + 1 = 16 \equiv 1 \not\equiv 0$$; for $$a \equiv 4$$: $$32 - 4 + 1 = 29 \equiv 4 \not\equiv 0$$.

For no value of $$a \pmod{5}$$ is the expression divisible by 5. Therefore, the variance is never a natural number for any $$a \in \mathbb{N}$$. The number of such values is $$\textbf{0}$$, so the answer is Option A: $$\textbf{0}$$.

We need to find the median of the data $$3, 5, 7, 2k, 12, 16, 21, 24$$ given that the mean deviation about the median is 6.

Since the data is in ascending order, we have $$7 \leq 2k \leq 12$$, which implies $$3.5 \leq k \leq 6$$.

For eight observations, the median is the average of the 4th and 5th values, so

$$M = \dfrac{2k + 12}{2} = k + 6$$

The mean deviation about the median is defined by

$$\text{M.D.} = \dfrac{1}{8}\sum_{i=1}^{8} |x_i - M|$$

Substituting $$M = k + 6$$ yields the following individual deviations:

$$|3 - (k+6)| = k + 3$$

$$|5 - (k+6)| = k + 1$$

$$|7 - (k+6)| = k - 1 \quad (\text{since } k \geq 3.5)$$

$$|2k - (k+6)| = |k - 6| = 6 - k \quad (\text{since } k \leq 6)$$

$$|12 - (k+6)| = 6 - k$$

$$|16 - (k+6)| = 10 - k$$

$$|21 - (k+6)| = 15 - k$$

$$|24 - (k+6)| = 18 - k$$

Their sum simplifies as follows:

$$ (k+3) + (k+1) + (k-1) + (6-k) + (6-k) + (10-k) + (15-k) + (18-k) = 58 - 2k $$

Setting the mean deviation equal to 6 gives

$$ \dfrac{58 - 2k}{8} = 6 \quad\Longrightarrow\quad 58 - 2k = 48 \quad\Longrightarrow\quad 2k = 10 \quad\Longrightarrow\quad k = 5 $$

Substituting back to find the median:

$$M = k + 6 = 5 + 6 = 11$$

We can verify that the data become $$3, 5, 7, 10, 12, 16, 21, 24$$ and the median is $$\dfrac{10 + 12}{2} = 11$$ $$\checkmark$$

The correct answer is Option D: $$11$$.

We need to find $$n$$ (odd) such that the mean deviation about the mean of $$1, 2, 3, \ldots, n$$ equals $$\frac{5(n+1)}{n}$$.

To begin, $$\bar{x} = \frac{n+1}{2}$$.

Next, letting $$n = 2m + 1$$ gives $$\bar{x} = m + 1$$.

The deviations $$|x_i - \bar{x}|$$ are: $$m, m-1, \ldots, 1, 0, 1, \ldots, m$$.

Hence, the sum of deviations is $$2(1 + 2 + \cdots + m) = 2 \cdot \frac{m(m+1)}{2} = m(m+1)$$.

Therefore, the mean deviation is $$\text{MD} = \frac{m(m+1)}{n} = \frac{m(m+1)}{2m+1}$$.

Equating this to $$\frac{5(n+1)}{n} = \frac{5(2m+2)}{2m+1} = \frac{10(m+1)}{2m+1}$$ and noting that $$m + 1 \neq 0$$ allows division by $$\frac{m+1}{2m+1}$$, which yields $$m = 10$$.

Therefore, $$n = 2m + 1 = 21$$.

The answer is $$\boxed{21}$$.

Given,

$$x_1=3,\qquad r=\frac12$$

Hence,

$$x_i=3\left(\frac12\right)^{i-1}$$

The new data is

$$y_i=(x_i-i)^2$$

Mean of new data is

$$\bar x=\frac1{20}\sum_{i=1}^{20}(x_i-i)^2$$

Expanding,

$$\bar x=\frac1{20}\left(\sum x_i^2+\sum i^2-2\sum ix_i\right)$$

Now,

$$x_i=3\left(\frac12\right)^{i-1}$$

Therefore,

$$x_i^2=9\left(\frac14\right)^{i-1}$$

Hence,

$$\sum_{i=1}^{20}x_i^2=9\sum_{i=1}^{20}\left(\frac14\right)^{i-1}$$

Using GP sum,

$$=9\cdot\frac{1-\left(\frac14\right)^{20}}{1-\frac14}$$

$$=12\left(1-\frac1{4^{20}}\right)$$

Also,

$$\sum_{i=1}^{20}i^2=\frac{20\cdot21\cdot41}{6}=2870$$

Now compute

$$\sum ix_i=3\sum_{i=1}^{20}\frac{i}{2^{i-1}}$$

Using

$$\sum_{i=1}^{n}ir^{i-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}$$

with

$$r=\frac12,\qquad n=20,$$

$$\sum_{i=1}^{20}\frac{i}{2^{i-1}} =\frac{1-\frac{21}{2^{20}}+\frac{20}{2^{21}}}{\left(\frac12\right)^2}$$

$$=4\left(1-\frac{11}{2^{20}}\right)$$

Hence,

$$\sum ix_i=12\left(1-\frac{11}{2^{20}}\right)$$

Therefore,

$$\bar x=\frac1{20}\left[12\left(1-\frac1{4^{20}}\right)+2870-24\left(1-\frac{11}{2^{20}}\right)\right]$$

$$=\frac1{20}\left(2858+\frac{264}{2^{20}}-\frac{12}{4^{20}}\right)$$

Since

$$0<\frac{264}{2^{20}}-\frac{12}{4^{20}}<1,$$

we get

$$142.9<\bar x<143$$

Hence,

$$\lfloor\bar x\rfloor=142$$

Therefore, the required answer is

$$\boxed{142}$$.

We have 40 observations with mean 30 and standard deviation 5. Two wrong observations (12 and 10) are removed. We need to find $$38\sigma^2$$ for the remaining 38 observations.

$$\bar{x} = 30 \implies \sum x_i = 40 \times 30 = 1200$$

$$\text{SD} = 5 \implies \text{Var} = 25$$

$$\text{Var} = \dfrac{\sum x_i^2}{40} - \bar{x}^2 = 25$$

$$\dfrac{\sum x_i^2}{40} = 25 + 900 = 925$$

$$\sum x_i^2 = 37000$$

$$\sum x_i' = 1200 - 12 - 10 = 1178$$

$$\sum (x_i')^2 = 37000 - 144 - 100 = 36756$$

$$\bar{x}' = \dfrac{1178}{38} = 31$$

$$\sigma^2 = \dfrac{\sum (x_i')^2}{38} - (\bar{x}')^2 = \dfrac{36756}{38} - 961 = 967.26... - 961$$

Let us compute exactly:

$$\dfrac{36756}{38} = \dfrac{36756}{38} = 967 + \dfrac{10}{38} = 967 + \dfrac{5}{19}$$

$$\sigma^2 = 967 + \dfrac{5}{19} - 961 = 6 + \dfrac{5}{19} = \dfrac{114 + 5}{19} = \dfrac{119}{19}$$

$$38\sigma^2 = 38 \times \dfrac{119}{19} = 2 \times 119 = 238$$

The correct answer is $$\boxed{238}$$.

Let the number of students failing be $$k$$.

Since a student fails if marks are less than $$50$$, in the worst case assume each failing student gets exactly $$49$$ marks.

The average marks of $$7$$ students is $$62$$.

Hence, total marks

$$=7\times62=434$$

Variance is $$20$$, so

$$\sum (x_i-62)^2=7\times20=140$$

For each failing student,

$$x_i=49$$

Therefore,

$$ (49-62)^2=169 $$

If $$k$$ students fail, then

$$169k\le140$$

Hence,

$$k\le\frac{140}{169}<1$$

Therefore, the maximum possible number of students who can fail is

$$\boxed{0}$$

Given: $$n = 15$$, mean $$\bar{x} = 8$$, standard deviation $$\sigma = 3$$.

$$\sum x_i = n \cdot \bar{x} = 15 \times 8 = 120$$

$$\sigma^2 = 9 = \frac{\sum x_i^2}{n} - \bar{x}^2 \implies \frac{\sum x_i^2}{15} = 9 + 64 = 73$$

$$\sum x_i^2 = 1095$$

The observation 20 was misread as 5. Correcting:

$$\sum x_i^{\text{correct}} = 120 - 5 + 20 = 135$$

$$\sum (x_i^{\text{correct}})^2 = 1095 - 25 + 400 = 1470$$

Correct mean: $$\bar{x}_c = \frac{135}{15} = 9$$

Correct variance:

$$\sigma_c^2 = \frac{\sum (x_i^{\text{correct}})^2}{n} - \bar{x}_c^2 = \frac{1470}{15} - 81 = 98 - 81 = 17$$

Hence the correct variance is $$\boxed{17}$$.

We are given 20 observations $$x_1, x_2, \ldots, x_{20}$$ with mean $$\bar{x} = 15$$ and variance $$\sigma^2 = 9$$. We know that $$\frac{1}{20}\sum x_i = 15$$ and $$\frac{1}{20}\sum x_i^2 - 15^2 = 9$$, so $$\frac{1}{20}\sum x_i^2 = 234$$.

We need the mean of $$(x_1+\alpha)^2, (x_2+\alpha)^2, \ldots, (x_{20}+\alpha)^2$$ to equal 178. This mean is:

$$\frac{1}{20}\sum(x_i+\alpha)^2 = \frac{1}{20}\sum(x_i^2 + 2\alpha x_i + \alpha^2) = \frac{1}{20}\sum x_i^2 + 2\alpha\cdot\frac{1}{20}\sum x_i + \alpha^2$$

$$= 234 + 2\alpha(15) + \alpha^2 = \alpha^2 + 30\alpha + 234$$

Setting this equal to 178: $$\alpha^2 + 30\alpha + 234 = 178$$, so $$\alpha^2 + 30\alpha + 56 = 0$$.

By the quadratic formula: $$\alpha = \frac{-30 \pm \sqrt{900 - 224}}{2} = \frac{-30 \pm \sqrt{676}}{2} = \frac{-30 \pm 26}{2}$$.

So $$\alpha = \frac{-30+26}{2} = -2$$ or $$\alpha = \frac{-30-26}{2} = -28$$. The maximum value of $$\alpha$$ is $$-2$$, and its square is $$(-2)^2 = 4$$.

Hence, the correct answer is 4.

We are given six observations 7, 10, 11, 15, $$a$$, $$b$$ with mean 10 and variance $$\frac{20}{3}$$.

From the mean: $$\frac{7 + 10 + 11 + 15 + a + b}{6} = 10$$, so $$43 + a + b = 60$$, giving $$a + b = 17$$.

From the variance: $$\frac{\sum x_i^2}{6} - \bar{x}^2 = \frac{20}{3}$$, so $$\frac{49 + 100 + 121 + 225 + a^2 + b^2}{6} = 100 + \frac{20}{3} = \frac{320}{3}.$$ $$495 + a^2 + b^2 = 640, \quad \text{hence } a^2 + b^2 = 145.$$

Now, $$(a + b)^2 = a^2 + 2ab + b^2$$, so $$289 = 145 + 2ab$$, giving $$ab = 72$$.

Therefore, $$(a - b)^2 = (a+b)^2 - 4ab = 289 - 288 = 1$$, so $$|a - b| = 1$$.

Let the two unknown observations be $$a$$ and $$b$$. The six observations are $$2, 4, 5, 7, a, b$$.

From the mean: $$\frac{2+4+5+7+a+b}{6} = 6.5$$, so $$18 + a + b = 39$$, giving $$a + b = 21$$.

From the variance formula $$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$$: $$10.25 = \frac{\sum x_i^2}{6} - (6.5)^2 \implies \frac{\sum x_i^2}{6} = 10.25 + 42.25 = 52.5 \implies \sum x_i^2 = 315.$$

The sum of squares of the known observations is $$4 + 16 + 25 + 49 = 94$$, so $$a^2 + b^2 = 315 - 94 = 221$$.

Now using $$(a+b)^2 = a^2 + 2ab + b^2$$: $$441 = 221 + 2ab$$, so $$ab = 110$$.

The values $$a$$ and $$b$$ are roots of $$t^2 - 21t + 110 = 0$$, which factors as $$(t-10)(t-11) = 0$$.

Therefore the two remaining observations are $$10$$ and $$11$$.

We have three observations $$a, b, c$$ with $$b = a + c$$, and the standard deviation of $$a + 2, b + 2, c + 2$$ is $$d$$.

Since adding a constant to each observation does not change the standard deviation, the standard deviation of $$a, b, c$$ is also $$d$$. The mean of $$a, b, c$$ is $$\bar{x} = \frac{a + b + c}{3} = \frac{a + (a+c) + c}{3} = \frac{2(a+c)}{3} = \frac{2b}{3}$$.

The variance is $$d^2 = \frac{(a - \bar{x})^2 + (b - \bar{x})^2 + (c - \bar{x})^2}{3}$$. Computing each deviation: $$a - \frac{2b}{3} = \frac{3a - 2b}{3} = \frac{3a - 2a - 2c}{3} = \frac{a - 2c}{3}$$, $$b - \frac{2b}{3} = \frac{b}{3} = \frac{a + c}{3}$$, and $$c - \frac{2b}{3} = \frac{3c - 2b}{3} = \frac{c - 2a}{3}$$.

So $$d^2 = \frac{1}{3} \cdot \frac{(a-2c)^2 + (a+c)^2 + (c-2a)^2}{9} = \frac{(a^2 - 4ac + 4c^2) + (a^2 + 2ac + c^2) + (4a^2 - 4ac + c^2)}{27} = \frac{6a^2 - 6ac + 6c^2}{27} = \frac{2(a^2 - ac + c^2)}{9}$$.

Therefore $$9d^2 = 2a^2 - 2ac + 2c^2$$. Now $$3(a^2 + c^2) - 9d^2 = 3a^2 + 3c^2 - 2a^2 + 2ac - 2c^2 = a^2 + 2ac + c^2 = (a + c)^2 = b^2$$.

Hence $$b^2 = 3(a^2 + c^2) - 9d^2$$.

In the original series of $$2n$$ observations, $$n$$ values equal $$a$$ and $$n$$ values equal $$-a$$. The original mean is $$\frac{n \cdot a + n \cdot (-a)}{2n} = 0$$.

After adding constant $$b$$ to each observation, the new values are $$a + b$$ ($$n$$ times) and $$-a + b$$ ($$n$$ times). The new mean is $$0 + b = b = 5$$, so $$b = 5$$.

Adding a constant does not change the standard deviation, so the standard deviation of the new set equals the standard deviation of the original set, which is 20. For the original data, the variance is $$\frac{n \cdot a^2 + n \cdot a^2}{2n} - 0^2 = a^2$$. Therefore the standard deviation is $$|a| = 20$$, giving $$a^2 = 400$$.

Hence $$a^2 + b^2 = 400 + 25 = 425$$.

We have 7 observations whose mean is given as $$\bar{x}=8$$. The definition of the mean says

$$\bar{x}=\dfrac{\sum_{i=1}^{7}x_i}{7}\;.$$

Substituting $$\bar{x}=8$$, we obtain

$$8=\dfrac{\sum_{i=1}^{7}x_i}{7}\;.$$

Multiplying both sides by 7,

$$\sum_{i=1}^{7}x_i = 7 \times 8 = 56\;.$$

Out of these 7 observations, two are specifically known: $$6$$ and $$8$$. Removing them from the total, the sum of the remaining 5 observations is

$$\sum_{i=1}^{5}y_i = 56-6-8 = 42\;.$$

Now we need their variance. First we recall the population variance formula (division by the number of observations):

$$\sigma^{2}=\dfrac{\sum x_i^{2}}{n}-\bar{x}^{2}\;.$$

The variance of all 7 observations is given as $$16$$, so with $$n=7$$ we write

$$16=\dfrac{\sum_{i=1}^{7}x_i^{2}}{7}-8^{2}\;.$$

Adding $$8^{2}=64$$ to both sides,

$$16+64=\dfrac{\sum_{i=1}^{7}x_i^{2}}{7}\;,$$

$$80=\dfrac{\sum_{i=1}^{7}x_i^{2}}{7}\;.$$

Multiplying by 7,

$$\sum_{i=1}^{7}x_i^{2}=7 \times 80 = 560\;.$$

Next, we subtract the squares of the two known observations to isolate the sum of squares of the remaining five. Their squares are $$6^{2}=36$$ and $$8^{2}=64$$, whose sum is $$36+64=100$$. Therefore,

$$\sum_{i=1}^{5}y_i^{2}=560-100=460\;.$$

For the five residual observations we already found that their sum is $$42$$, so their mean is

$$\bar{y}=\dfrac{42}{5}=8.4\;.$$

Using the same variance formula for these five observations (now $$n=5$$), we have

$$\sigma_{y}^{2}=\dfrac{\sum_{i=1}^{5}y_i^{2}}{5}-\bar{y}^{2}\;.$$

Substituting the values,

$$\sigma_{y}^{2}=\dfrac{460}{5}-(8.4)^{2}\;.$$

Computing each term:

$$\dfrac{460}{5}=92,\qquad (8.4)^{2}=70.56\;.$$

Hence,

$$\sigma_{y}^{2}=92-70.56=21.44\;.$$

Now we change $$21.44$$ to an exact fraction. Observe that

$$21.44 = 21 + 0.44 = 21 + \dfrac{44}{100} = 21 + \dfrac{11}{25}= \dfrac{525}{25}+\dfrac{11}{25}= \dfrac{536}{25}\;.$$

Thus the variance of the remaining five observations is

$$\boxed{\dfrac{536}{25}}\;.$$

Hence, the correct answer is Option D.

First we write down the information given for the original frequency distribution.

The values are $$x_1 = 2,\;x_2 = 6,\;x_3 = 8,\;x_4 = 9$$ and the corresponding frequencies are $$f_1 = 4,\;f_2 = 4,\;f_3 = \alpha,\;f_4 = \beta.$$

We are told that the mean of this distribution is $$6.$$ By definition, the mean (denoted by $$\bar{x}$$) of a discrete frequency distribution is

$$\bar{x} \;=\; \dfrac{\displaystyle\sum_{i=1}^{4} f_i x_i}{\displaystyle\sum_{i=1}^{4} f_i}.$$

Substituting the known numbers we obtain

$$6 \;=\; \dfrac{4 \cdot 2 \;+\; 4 \cdot 6 \;+\; \alpha \cdot 8 \;+\; \beta \cdot 9}{4 + 4 + \alpha + \beta}.$$

Simplifying the numerator and denominator separately, we have

Numerator: $$4\cdot 2 = 8,\;\;4\cdot 6 = 24$$ so $$8 + 24 = 32.$$ Hence the numerator is $$32 + 8\alpha + 9\beta.$$

Denominator: $$4 + 4 = 8,$$ so the total frequency is $$8 + \alpha + \beta.$$

Therefore the mean equation becomes

$$6 \;=\; \dfrac{32 + 8\alpha + 9\beta}{8 + \alpha + \beta}.$$

Cross-multiplying gives

$$32 + 8\alpha + 9\beta \;=\; 6\,(8 + \alpha + \beta).$$

Expanding the right-hand side, $$6(8 + \alpha + \beta) = 48 + 6\alpha + 6\beta.$$

Now we bring all terms to one side:

$$32 + 8\alpha + 9\beta - 48 - 6\alpha - 6\beta = 0.$$

Combining like terms,

$$(8\alpha - 6\alpha) + (9\beta - 6\beta) + (32 - 48) = 0,$$ $$2\alpha + 3\beta - 16 = 0,$$ so

$$2\alpha + 3\beta = 16. \qquad (1)$$

Next we use the given variance. For a frequency distribution, the variance $$\sigma^{2}$$ is given by

$$\sigma^{2} \;=\; \dfrac{\displaystyle\sum_{i=1}^{4} f_i\,(x_i - \bar{x})^{2}}{\displaystyle\sum_{i=1}^{4} f_i}.$$

We are told that $$\sigma^{2} = 6.8.$$ Since the mean is $$\bar{x}=6,$$ we compute each squared deviation:

For $$x_1 = 2: (2-6)^2 = (-4)^2 = 16,$$ for $$x_2 = 6: (6-6)^2 = 0,$$ for $$x_3 = 8: (8-6)^2 = 2^2 = 4,$$ for $$x_4 = 9: (9-6)^2 = 3^2 = 9.$$

Multiplying by their frequencies gives

$$f_1(2-6)^2 = 4 \cdot 16 = 64,$$ $$f_2(6-6)^2 = 4 \cdot 0 = 0,$$ $$f_3(8-6)^2 = \alpha \cdot 4 = 4\alpha,$$ $$f_4(9-6)^2 = \beta \cdot 9 = 9\beta.$$

Hence the numerator of the variance expression is

$$64 + 4\alpha + 9\beta.$$

The denominator is the same total frequency $$8 + \alpha + \beta.$$

Thus the variance equation is

$$6.8 \;=\; \dfrac{64 + 4\alpha + 9\beta}{8 + \alpha + \beta}.$$

Cross-multiplying once again,

$$64 + 4\alpha + 9\beta \;=\; 6.8\,(8 + \alpha + \beta).$$

Compute $$6.8 \times 8 = 54.4,$$ so the right side is $$54.4 + 6.8\alpha + 6.8\beta.$$

Bringing all terms to the left gives

$$64 + 4\alpha + 9\beta - 54.4 - 6.8\alpha - 6.8\beta = 0.$$

Combining like terms,

$$(4\alpha - 6.8\alpha) + (9\beta - 6.8\beta) + (64 - 54.4) = 0,$$ $$-2.8\alpha + 2.2\beta + 9.6 = 0.$$

To clear the decimals, multiply every term by $$10$$:

$$-28\alpha + 22\beta + 96 = 0.$$

Dividing by $$2$$ for simplicity gives

$$-14\alpha + 11\beta + 48 = 0,$$ or equivalently

$$11\beta - 14\alpha = -48. \qquad (2)$$

Now we solve the simultaneous linear equations (1) and (2). From equation (1) we isolate $$\alpha$$:

$$2\alpha + 3\beta = 16 \;\;\Rightarrow\;\; 2\alpha = 16 - 3\beta \;\;\Rightarrow\;\; \alpha = 8 - \dfrac{3}{2}\beta.$$

Substituting this value of $$\alpha$$ into equation (2):

$$11\beta - 14\Bigl(8 - \dfrac{3}{2}\beta\Bigr) = -48.$$

First expand the term inside:

$$14\Bigl(8 - \dfrac{3}{2}\beta\Bigr) = 14 \cdot 8 - 14 \cdot \dfrac{3}{2}\beta = 112 - 21\beta.$$

Hence equation (2) becomes

$$11\beta - \bigl(112 - 21\beta\bigr) = -48.$$

Removing the parentheses carefully:

$$11\beta - 112 + 21\beta = -48.$$

Combine the $$\beta$$ terms:

$$(11\beta + 21\beta) - 112 = -48,$$ $$32\beta - 112 = -48.$$

Add $$112$$ to both sides:

$$32\beta = 64 \;\;\Rightarrow\;\; \beta = \dfrac{64}{32} = 2.$$

Now we substitute $$\beta = 2$$ back into $$\alpha = 8 - \dfrac{3}{2}\beta$$:

$$\alpha = 8 - \dfrac{3}{2}\times 2 = 8 - 3 = 5.$$

Thus we have determined the missing frequencies:

$$\alpha = 5,\quad \beta = 2.$$

Next, the problem states that the value $$x_3$$ is changed from $$8$$ to $$7.$$ Therefore the revised distribution is

$$x: 2,\;6,\;7,\;9$$ with the same frequencies $$4,\;4,\;5,\;2$$ respectively.

We now compute the mean of this new data set. The total frequency remains

$$N = 4 + 4 + 5 + 2 = 15.$$

The sum of the products $$f_i x_i$$ for the new data is

$$\begin{aligned} 4\cdot 2 &= 8,\\ 4\cdot 6 &= 24,\\ 5\cdot 7 &= 35,\\ 2\cdot 9 &= 18. \end{aligned}$$

Adding these contributions together,

$$8 + 24 = 32,\quad 32 + 35 = 67,\quad 67 + 18 = 85.$$

Hence the new mean $$\bar{x}_{\text{new}}$$ is

$$\bar{x}_{\text{new}} \;=\; \dfrac{85}{15} = \dfrac{17}{3}.$$

This value simplifies to $$5\dfrac{2}{3},$$ but writing it as an exact fraction clearly matches the choice given in the options.

Hence, the correct answer is Option C.

We have $$n = 20$$ observations whose mean and standard deviation were first calculated (with the mistaken value) as $$\bar x = 10$$ and $$s = 2.5$$ respectively.

The mean formula is stated as  $$\bar x = \dfrac{\sum x_i}{n}\,.$$

So for the incorrect data the total of all observations is

$$\sum x_i = \bar x \times n = 10 \times 20 = 200.$$

One observation was taken as $$25$$ instead of the correct value $$35$$. Replacing this in the total, we get the correct sum

$$\sum x_i^{(\text{correct})}= 200 - 25 + 35 = 210.$$

Hence the correct mean is

$$\alpha = \dfrac{210}{20}=10.5.$$

For the standard deviation we first recall the relation

$$s^2 = \dfrac{\sum x_i^{2}}{n} - \bar x^{\,2}.$$

With the wrong data we have $$s = 2.5$$ and $$\bar x = 10$$, so

$$2.5^{2}= \dfrac{\sum x_i^{2}}{20} - 10^{2}.$$

This gives

$$6.25 = \dfrac{\sum x_i^{2}}{20} - 100,$$

$$\dfrac{\sum x_i^{2}}{20} = 106.25,$$

$$\sum x_i^{2}= 106.25 \times 20 = 2125.$$

The mistaken squared term $$25^{2}=625$$ must be replaced by the correct squared term $$35^{2}=1225$$. Therefore the corrected sum of squares is

$$\sum x_i^{2\,(\text{correct})}= 2125 - 625 + 1225 = 2725.$$

Substituting in the variance formula with the corrected mean $$\alpha = 10.5$$, we get

$$\sigma^{2}= \dfrac{2725}{20} - (10.5)^{2}.$$

The term $$\dfrac{2725}{20}=136.25$$ and $$(10.5)^{2}=110.25,$$ so

$$\sigma^{2}=136.25 - 110.25 = 26.$$

Thus the corrected standard deviation is $$\sigma = \sqrt{26}.$$ By the wording of the question $$\sqrt{\beta} = \sigma,$$ hence $$\beta = 26.$$

We have found $$\alpha = 10.5$$ and $$\beta = 26,$$ so the ordered pair is $$(\alpha, \beta) = (10.5, 26).$$

Hence, the correct answer is Option A.

We start with the eight observations 6, 10, 7, 13, $$a$$, 12, $$b$$, 12. The number of observations is $$n = 8$$.

The mean is given to be $$9$$. For any data set, the mean $$\bar x$$ is obtained by the formula $$\bar x = \dfrac{\text{sum of all observations}}{n}.$$ So we have

$$9 = \dfrac{6 + 10 + 7 + 13 + a + 12 + b + 12}{8}.$$

Adding the known numbers first, $$6 + 10 + 7 + 13 + 12 + 12 = 60.$$ Hence

$$9 = \dfrac{60 + a + b}{8}.$$

Multiplying both sides by $$8$$, $$72 = 60 + a + b.$$ So

$$a + b = 72 - 60 = 12.$$

Next, the variance is given as $$\dfrac{37}{4}$$. For a population of size $$n$$, the variance $$\sigma^{2}$$ is defined as $$\sigma^{2} = \dfrac{\sum (x - \bar x)^2}{n}.$$ Therefore

$$\dfrac{37}{4} = \dfrac{(6-9)^2 + (10-9)^2 + (7-9)^2 + (13-9)^2 + (a-9)^2 + (12-9)^2 + (b-9)^2 + (12-9)^2}{8}.$$

We now evaluate the squared deviations of the known numbers:

$$\begin{aligned} (6-9)^2 &= (-3)^2 = 9,\\ (10-9)^2 &= 1^2 = 1,\\ (7-9)^2 &= (-2)^2 = 4,\\ (13-9)^2 &= 4^2 = 16,\\ (12-9)^2 &= 3^2 = 9,\\ (12-9)^2 &= 3^2 = 9. \end{aligned}$$

Adding these six values, $$9 + 1 + 4 + 16 + 9 + 9 = 48.$$

Substituting into the variance formula and multiplying both sides by $$8$$ gives

$$8 \times \dfrac{37}{4} = 48 + (a-9)^2 + (b-9)^2.$$

Since $$8 \times \dfrac{37}{4} = 2 \times 37 = 74$$, we obtain

$$(a-9)^2 + (b-9)^2 = 74 - 48 = 26.$$

Let us expand the left side. We know the algebraic identity $$(x-9)^2 = x^2 - 18x + 81.$$ Applying this to both $$a$$ and $$b$$,

$$(a-9)^2 + (b-9)^2 = (a^2 - 18a + 81) + (b^2 - 18b + 81) = a^2 + b^2 - 18(a + b) + 162.$$

We already found $$a + b = 12$$, so substitute:

$$a^2 + b^2 - 18 \times 12 + 162 = a^2 + b^2 - 216 + 162 = a^2 + b^2 - 54.$$

Set this equal to $$26$$ (from the variance condition):

$$a^2 + b^2 - 54 = 26 \quad\Longrightarrow\quad a^2 + b^2 = 80.$$

To find $$(a-b)^2$$, recall the identity $$(a-b)^2 = a^2 + b^2 - 2ab.$$ We already know $$a^2 + b^2 = 80$$, but we still need $$ab$$. Using another standard identity, $$(a + b)^2 = a^2 + b^2 + 2ab.$$

Substituting $$a + b = 12$$ and $$a^2 + b^2 = 80$$, we have

$$12^2 = 144 = 80 + 2ab.$$

Hence $$2ab = 144 - 80 = 64 \quad\Longrightarrow\quad ab = 32.$$

Now apply the identity for $$(a-b)^2$$:

$$(a-b)^2 = 80 - 2 \times 32 = 80 - 64 = 16.$$

Hence, the correct answer is Option D.

Let us denote the size, mean and standard deviation of the first sample by $$n_1,\;\mu_1,\;\sigma_1$$ and of the second sample by $$n_2,\;\mu_2,\;\sigma_2.$$ The symbols $$n,\;\mu,\;\sigma$$ will represent the corresponding quantities for the whole group.

We have for the first sample $$n_1 = 100,\;\mu_1 = 15,\;\sigma_1 = 3.$$ Therefore the variance of the first sample is $$\sigma_1^{2} = 3^{2} = 9.$$

The whole group consists of $$n = 250$$ items with mean $$\mu = 15.6$$ and standard deviation $$\sigma = \sqrt{13.44},$$ so the variance of the whole group is $$\sigma^{2} = 13.44.$$

Because the whole group contains 250 items and the first sample contains 100 items, the second sample must contain

$$n_2 = n - n_1 = 250 - 100 = 150$$

items.

First, we compute the mean of the second sample. The formula for the combined mean of two groups is

$$\mu = \frac{n_1\mu_1 + n_2\mu_2}{n_1 + n_2}.$$

Solving this for $$\mu_2$$ gives

$$\mu_2 = \frac{(n_1 + n_2)\mu - n_1\mu_1}{n_2}.$$

Substituting the known numbers, we obtain

$$\mu_2 = \frac{250 \times 15.6 - 100 \times 15}{150} = \frac{3900 - 1500}{150} = \frac{2400}{150} = 16.$$

Now we move to the variance. The formula that expresses the variance of the combined group in terms of the variances and means of the two sub-groups is

$$\sigma^{2} \;=\; \frac{n_1\bigl[\sigma_1^{2} + (\mu_1 - \mu)^{2}\bigr] \;+\; n_2\bigl[\sigma_2^{2} + (\mu_2 - \mu)^{2}\bigr]} {n_1 + n_2}.$$

We already know $$\sigma^{2} = 13.44,$$ so we substitute every known quantity and keep $$\sigma_2^{2}$$ as the unknown.

First compute the bracket for the first sample:

$$\sigma_1^{2} + (\mu_1 - \mu)^{2} = 9 + (15 - 15.6)^{2} = 9 + (-0.6)^{2} = 9 + 0.36 = 9.36.$$

Multiplying by $$n_1$$ gives

$$n_1\bigl[\sigma_1^{2} + (\mu_1 - \mu)^{2}\bigr] = 100 \times 9.36 = 936.$$

Next, form the bracket for the second sample, leaving $$\sigma_2^{2}$$ unknown:

$$\sigma_2^{2} + (\mu_2 - \mu)^{2} = \sigma_2^{2} + (16 - 15.6)^{2} = \sigma_2^{2} + 0.4^{2} = \sigma_2^{2} + 0.16.$$

Multiplying by $$n_2$$ gives

$$n_2\bigl[\sigma_2^{2} + (\mu_2 - \mu)^{2}\bigr] = 150\bigl(\sigma_2^{2} + 0.16\bigr) = 150\sigma_2^{2} + 24.$$

Putting all these into the combined variance formula, we get

$$13.44 = \frac{936 + 150\sigma_2^{2} + 24}{250}.$$

Simplify the numerator inside the fraction:

$$936 + 24 = 960,$$

so the equation becomes

$$13.44 = \frac{960 + 150\sigma_2^{2}}{250}.$$

Now multiply both sides by 250 to clear the denominator:

$$13.44 \times 250 = 960 + 150\sigma_2^{2}.$$

Calculating the left-hand product, we get

$$13.44 \times 250 = 3360,$$

so

$$3360 = 960 + 150\sigma_2^{2}.$$

Subtract 960 from both sides:

$$3360 - 960 = 150\sigma_2^{2},$$

$$2400 = 150\sigma_2^{2}.$$

Finally, divide by 150:

$$\sigma_2^{2} = \frac{2400}{150} = 16.$$

The variance of the second sample is therefore $$16,$$ and its standard deviation is the square root of this value:

$$\sigma_2 = \sqrt{16} = 4.$$

Hence, the correct answer is Option C.

Let the $$3n$$ numbers be $$x_1, x_2, \ldots, x_{3n}$$ with mean of the first $$2n$$ numbers equal to $$6$$ and mean of the remaining $$n$$ numbers equal to $$3$$. The overall mean is $$\bar{x} = \frac{2n \cdot 6 + n \cdot 3}{3n} = \frac{15n}{3n} = 5$$.

The variance is given as $$4$$, so $$\frac{1}{3n}\sum_{i=1}^{3n} x_i^2 - \bar{x}^2 = 4$$, giving $$\frac{1}{3n}\sum x_i^2 = 29$$, hence $$\sum x_i^2 = 87n$$.

Now the new set is formed by adding $$1$$ to each of the first $$2n$$ numbers and subtracting $$1$$ from each of the remaining $$n$$ numbers. The new values are $$y_i = x_i + 1$$ for $$i = 1, \ldots, 2n$$ and $$y_i = x_i - 1$$ for $$i = 2n+1, \ldots, 3n$$.

The new sum is $$\sum y_i = \sum_{i=1}^{2n}(x_i+1) + \sum_{i=2n+1}^{3n}(x_i-1) = \sum x_i + 2n - n = 15n + n = 16n$$. The new mean is $$\bar{y} = \frac{16n}{3n} = \frac{16}{3}$$.

The new sum of squares is $$\sum y_i^2 = \sum_{i=1}^{2n}(x_i+1)^2 + \sum_{i=2n+1}^{3n}(x_i-1)^2 = \sum_{i=1}^{2n}(x_i^2 + 2x_i + 1) + \sum_{i=2n+1}^{3n}(x_i^2 - 2x_i + 1)$$.

This equals $$\sum x_i^2 + 2\sum_{i=1}^{2n}x_i - 2\sum_{i=2n+1}^{3n}x_i + 2n + n = 87n + 2(12n) - 2(3n) + 3n = 87n + 24n - 6n + 3n = 108n$$.

The new variance is $$k = \frac{\sum y_i^2}{3n} - \bar{y}^2 = \frac{108n}{3n} - \frac{256}{9} = 36 - \frac{256}{9} = \frac{324 - 256}{9} = \frac{68}{9}$$.

Therefore $$9k = 9 \cdot \frac{68}{9} = 68$$.

For the first set: $$n_1 = 10$$, $$\bar{x}_1 = 2$$, $$\sigma_1^2 = 2$$, so $$\sum x_i = 20$$ and $$\sum x_i^2 = n_1(\sigma_1^2 + \bar{x}_1^2) = 10(2+4) = 60$$.

For the second set: $$n_2 = n$$, $$\bar{x}_2 = 3$$, $$\sigma_2^2 = 1$$, so $$\sum y_j = 3n$$ and $$\sum y_j^2 = n(1+9) = 10n$$.

The combined mean is $$\bar{x} = \frac{20 + 3n}{10 + n}$$.

The combined variance is $$\sigma^2 = \frac{\sum x_i^2 + \sum y_j^2}{10+n} - \bar{x}^2 = \frac{60 + 10n}{10+n} - \left(\frac{20+3n}{10+n}\right)^2$$.

Setting this equal to $$\frac{17}{9}$$: $$\frac{(60+10n)(10+n) - (20+3n)^2}{(10+n)^2} = \frac{17}{9}$$.

Expanding the numerator: $$(60+10n)(10+n) = 10n^2 + 160n + 600$$ and $$(20+3n)^2 = 9n^2 + 120n + 400$$.

Numerator = $$n^2 + 40n + 200$$.

So $$9(n^2 + 40n + 200) = 17(10+n)^2 = 17(n^2 + 20n + 100)$$.

$$9n^2 + 360n + 1800 = 17n^2 + 340n + 1700$$, giving $$8n^2 - 20n - 100 = 0$$, i.e., $$2n^2 - 5n - 25 = 0$$.

$$n = \frac{5 \pm \sqrt{25 + 200}}{4} = \frac{5 \pm 15}{4}$$. Since $$n > 0$$, $$n = \frac{20}{4} = 5$$.

The answer is $$5$$.

The frequency distribution is: Class 0-6 (frequency $$a$$), 6-12 (frequency $$b$$), 12-18 (frequency 12), 18-24 (frequency 9), 24-30 (frequency 5).

The total frequency is $$N = a + b + 12 + 9 + 5 = a + b + 26$$.

The class midpoints are 3, 9, 15, 21, 27. Using the mean formula: $$\text{Mean} = \frac{3a + 9b + 15(12) + 21(9) + 27(5)}{N} = \frac{3a + 9b + 180 + 189 + 135}{a + b + 26} = \frac{3a + 9b + 504}{a + b + 26}$$.

Setting this equal to $$\frac{309}{22}$$ and cross-multiplying: $$22(3a + 9b + 504) = 309(a + b + 26)$$.

Expanding: $$66a + 198b + 11088 = 309a + 309b + 8034$$.

Simplifying: $$243a + 111b = 3054$$. Dividing by 3: $$81a + 37b = 1018$$. $$-(1)$$

For the median, the cumulative frequencies are: up to 6 is $$a$$, up to 12 is $$a + b$$, up to 18 is $$a + b + 12$$, up to 24 is $$a + b + 21$$, up to 30 is $$a + b + 26$$.

Since the median is 14, it falls in the class 12-18 (with frequency 12). Using the median formula: $$\text{Median} = L + \frac{\frac{N}{2} - F}{f} \times h$$, where $$L = 12$$, $$F = a + b$$ (cumulative frequency before the median class), $$f = 12$$, and $$h = 6$$.

Substituting: $$14 = 12 + \frac{\frac{N}{2} - (a+b)}{12} \times 6$$.

So $$2 = \frac{N/2 - (a+b)}{2}$$, which gives $$N/2 - (a+b) = 4$$. $$-(2)$$

Since $$N = a + b + 26$$, we get $$\frac{a + b + 26}{2} - (a + b) = 4$$, so $$a + b + 26 - 2(a+b) = 8$$, giving $$a + b = 18$$. $$-(3)$$

Substituting $$b = 18 - a$$ into equation $$(1)$$: $$81a + 37(18 - a) = 1018$$.

$$81a + 666 - 37a = 1018$$, so $$44a = 352$$, giving $$a = 8$$.

From $$(3)$$: $$b = 18 - 8 = 10$$.

Therefore $$(a - b)^2 = (8 - 10)^2 = (-2)^2 = 4$$.

We are given the consecutive natural numbers $$1,2,3,\dots ,n$$ and told that their variance equals $$14$$. Because every positive integer from $$1$$ to $$n$$ appears exactly once, the total number of observations is also $$n$$. Throughout this solution we shall regard the data set as the whole population, so we use the population variance formula

$$\sigma^{2}= \dfrac{1}{N}\sum_{i=1}^{N}(x_{i}-\mu)^{2},$$

where $$N$$ is the number of observations, $$x_{i}$$ denotes an individual observation and $$\mu$$ denotes the mean (average) of all the observations.

First we determine the mean. The sum of the first $$n$$ natural numbers is well known to be

$$\sum_{i=1}^{n} i=\dfrac{n(n+1)}{2}.$$

Dividing this sum by the total number of observations $$n$$ gives the mean

$$\mu=\dfrac{\dfrac{n(n+1)}{2}}{n}= \dfrac{n+1}{2}.$$

Next we require the second moment, that is, the mean of the squares. The sum of the squares of the first $$n$$ natural numbers is

$$\sum_{i=1}^{n} i^{2}= \dfrac{n(n+1)(2n+1)}{6}.$$

Hence the mean of the squares, often written as $$E[X^{2}]$$, equals

$$E[X^{2}]=\dfrac{1}{n}\sum_{i=1}^{n} i^{2}= \dfrac{1}{n}\cdot\dfrac{n(n+1)(2n+1)}{6}= \dfrac{(n+1)(2n+1)}{6}.$$

The variance can be computed by the identity

$$\sigma^{2}=E[X^{2}] - \mu^{2}.$$

Substituting the expressions we have just obtained, we write

$$\sigma^{2}= \dfrac{(n+1)(2n+1)}{6} - \left(\dfrac{n+1}{2}\right)^{2}.$$

Because it is given that the variance equals $$14$$, we set the above expression equal to $$14$$:

$$\dfrac{(n+1)(2n+1)}{6} - \left(\dfrac{n+1}{2}\right)^{2}=14.$$

To simplify, notice that each term contains the factor $$(n+1).$$ We write

$$\dfrac{(n+1)(2n+1)}{6} - \dfrac{(n+1)^{2}}{4}=14.$$

Factorising $$(n+1)$$ from both numerators gives

$$(n+1)\left[\dfrac{2n+1}{6}-\dfrac{n+1}{4}\right]=14.$$

We now combine the expressions in the square brackets by taking a common denominator of $$12$$:

$$\dfrac{2n+1}{6}= \dfrac{4n+2}{12},\qquad \dfrac{n+1}{4}= \dfrac{3n+3}{12}.$$

Subtracting inside the brackets, we obtain

$$\dfrac{4n+2}{12}-\dfrac{3n+3}{12}= \dfrac{4n+2-3n-3}{12}= \dfrac{n-1}{12}.$$

Therefore

$$(n+1)\left(\dfrac{n-1}{12}\right)=14.$$

Multiplying both sides by $$12$$ to clear the denominator, we have

$$(n+1)(n-1)=14\times12=168.$$

Recognising the left side as the difference of squares, we write

$$n^{2}-1=168.$$

Adding $$1$$ to both sides produces

$$n^{2}=169.$$

Taking the positive square root (because $$n$$ is a natural number) yields

$$n=13.$$

Finally, the question stipulates that $$n$$ must be odd; $$13$$ is indeed odd, so our solution is fully consistent.

So, the answer is $$13$$.

We are given ten numbers that follow the pattern $$7 \times 8,\; 10 \times 10,\; 13 \times 12,\; 16 \times 14,\; \ldots$$ and we have to find their mean.

First we observe the pattern in each factor:

• The first factor starts at $$7$$ and increases by $$3$$ each time: $$7,\; 10,\; 13,\; 16,\; \ldots$$

• The second factor starts at $$8$$ and increases by $$2$$ each time: $$8,\; 10,\; 12,\; 14,\; \ldots$$

So for the $$n^{\text{th}}$$ term (counting the first term as $$n=1$$) we can write

$$a_n=\bigl(7+3(n-1)\bigr)\bigl(8+2(n-1)\bigr).$$

Because we need ten terms, it will be convenient to let $$n$$ run from $$0$$ to $$9$$ instead of $$1$$ to $$10$$; that re-indexes nothing but simplifies the algebra. Thus we define

$$A_n=\bigl(7+3n\bigr)\bigl(8+2n\bigr),\qquad n=0,1,2,\ldots,9.$$

We now expand each product algebraically:

$$\begin{aligned} A_n&=\bigl(7+3n\bigr)\bigl(8+2n\bigr)\\ &=7\cdot8+7\cdot2n+3n\cdot8+3n\cdot2n\\ &=56+14n+24n+6n^{2}\\ &=56+38n+6n^{2}. \end{aligned}$$

The sum of all ten terms is therefore

$$S=\sum_{n=0}^{9}A_n=\sum_{n=0}^{9}\bigl(56+38n+6n^{2}\bigr).$$

We separate the sum into three simpler sums:

$$S=56\sum_{n=0}^{9}1+38\sum_{n=0}^{9}n+6\sum_{n=0}^{9}n^{2}.$$

We now evaluate each of these standard sums one by one.

1. The sum of ten ones:

$$\sum_{n=0}^{9}1=10.$$

2. The sum of the first nine positive integers (starting from zero):
The formula for the sum of the first $$m$$ integers is $$\dfrac{m(m+1)}{2}.$$ Here $$m=9$$, so

$$\sum_{n=0}^{9}n=\frac{9\cdot10}{2}=45.$$

3. The sum of the squares of the first nine positive integers (starting from zero):
The formula for the sum of squares is $$\dfrac{m(m+1)(2m+1)}{6}.$$ Again $$m=9$$, hence

$$\sum_{n=0}^{9}n^{2}=\frac{9\cdot10\cdot19}{6}=285.$$

Substituting these values back into $$S$$ we get

$$\begin{aligned} S&=56(10)+38(45)+6(285)\\ &=560+1710+1710\\ &=3980. \end{aligned}$$

The mean (average) of the ten numbers is the total sum divided by 10:

$$\text{Mean}=\frac{S}{10}=\frac{3980}{10}=398.$$

So, the answer is $$398$$.

We have been given a grouped frequency distribution with class-interval width $$h = 10$$. The classes and their (as yet unknown) frequencies are

10 - 20 : $$\alpha$$,    20 - 30 : 110,    30 - 40 : 54,    40 - 50 : 30,    50 - 60 : $$\beta$$.

The total number of observations is stated to be $$N = 584$$ and the median is $$45$$.

For a continuous (grouped) distribution, the median is found by the formula

$$\text{Median} = L + \frac{\dfrac{N}{2} - \text{c.f.\,(prev)}}{f_m}\;h,$$

where $$L$$ is the lower class boundary of the median class, $$\text{c.f.\,(prev)}$$ is the cumulative frequency up to (but not including) the median class, $$f_m$$ is the frequency of the median class, and $$h$$ is the class width.

Since the median value $$45$$ lies in the interval $$40\!-\!50$$, that class is the median class. Hence $$L = 40$$, $$f_m = 30$$ and $$h = 10$$.

The cumulative frequency just before the median class is the sum of the first three class frequencies:

$$\text{c.f.\,(prev)} = \alpha + 110 + 54 = \alpha + 164.$$

Substituting all these quantities into the median formula gives

$$45 = 40 + \frac{\dfrac{584}{2} - (\alpha + 164)}{30}\;10.$$

Simplifying step by step:

$$45 - 40 = \frac{292 - (\alpha + 164)}{30}\;10$$ $$5 = \frac{292 - \alpha - 164}{30}\;10$$ $$5 = \frac{128 - \alpha}{3}$$ $$5 \times 3 = 128 - \alpha$$ $$15 = 128 - \alpha$$ $$\alpha = 128 - 15$$ $$\alpha = 113.$$

Now we use the fact that the sum of all frequencies equals $$584$$:

$$\alpha + 110 + 54 + 30 + \beta = 584.$$

Substituting $$\alpha = 113$$ we get

$$113 + 110 + 54 + 30 + \beta = 584.$$

Adding the known numbers:

$$307 + \beta = 584$$ $$\beta = 584 - 307$$ $$\beta = 277.$$

Finally, we require the absolute difference between $$\alpha$$ and $$\beta$$:

$$|\alpha - \beta| = |113 - 277| = |-164| = 164.$$

Hence, the correct answer is Option 164.

We have four original numbers $$3,\,7,\,x,\,y$$ whose mean is given to be $$5$$. The definition of mean says

$$\bar{x}= \frac{3+7+x+y}{4}=5.$$

Multiplying by $$4$$ we get

$$3+7+x+y = 20.$$

Simplifying,

$$x+y = 10.$$

Now the variance of these four numbers is given as $$10$$. For a population of four numbers, variance is defined by the formula

$$\sigma^{2}= \frac{1}{4}\left[(3-\bar{x})^{2}+(7-\bar{x})^{2}+(x-\bar{x})^{2}+(y-\bar{x})^{2}\right].$$

Here $$\bar{x}=5$$, so we substitute:

$$10 = \frac{1}{4}\Big[(3-5)^{2} + (7-5)^{2} + (x-5)^{2} + (y-5)^{2}\Big].$$

Calculating the first two squares,

$$(3-5)^{2}=(-2)^{2}=4,\qquad (7-5)^{2}=2^{2}=4.$$

Hence

$$10 = \frac{1}{4}\Big[4 + 4 + (x-5)^{2} + (y-5)^{2}\Big].$$

Multiplying both sides by $$4$$,

$$40 = 8 + (x-5)^{2} + (y-5)^{2}.$$

So

$$(x-5)^{2} + (y-5)^{2} = 32.$$

From the mean calculation we already have $$y = 10 - x$$. Substituting this into the variance equation,

$$(x-5)^{2} + \big[(10 - x) - 5\big]^{2} = 32.$$

The second square simplifies as

$$(10 - x) - 5 = 5 - x,$$

and $$\big(5 - x\big)^{2} = (x - 5)^{2}$$ (since squaring removes the sign). Therefore

$$(x-5)^{2} + (x-5)^{2} = 32,$$

or

$$2(x-5)^{2} = 32.$$

Dividing by $$2$$,

$$(x-5)^{2} = 16.$$

Taking square roots,

$$x-5 = \pm 4 \;\;\Longrightarrow\;\; x = 5 \pm 4.$$

Thus $$x = 9$$ or $$x = 1$$. We are told $$x > y$$, and recall $$x + y = 10$$, so

• If $$x = 9$$, then $$y = 1$$, giving $$x > y$$ (acceptable).
• If $$x = 1$$, then $$y = 9$$, which violates $$x > y$$.

Hence the only valid pair is $$x = 9,\,y = 1$$.

Now we form the new set of numbers: $$3 + 2x,\; 7 + 2y,\; x + y,\; x - y.$$

Substituting $$x = 9,\,y = 1$$:

$$3 + 2x = 3 + 2\cdot9 = 21,$$ $$7 + 2y = 7 + 2\cdot1 = 9,$$ $$x + y = 9 + 1 = 10,$$ $$x - y = 9 - 1 = 8.$$

The mean of these four numbers is

$$\text{Mean} = \frac{21 + 9 + 10 + 8}{4} = \frac{48}{4} = 12.$$

So, the answer is $$12$$.

From $$\sum_{i=1}^{18}(X_i - \alpha) = 36$$, we get $$\sum X_i - 18\alpha = 36$$, so the mean is $$\bar{X} = \frac{\sum X_i}{18} = \alpha + 2$$.

Expanding $$\sum_{i=1}^{18}(X_i - \beta)^2 = \sum_{i=1}^{18}[(X_i - \bar{X}) + (\bar{X} - \beta)]^2 = \sum(X_i - \bar{X})^2 + 18(\bar{X} - \beta)^2$$.

Since the standard deviation is 1, the variance is 1, so $$\sum(X_i - \bar{X})^2 = 18 \cdot 1 = 18$$.

Substituting into the given equation: $$90 = 18 + 18(\bar{X} - \beta)^2$$, which gives $$(\bar{X} - \beta)^2 = 4$$, so $$\bar{X} - \beta = \pm 2$$.

Since $$\bar{X} = \alpha + 2$$, we have $$\alpha + 2 - \beta = \pm 2$$, giving $$\alpha - \beta = 0$$ or $$\alpha - \beta = -4$$. Since $$\alpha$$ and $$\beta$$ are distinct, $$|\alpha - \beta| = 4$$.

Let us first interpret the data. There are in all 50 candidates. Out of these, the number of boys is 20 and the number of girls is 30.

We are told that the average (mean) marks of the boys is $$\mu_1 = 12$$ and the variance of the boys’ marks is $$\sigma_1^{2}=2.$$ Likewise, the variance of the girls’ marks is given to be $$\sigma_2^{2}=2,$$ while their mean is presently unknown; let us denote it by $$\mu_2 = \mu.$$ The overall mean (average of all 50 candidates taken together) is given as $$\bar{x}=15.$$

We begin by finding the girls’ mean $$\mu.$$ Using the definition of mean, the total marks obtained by all 50 candidates equals the sum of total marks of boys and girls. Hence

$$\underbrace{50}_{\text{total students}}\times\underbrace{15}_{\text{overall mean}} \;=\; \underbrace{20}_{\text{boys}}\times\underbrace{12}_{\text{boys’ mean}} \;+\; \underbrace{30}_{\text{girls}}\times\mu.$$

Carrying out the multiplication, we have

$$750 \;=\; 240 \;+\; 30\mu.$$

Now subtract 240 from both sides:

$$750 - 240 \;=\; 30\mu.$$ $$510 \;=\; 30\mu.$$

Divide both sides by 30:

$$\mu \;=\; \frac{510}{30} \;=\; 17.$$

Thus the average marks of the girls is $$\mu_2 = 17.$$

Next we compute the variance of marks for all 50 candidates together. For two groups, the formula for the combined variance $$\sigma^{2}$$ is

$$\sigma^{2} \;=\; \frac{n_1\Big(\sigma_1^{2} + (\mu_1 - \bar{x})^{2}\Big) \;+\; n_2\Big(\sigma_2^{2} + (\mu_2 - \bar{x})^{2}\Big)}{n_1 + n_2},$$

where $$n_1,\,\mu_1,\,\sigma_1^{2}$$ refer to the first group (boys) and $$n_2,\,\mu_2,\,\sigma_2^{2}$$ refer to the second group (girls), while $$\bar{x}$$ is the overall mean.

Substituting the known numbers:

$$\sigma^{2} \;=\; \frac{20\Big(2 + (12 - 15)^{2}\Big) + 30\Big(2 + (17 - 15)^{2}\Big)}{20 + 30}.$$

Let us evaluate each bracket step by step.

For boys: $$(12 - 15)^{2} = (-3)^{2} = 9,$$ so $$2 + 9 = 11.$$ Multiplying by the number of boys: $$20 \times 11 = 220.$$

For girls: $$(17 - 15)^{2} = (2)^{2} = 4,$$ so $$2 + 4 = 6.$$ Multiplying by the number of girls: $$30 \times 6 = 180.$$

Add these two totals:

$$220 + 180 = 400.$$

Finally, divide by the total number of candidates (50):

$$\sigma^{2} = \frac{400}{50} = 8.$$

Thus the variance of marks of the entire set of 50 candidates is $$\sigma^{2}=8.$$

We are asked to find $$\mu + \sigma^{2}.$$ We already have $$\mu = 17$$ and $$\sigma^{2}=8,$$ therefore

$$\mu + \sigma^{2} = 17 + 8 = 25.$$

So, the answer is $$25.$$

We have 10 natural numbers: nine 1's and one value $$k$$. The mean is $$\mu = \frac{9 + k}{10}$$.

The variance is $$\sigma^2 = \frac{1}{10}\left[9(1 - \mu)^2 + (k - \mu)^2\right]$$. We compute each deviation: $$1 - \mu = 1 - \frac{9 + k}{10} = \frac{1 - k}{10}$$ and $$k - \mu = k - \frac{9 + k}{10} = \frac{9(k - 1)}{10}$$.

Substituting: $$\sigma^2 = \frac{1}{10}\left[9 \cdot \frac{(k-1)^2}{100} + \frac{81(k-1)^2}{100}\right] = \frac{1}{10} \cdot \frac{(k-1)^2}{100}(9 + 81) = \frac{90(k-1)^2}{1000} = \frac{9(k-1)^2}{100}$$.

We need $$\frac{9(k-1)^2}{100} < 10$$, which gives $$(k-1)^2 < \frac{1000}{9} \approx 111.11$$. Since $$k$$ is a natural number, $$k - 1 \leq 10$$, i.e., $$k \leq 11$$.

Verification: for $$k = 11$$, the variance is $$\frac{9 \times 100}{100} = 9 < 10$$. For $$k = 12$$, the variance is $$\frac{9 \times 121}{100} = 10.89 > 10$$.

The maximum possible value of $$k$$ is $$11$$.

We have a discrete random variable $$X$$ that can take the five values $$-2,\,-1,\,3,\,4,\,6$$ with respective probabilities $$\dfrac15,\,a,\,\dfrac13,\,\dfrac15,\,b$$. Because these probabilities must add up to $$1$$, we first write

$$\dfrac15 + a + \dfrac13 + \dfrac15 + b = 1.$$

The two fractions on the left that are already known combine as

$$\dfrac15 + \dfrac15 = \dfrac25,$$

and using a common denominator of $$15$$ we get

$$\dfrac25 + \dfrac13 = \dfrac{6}{15} + \dfrac{5}{15} = \dfrac{11}{15}.$$

So the probability-sum equation becomes

$$a + b + \dfrac{11}{15} = 1 \quad\Longrightarrow\quad a + b = 1 - \dfrac{11}{15} = \dfrac{4}{15}. \quad -(1)$$

Next, the mean (expected value) is given to be $$2.3$$. The formula for the mean of a discrete variable is

$$E[X] = \sum x_i P(X = x_i).$$

Substituting all values and probabilities we obtain

$$E[X] = (-2)\!\left(\dfrac15\right) + (-1)\,a + 3\!\left(\dfrac13\right) + 4\!\left(\dfrac15\right) + 6b.$$

We now simplify each product:

$$(-2)\!\left(\dfrac15\right) = -\dfrac25,\qquad 3\!\left(\dfrac13\right) = 1,\qquad 4\!\left(\dfrac15\right) = \dfrac45.$$

Hence

$$E[X] = -\dfrac25 - a + 1 + \dfrac45 + 6b.$$

Combining the three constant fractions gives

$$-\dfrac25 + 1 + \dfrac45 = \dfrac{-2 + 5 + 4}{5} = \dfrac75 = 1.4.$$

Therefore

$$1.4 - a + 6b = 2.3.$$

Writing $$2.3$$ as the fraction $$\dfrac{23}{10}$$ and $$1.4$$ as $$\dfrac{14}{10},$$ we get

$$-\!a + 6b = \dfrac{23}{10} - \dfrac{14}{10} = \dfrac{9}{10}. \quad -(2)$$

We now solve the simultaneous linear equations (1) and (2):

From (1)  $$a = \dfrac{4}{15} - b.$$

Substituting this into (2), we have

$$-\!\left(\dfrac{4}{15} - b\right) + 6b = \dfrac{9}{10}.$$

Simplifying inside the brackets and then combining terms:

$$-\dfrac{4}{15} + b + 6b = \dfrac{9}{10}$$

$$-\dfrac{4}{15} + 7b = \dfrac{9}{10}$$

$$7b = \dfrac{9}{10} + \dfrac{4}{15}.$$

Adding the two fractions on the right using the common denominator $$30$$, we get

$$\dfrac{9}{10} = \dfrac{27}{30},\qquad \dfrac{4}{15} = \dfrac{8}{30},\qquad \dfrac{27}{30} + \dfrac{8}{30} = \dfrac{35}{30} = \dfrac76.$$

Thus

$$7b = \dfrac76 \quad\Longrightarrow\quad b = \dfrac16.$$

Using $$a + b = \dfrac{4}{15},$$ we now find

$$a = \dfrac{4}{15} - \dfrac16 = \dfrac{8}{30} - \dfrac{5}{30} = \dfrac{3}{30} = \dfrac1{10}.$$

With both unknown probabilities known, we can compute the variance. First, we need $$E[X^2].$$ The required formula is

$$E[X^2] = \sum x_i^2 P(X = x_i).$$

We list each term:

$$(-2)^2\!\left(\dfrac15\right) = 4\!\left(\dfrac15\right) = \dfrac45,$$

$$( -1)^2 a = 1\!\left(\dfrac1{10}\right) = \dfrac1{10},$$

$$3^2\!\left(\dfrac13\right) = 9\!\left(\dfrac13\right) = 3,$$

$$4^2\!\left(\dfrac15\right) = 16\!\left(\dfrac15\right) = \dfrac{16}{5},$$

$$6^2 b = 36\!\left(\dfrac16\right) = 6.$$

Adding them together:

$$E[X^2] = \dfrac45 + \dfrac1{10} + 3 + \dfrac{16}{5} + 6.$$

Converting each to the denominator $$10$$ for ease of addition, we have

$$\dfrac45 = \dfrac8{10},\quad \dfrac{16}{5} = \dfrac{32}{10},\quad 3 = \dfrac{30}{10},\quad 6 = \dfrac{60}{10}.$$

So

$$E[X^2] = \dfrac8{10} + \dfrac1{10} + \dfrac{30}{10} + \dfrac{32}{10} + \dfrac{60}{10} = \dfrac{131}{10}.$$

Now, the variance formula is

$$\sigma^2 = E[X^2] - \bigl(E[X]\bigr)^2.$$

We know $$E[X] = 2.3 = \dfrac{23}{10}.$$ Therefore

$$\sigma^2 = \dfrac{131}{10} - \left(\dfrac{23}{10}\right)^2 = \dfrac{131}{10} - \dfrac{529}{100}.$$

Writing $$\dfrac{131}{10} = \dfrac{1310}{100},$$ we obtain

$$\sigma^2 = \dfrac{1310}{100} - \dfrac{529}{100} = \dfrac{781}{100}.$$

Finally, multiplying by $$100$$ as required, we get

$$100\sigma^2 = 100 \times \dfrac{781}{100} = 781.$$

So, the answer is $$781$$.

We have the finite set $$X=\{1,2,3,\dots ,17\}$$. Its size is clearly $$|X|=17$$, so whenever we speak of the “mean’’ and “variance’’ of the elements of $$X$$ we are using the uniform distribution in which each element appears with probability $$\tfrac1{17}$$.

First we calculate the mean of the elements of $$X$$. The sum of the first $$n$$ natural numbers is given by the well-known formula $$\sum_{x=1}^{n} x=\frac{n(n+1)}2.$$ Substituting $$n=17$$ we obtain $$\sum_{x=1}^{17} x=\frac{17\cdot18}2=153.$$ Hence the mean of $$X$$ is $$\mu_X=\frac1{17}\sum_{x=1}^{17}x=\frac{153}{17}=9.$$

Next we need the variance of the numbers $$1,2,\dots ,17$$. By definition, $$\sigma_X^{2}=\frac1{17}\sum_{x=1}^{17}(x-\mu_X)^2.$$ A convenient way to compute this is to use the identity $$\sigma_X^{2}=\frac1{17}\sum_{x=1}^{17}x^{2}-\mu_X^{2}.$$ The sum of squares formula $$\sum_{x=1}^{n}x^{2}=\frac{n(n+1)(2n+1)}6$$ gives, for $$n=17$$, $$\sum_{x=1}^{17}x^{2}=\frac{17\cdot18\cdot35}6.$$ Evaluating step by step, we see $$17\cdot18=306,\qquad 306\cdot35=10\,710,$$ and finally $$\frac{10\,710}{6}=1\,785.$$ Therefore $$\frac1{17}\sum_{x=1}^{17}x^{2}=\frac{1\,785}{17}=105.$$ Now subtracting the square of the mean, $$\sigma_X^{2}=105-9^{2}=105-81=24.$$

We are told that every element of $$Y$$ is obtained from some element of $$X$$ via the linear transformation $$Y=aX+b,$$ where $$a,b\in\mathbb{R}$$ with $$a>0$$. For any random variable, a linear change of scale and origin obeys the following standard formulas:

• Mean transformation: $$\mu_Y=a\mu_X+b.$$ • Variance transformation: $$\sigma_Y^{2}=a^{2}\sigma_X^{2}.$$

The problem states that the mean and variance of the set $$Y$$ are $$17$$ and $$216$$ respectively. Writing these facts with the above formulas we get the two equations $$a\mu_X+b=17,\qquad a^{2}\sigma_X^{2}=216.$$

Since we have already found $$\mu_X=9$$ and $$\sigma_X^{2}=24$$, we substitute:

From the mean condition: $$a\cdot9+b=17\quad\Longrightarrow\quad 9a+b=17. \quad -(1)$$

From the variance condition: $$a^{2}\cdot24=216\quad\Longrightarrow\quad a^{2}=\frac{216}{24}=9\quad\Longrightarrow\quad a=\pm3.$$ But the question explicitly requires $$a>0$$, therefore $$a=3.$$

Putting $$a=3$$ into equation (1): $$9(3)+b=17\quad\Longrightarrow\quad 27+b=17\quad\Longrightarrow\quad b=17-27=-10.$$

Finally, the quantity we are asked for is $$a+b=3+(-10)=-7.$$

Hence, the correct answer is Option B.

We have a set of $$n = 20$$ observations whose (original) mean is given to be $$\bar{x}=10$$ and whose variance is given to be $$\sigma^{2}=4$$.

First, from the mean we find the total (aggregate) of all the observations. Using the relation $$\bar{x}=\dfrac{\sum x_{i}}{n}$$ we get $$\sum x_{i}=n\bar{x}=20 \times 10 = 200.$$

Next, we use the formula that connects variance with the sum of squares of the observations:

$$\sigma^{2}=\dfrac{\sum x_{i}^{2}}{n}-\bar{x}^{2}.$$

Substituting the known values,

$$4 = \dfrac{\sum x_{i}^{2}}{20} - (10)^{2}.$$

Rearranging,

$$\dfrac{\sum x_{i}^{2}}{20} = 4 + 100 = 104.$$ Hence, $$\sum x_{i}^{2} = 104 \times 20 = 2080.$$

Now the re-checking shows that one observation was wrongly recorded as $$9$$; the correct value is $$11$$. We therefore make the following corrections:

1. Corrected sum of observations

Old total $$=200$$. We remove the wrong value and insert the right value:

$$\sum x_{i}^{(\text{new})}=200 - 9 + 11 = 202.$$

2. Corrected mean

$$\bar{x}_{\text{new}} = \dfrac{202}{20} = 10.1.$$

3. Corrected sum of squares of observations

Old sum of squares $$=2080$$. Replace $$9^{2}$$ by $$11^{2}$$:

$$\sum x_{i}^{2\;(\text{new})}=2080 - 9^{2} + 11^{2} = 2080 - 81 + 121 = 2080 + 40 = 2120.$$

4. Corrected variance

Again applying the variance formula,

$$\sigma_{\text{new}}^{2}= \dfrac{\sum x_{i}^{2\;(\text{new})}}{n} - \bigl(\bar{x}_{\text{new}}\bigr)^{2}.$$

Substituting,

$$\sigma_{\text{new}}^{2}= \dfrac{2120}{20} - (10.1)^{2}.$$

We calculate each term: $$\dfrac{2120}{20}=106, \qquad (10.1)^{2}=102.01.$$

Therefore,

$$\sigma_{\text{new}}^{2}=106 - 102.01 = 3.99.$$

Hence, the correct answer is Option A.

We are told that the ten numbers $$x_1,x_2,\ldots ,x_{10}$$ satisfy two conditions:

1. $$\displaystyle \sum_{i=1}^{10}(x_i-5)=10$$

2. $$\displaystyle \sum_{i=1}^{10}(x_i-5)^2=40$$

First we convert the given information into the usual sums of the $$x_i$$.

From the first condition we have

$$$ \sum_{i=1}^{10}(x_i-5)=\sum_{i=1}^{10}x_i-\sum_{i=1}^{10}5 =\sum_{i=1}^{10}x_i-50=10. $$$

So

$$$ \sum_{i=1}^{10}x_i = 50+10 = 60. $$$

The mean of the original observations is therefore

$$$ \overline{x}= \frac{\sum_{i=1}^{10}x_i}{10}=\frac{60}{10}=6. $$$

Now we form the shifted observations

$$$ y_i = x_i-3\qquad (i=1,2,\ldots,10). $$$

The mean $$\mu$$ of the $$y_i$$ is just the old mean minus $$3$$ because subtracting a constant from every observation moves the mean by the same constant:

$$$ \mu = \overline{x}-3 = 6-3 = 3. $$$

Next we determine the variance $$\lambda$$ of the $$y_i$$. A key fact is that adding or subtracting a constant does not change variance. Nevertheless, we shall verify this directly from the data.

By definition, for our ten observations the variance is

$$$ \lambda=\frac1{10}\sum_{i=1}^{10}(y_i-\mu)^2. $$$

Because $$y_i=x_i-3$$ and $$\mu=3$$, we have $$y_i-\mu = (x_i-3)-3 = x_i-6$$. Hence

$$$ \lambda=\frac1{10}\sum_{i=1}^{10}(x_i-6)^2. $$$

We already know $$\displaystyle\sum (x_i-5)^2$$, so we convert $$(x_i-6)^2$$ in terms of $$(x_i-5)$$. Observe that

$$$ x_i-6=(x_i-5)-1. $$$

Squaring and summing gives

$$$ \sum_{i=1}^{10}(x_i-6)^2 =\sum_{i=1}^{10}\left[(x_i-5)-1\right]^2 =\sum_{i=1}^{10}\Big[(x_i-5)^2-2(x_i-5)+1\Big]. $$$

We split this sum term by term:

$$$ \sum_{i=1}^{10}(x_i-6)^2 =\underbrace{\sum_{i=1}^{10}(x_i-5)^2}_{=40} -2\underbrace{\sum_{i=1}^{10}(x_i-5)}_{=10} +\underbrace{\sum_{i=1}^{10}1}_{=10}. $$$

Substituting the known values, we obtain

$$$ \sum_{i=1}^{10}(x_i-6)^2 = 40 -2(10) + 10 = 40 - 20 + 10 = 30. $$$

Therefore the variance is

$$$ \lambda = \frac1{10}\times 30 = 3. $$$

We have found

$$$ (\mu,\lambda) = (3,3). $$$

Hence, the correct answer is Option A.

We have a grouped (but discrete) frequency distribution whose observed values satisfy $$0<x_1<x_2<\dots<x_{15}=10.$$ Thus every variate value lies strictly between a positive number (call it $$m=x_1$$) and $$10$$. Hence each $$x_i$$ is contained in the closed interval $$[\,m,\,10\,]$$ where $$0<m<10.$$

Let the total frequency be $$N=\displaystyle\sum_{i=1}^{15}f_i,$$ which is given to be positive. The (arithmetic) mean of the distribution is

$$\bar x \;=\;\dfrac{\sum_{i=1}^{15}f_i\,x_i}{\sum_{i=1}^{15}f_i} \;=\;\dfrac{\sum f_i\,x_i}{N}.$$

The population standard deviation (about the mean) is defined by the well-known formula

$$\sigma \;=\;\sqrt{\dfrac{\sum_{i=1}^{15}f_i\,(x_i-\bar x)^2}{\sum_{i=1}^{15}f_i}} \;=\;\sqrt{\dfrac{\sum f_i\,(x_i-\bar x)^2}{N}}.$$

To find an upper bound on $$\sigma$$ we invoke an elementary but very useful inequality for any set of numbers lying in a fixed interval. Let the interval be $$[a,b]$$ and denote the mean by $$\mu.$$ Since every observation satisfies $$a\le x\le b,$$ we can write

$$x-\mu \;\le\; b-\mu\quad\text{and}\quad \mu-x \;\le\; \mu-a.$$

Consequently the squared deviation satisfies

$$0\;\le\;(x-\mu)^2\;\le\;\max\{(b-\mu)^2,\,(\mu-a)^2\}.$$ Because $$\mu$$ itself is inside the interval $$[a,b],$$ both distances $$b-\mu$$ and $$\mu-a$$ are at most the total length $$b-a$$. Furthermore, at least one of them is not larger than half that length; precisely, we always have

$$\max\{\,b-\mu,\,\mu-a\,\}\;\le\;\dfrac{b-a}{2}.$$

Squaring this inequality gives

$$(x-\mu)^2\;\le\;\left(\dfrac{b-a}{2}\right)^2.$$

Averaging the above inequality over all observations can never increase the right-hand side; therefore

$$\sigma^2 =\dfrac{\sum f_i\,(x_i-\mu)^2}{N} \;\le\;\left(\dfrac{b-a}{2}\right)^2,$$ or equivalently

$$\sigma\;\le\;\dfrac{b-a}{2}.$$

This result—which is sometimes quoted as “the variance of any sample contained in an interval of length $$L$$ cannot exceed $$\dfrac{L^2}{4}$$”—tells us that the standard deviation never exceeds half the range.

Now we apply the inequality to the present question. Our lower bound is $$a=m$$ and our upper bound is $$b=10.$$ Therefore

$$\sigma\;\le\;\dfrac{\,10-m\,}{2}.$$

Because $$m>0,$$ we have $$10-m<10,$$ and hence the right-hand side is strictly less than $$\dfrac{10}{2}=5.$$ Putting it plainly,

$$\sigma\;<\;5.$$

Thus no matter how the actual frequencies $$f_1,f_2,\dots,f_{15}$$ are chosen (provided they are all positive), the standard deviation of the distribution can never reach $$5,$$ let alone exceed it.

Let us compare this strict upper bound with the four numerical options offered in the question:

Option A: $$4 \quad(\text{possible, because }4<5)$$
Option B: $$1 \quad(\text{certainly possible})$$
Option C: $$6 \quad(\text{impossible, because }6>5)$$
Option D: $$2 \quad(\text{possible})$$

Only the value $$6$$ violates the theoretical ceiling of $$5$$, so it is the unique value that the standard deviation cannot take.

Hence, the correct answer is Option C.

We have the $$n$$ observations $$x_1,\,x_2,\,\ldots ,x_n$$ with two given relations

$$\sum_{i=1}^{n}(x_i-a)=n\quad\text{and}\quad\sum_{i=1}^{n}(x_i-a)^2=na,$$

where $$n\gt 1,\;a\gt 1.$$

First we extract the arithmetic mean of the observations. Writing the first relation explicitly,

$$\sum_{i=1}^{n}(x_i-a)=\sum_{i=1}^{n}x_i-\sum_{i=1}^{n}a =\sum_{i=1}^{n}x_i-na=n.$$

So

$$\sum_{i=1}^{n}x_i=na+n=n(a+1).$$

Dividing by the number of observations gives the mean

$$\overline x=\frac1n\sum_{i=1}^{n}x_i =\frac{n(a+1)}{n}=a+1.$$

Now we need the standard deviation. By definition, the variance is

$$\sigma^2=\frac1n\sum_{i=1}^{n}(x_i-\overline x)^2.$$

Because $$\overline x=a+1,$$ we must convert the given sum $$\sum_{i=1}^{n}(x_i-a)^2$$ to the required sum $$\sum_{i=1}^{n}(x_i-(a+1))^2.$$

For each term we note the algebraic identity

$$(x_i-(a+1))^2=\bigl[(x_i-a)-1\bigr]^2 =(x_i-a)^2-2(x_i-a)+1.$$

Summing this expression from $$i=1$$ to $$n$$ yields

$$\sum_{i=1}^{n}(x_i-(a+1))^2 =\sum_{i=1}^{n}(x_i-a)^2 -2\sum_{i=1}^{n}(x_i-a) +\sum_{i=1}^{n}1.$$

Substituting the given values step by step:

$$\sum_{i=1}^{n}(x_i-a)^2=na,$$

$$\sum_{i=1}^{n}(x_i-a)=n,$$

$$\sum_{i=1}^{n}1=n.$$

Hence

$$\sum_{i=1}^{n}(x_i-(a+1))^2 =na-2n+n =n(a-1).$$

So the variance is

$$\sigma^2=\frac1n\sum_{i=1}^{n}(x_i-\overline x)^2 =\frac1n\bigl[n(a-1)\bigr] =a-1.$$

Taking the positive square root (because standard deviation is non-negative) gives

$$\sigma=\sqrt{\sigma^2}=\sqrt{a-1}.$$

Hence, the correct answer is Option D.

We have ten observations $$x_1,x_2,\dots ,x_{10}$$ of the random variable $$X$$. The statements given are

$$\sum_{i=1}^{10}(x_i-p)=3 \qquad\text{and}\qquad \sum_{i=1}^{10}(x_i-p)^2=9,$$

where $$p\neq 0,\;p\in\mathbb R.$$ Our task is to find the standard deviation of the ten observations.

First recall that the (population) mean $$\mu$$ of the observations is defined by the formula

$$\mu=\frac1{10}\sum_{i=1}^{10}x_i.$$

To express $$\mu$$ in terms of the known quantity $$p$$, we write every $$x_i$$ as $$(x_i-p)+p$$ and sum:

$$\sum_{i=1}^{10}x_i=\sum_{i=1}^{10}\bigl[(x_i-p)+p\bigr] =\sum_{i=1}^{10}(x_i-p)+10p =3+10p.$$

Hence the mean is

$$\mu=\frac{3+10p}{10}=p+\frac{3}{10}.$$

Now the (population) variance $$\sigma^{2}$$ is defined by

$$\sigma^{2}=\frac1{10}\sum_{i=1}^{10}(x_i-\mu)^2.$$

We already know $$\sum_{i=1}^{10}(x_i-p)^2$$, so we convert the squared deviations from $$\mu$$ to squared deviations from $$p$$ using the identity

$$(x_i-\mu)^2=\bigl[(x_i-p)-( \mu-p)\bigr]^2.$$ Expanding gives

$$(x_i-\mu)^2=(x_i-p)^2-2(\mu-p)(x_i-p)+(\mu-p)^2.$$

Summing this expression over all ten indices, we obtain

$$\sum_{i=1}^{10}(x_i-\mu)^2 =\sum_{i=1}^{10}(x_i-p)^2-2(\mu-p)\sum_{i=1}^{10}(x_i-p)+10(\mu-p)^2.$$

Every term on the right is known: we have $$\sum_{i=1}^{10}(x_i-p)^2=9,$$ $$\sum_{i=1}^{10}(x_i-p)=3,$$ and $$\mu-p=\frac{3}{10}.$$ Substituting these values, we find

$$\sum_{i=1}^{10}(x_i-\mu)^2 =9-2\left(\frac{3}{10}\right)\!(3)+10\left(\frac{3}{10}\right)^{\!2}.$$

Calculating term by term,

$$-2\left(\frac{3}{10}\right)(3)=-\frac{18}{10}=-\frac{9}{5},$$

$$10\left(\frac{3}{10}\right)^2=10\left(\frac{9}{100}\right)=\frac{90}{100}=\frac{9}{10}.$$

So

$$\sum_{i=1}^{10}(x_i-\mu)^2 =9-\frac{9}{5}+\frac{9}{10} =9-\frac{18}{10}+\frac{9}{10} =9-\frac{9}{10} =\frac{81}{10} =8.1.$$

Therefore the variance is

$$\sigma^{2}=\frac1{10}\sum_{i=1}^{10}(x_i-\mu)^2 =\frac{8.1}{10}=0.81.$$

Taking the square root gives the standard deviation:

$$\sigma=\sqrt{0.81}=0.9=\frac{9}{10}.$$

Hence, the correct answer is Option C.

We are given 10 observations whose original mean is $$\mu = 20$$ and whose original standard deviation is $$\sigma = 2$$.

Each observation $$x$$ is transformed to a new value $$y$$ according to the linear rule

$$y = p\,x - q,$$

where $$p \neq 0$$ and $$q \neq 0$$. We now examine how the mean and standard deviation change under such a transformation.

Formula for the mean after a linear transformation:
If $$y = p\,x + c,$$ then $$\mu_y = p\,\mu_x + c.$$ In our case the constant added is $$c = -q,$$ so

$$\mu_{\text{new}} = p\,\mu - q.$$

Formula for the standard deviation after a linear transformation:
If $$y = p\,x + c,$$ then $$\sigma_y = |p|\,\sigma_x,$$ because adding or subtracting a constant does not affect the spread, while multiplication by $$p$$ scales every deviation by the factor $$|p|$$.

The problem states that the new mean and the new standard deviation are both half of their original values. Therefore

$$\mu_{\text{new}} = \tfrac{1}{2}\,\mu = \tfrac{1}{2}\times 20 = 10,$$

$$\sigma_{\text{new}} = \tfrac{1}{2}\,\sigma = \tfrac{1}{2}\times 2 = 1.$$

We now set up the two equations that arise from these conditions.

From the mean condition:
$$p\,\mu - q = 10.$$
Substituting $$\mu = 20$$ gives

$$20\,p - q = 10 \quad\Longrightarrow\quad q = 20\,p - 10. \quad -(1)$$

From the standard-deviation condition:
$$|p|\,\sigma = 1.$$
Substituting $$\sigma = 2$$ gives

$$|p| \times 2 = 1 \quad\Longrightarrow\quad |p| = \frac{1}{2}.$$

This magnitude equation yields two possibilities for $$p$$:

$$p = \frac{1}{2}\quad\text{or}\quad p = -\frac{1}{2}.$$

We substitute each value of $$p$$ into equation (1) to find $$q$$.

Case 1: $$p = \tfrac{1}{2}$$
$$q = 20\left(\tfrac{1}{2}\right) - 10 = 10 - 10 = 0.$$
But the question stipulates $$q \neq 0,$$ so this case is rejected.

Case 2: $$p = -\tfrac{1}{2}$$
$$q = 20\left(-\tfrac{1}{2}\right) - 10 = -10 - 10 = -20.$$
Here $$q \neq 0,$$ so this value is admissible.

Thus the required value of $$q$$ is $$-20$$.

Hence, the correct answer is Option C.

We have a total of 7 observations whose arithmetic mean is given to be $$8$$. By definition, the mean $$\bar x$$ of $$n$$ observations satisfies the relation

$$\bar x=\dfrac{\sum_{i=1}^{n}x_i}{n}\;.$$

Putting $$\bar x=8$$ and $$n=7$$ we obtain

$$\sum_{i=1}^{7}x_i \;=\; 7 \times 8 \;=\; 56.$$

Five of the observations are already known, namely $$2,4,10,12,14$$. Let the remaining two observations be denoted by $$a$$ and $$b$$. Summing the known five values,

$$2+4+10+12+14 \;=\; 42.$$

Substituting this in the previous result gives

$$42 + a + b \;=\; 56 \;\;\Longrightarrow\;\; a + b = 14.$$

Next, we use the information about the variance. For a set of $$n$$ observations, the (population) variance $$\sigma^{2}$$ is defined as

$$\sigma^{2}= \dfrac{\sum_{i=1}^{n}(x_i-\bar x)^{2}}{n}.$$

The question states that the variance equals $$16$$, so

$$16 \;=\; \dfrac{\sum_{i=1}^{7}(x_i-\bar x)^{2}}{7}\;.$$

Multiplying by $$7$$ we get the total of squared deviations:

$$\sum_{i=1}^{7}(x_i-\bar x)^{2}=16 \times 7 = 112.$$

A very convenient identity connects this quantity with the sum of squares of the observations:

$$\sum_{i=1}^{n}(x_i-\bar x)^{2}=\sum_{i=1}^{n}x_i^{2}-\dfrac{\left(\sum_{i=1}^{n}x_i\right)^{2}}{n}.$$

We already know $$\sum_{i=1}^{7}x_i=56$$, so

$$112 = \sum_{i=1}^{7}x_i^{2}-\dfrac{56^{2}}{7}.$$

Calculating the second term on the right,

$$\dfrac{56^{2}}{7}=\dfrac{3136}{7}=448.$$

Substituting this back, we have

$$112 = \sum_{i=1}^{7}x_i^{2}-448 \;\;\Longrightarrow\;\; \sum_{i=1}^{7}x_i^{2}=112+448=560.$$

Now we list the squares of the five known observations:

$$2^{2}=4,\;4^{2}=16,\;10^{2}=100,\;12^{2}=144,\;14^{2}=196.$$

Adding them gives

$$4+16+100+144+196 = 460.$$

Hence, the sum of the squares of the two unknown observations is

$$a^{2}+b^{2}=560-460=100.$$

At this point we have the simultaneous relations

$$a+b = 14,\qquad a^{2}+b^{2}=100.$$

To find the absolute difference $$|a-b|$$, we use the identity

$$a^{2}+b^{2}=(a+b)^{2}-2ab.$$

Substituting the known values,

$$100 = 14^{2} - 2ab \;\;\Longrightarrow\;\; 100 = 196 - 2ab$$

and therefore

$$2ab = 196 - 100 = 96 \;\;\Longrightarrow\;\; ab = 48.$$

Another standard identity is

$$(a-b)^{2} = (a+b)^{2} - 4ab.$$

Substituting $$a+b = 14$$ and $$ab = 48$$ we obtain

$$(a-b)^{2} = 14^{2} - 4 \times 48 = 196 - 192 = 4.$$

Taking the positive square root gives the absolute difference:

$$|a-b| = \sqrt{4} = 2.$$

Hence, the correct answer is Option C.

We have a total of 8 observations whose mean is given as $$\mu = 10$$. The definition of mean states

$$\text{Mean }(\mu)=\dfrac{\text{Sum of all observations}}{\text{Number of observations}}.$$

So

$$10=\dfrac{\text{Sum of all 8 observations}}{8}.$$

Multiplying both sides by 8, we get

$$\text{Sum of all 8 observations}=10 \times 8 = 80.$$

Out of these 8 observations, six are already known: $$5,\,7,\,10,\,12,\,14,\,15.$$ Let us add them:

$$5+7+10+12+14+15 = 63.$$

Hence the sum of the remaining two unknown observations, say $$a$$ and $$b$$, must satisfy

$$a + b = 80 - 63 = 17.$$

Now we use the information about variance. The variance of the 8 observations is given as $$\sigma^{2}=13.5.$$ For an entire population (which is the usual convention in such questions) the variance formula is

$$\sigma^{2}=\dfrac{1}{n}\sum_{i=1}^{n}x_i^{2}-\mu^{2},$$

where $$n=8$$ and $$\mu=10.$$ Substituting, we have

$$13.5=\dfrac{1}{8}\sum_{i=1}^{8}x_i^{2}-10^{2}.$$

First move $$10^{2}=100$$ to the left:

$$13.5 + 100 = \dfrac{1}{8}\sum_{i=1}^{8}x_i^{2}.$$

So

$$113.5 = \dfrac{1}{8}\sum_{i=1}^{8}x_i^{2}.$$

Multiplying by 8 gives

$$\sum_{i=1}^{8}x_i^{2} = 113.5 \times 8 = 908.$$

Next we find the sum of squares of the six known observations:

$$5^{2}+7^{2}+10^{2}+12^{2}+14^{2}+15^{2} = 25+49+100+144+196+225 = 739.$$

Therefore the sum of squares of the remaining two observations must be

$$a^{2}+b^{2}=908-739=169.$$

We now have the simultaneous relations

$$a + b = 17, \qquad a^{2}+b^{2}=169.$$

To find $$|a-b|$$ we use the algebraic identity

$$(a+b)^{2}=a^{2}+b^{2}+2ab.$$

Substituting the known values, we get

$$17^{2}=169+2ab.$$

Since $$17^{2}=289,$$ this becomes

$$289 = 169 + 2ab \;\Longrightarrow\; 2ab = 289 - 169 = 120 \;\Longrightarrow\; ab = 60.$$

Now we apply another standard identity:

$$(a-b)^{2} = (a+b)^{2} - 4ab.$$

Substituting $$a+b = 17$$ and $$ab = 60$$ yields

$$(a-b)^{2} = 17^{2} - 4 \times 60 = 289 - 240 = 49.$$

Taking the positive square root (because absolute value is always non-negative), we obtain

$$|a-b| = \sqrt{49} = 7.$$

Hence, the correct answer is Option D.

We have the five data values $$3,\;5,\;7,\;a,\;b$$ whose mean is given to be $$5$$.

The formula for the mean of $$n$$ numbers is $$$\text{Mean}=\dfrac{\text{Sum of the observations}}{n}.$$$ So, using $$n=5$$, we write

$$\dfrac{3+5+7+a+b}{5}=5.$$

Multiplying both sides by $$5$$ gives

$$3+5+7+a+b=25.$$

Adding the known numbers, $$3+5+7=15$$, so

$$15+a+b=25.$$

Subtracting $$15$$ from both sides, we obtain the first relation

$$a+b=10.\qquad(1)$$

Next, the standard deviation is given as $$2$$. For a population of size $$n$$, the standard deviation $$\sigma$$ satisfies

$$\sigma=\sqrt{\dfrac{\displaystyle\sum_{i=1}^{n}(x_i-\mu)^2}{n}},$$

where $$\mu$$ is the mean. Squaring both sides yields

$$\sigma^2=\dfrac{\displaystyle\sum_{i=1}^{n}(x_i-\mu)^2}{n}.$$

Here $$\sigma=2$$ and $$n=5$$, so $$\sigma^2=4$$. Hence

$$$\dfrac{(3-5)^2+(5-5)^2+(7-5)^2+(a-5)^2+(b-5)^2}{5}=4.$$$

Calculating the first three squares:

$$$\begin{aligned} (3-5)^2 &= (-2)^2 = 4,\\ (5-5)^2 &= 0^2 = 0,\\ (7-5)^2 &= 2^2 = 4. \end{aligned}$$$

The numerator becomes $$$4+0+4+(a-5)^2+(b-5)^2 = 8+(a-5)^2+(b-5)^2$$$. Therefore,

$$\dfrac{8+(a-5)^2+(b-5)^2}{5}=4.$$

Multiplying both sides by $$5$$ gives

$$8+(a-5)^2+(b-5)^2=20.$$

Subtracting $$8$$ from both sides, we get

$$(a-5)^2+(b-5)^2=12.\qquad(2)$$

Now we expand the squares in (2):

$$$\begin{aligned} (a-5)^2+(b-5)^2 &= \bigl(a^2-10a+25\bigr)+\bigl(b^2-10b+25\bigr)\\ &= a^2+b^2-10(a+b)+50. \end{aligned}$$$

Using the earlier result (1), $$a+b=10$$, so

$$$a^2+b^2-10\cdot10+50 = a^2+b^2-100+50 = a^2+b^2-50.$$$

Equation (2) says this expression equals $$12$$, hence

$$a^2+b^2-50 = 12.$$

Adding $$50$$ to both sides gives

$$a^2+b^2 = 62.\qquad(3)$$

Next, we recall the algebraic identity

$$(a+b)^2 = a^2 + 2ab + b^2.$$

Substituting $$a+b=10$$ and $$a^2+b^2=62$$ into this identity yields

$$10^2 = 62 + 2ab.$$

Calculating $$10^2$$ gives $$100$$, so

$$100 = 62 + 2ab.$$

Subtracting $$62$$ from both sides,

$$38 = 2ab.$$

Dividing by $$2$$, we find

$$ab = 19.\qquad(4)$$

The sum and product of $$a$$ and $$b$$ have now been obtained as

$$$S = a+b = 10, \quad P = ab = 19.$$$

A monic quadratic equation whose roots are $$a$$ and $$b$$ is written as

$$x^2 - Sx + P = 0.$$

Substituting the values of $$S$$ and $$P$$, we arrive at

$$x^2 - 10x + 19 = 0.$$

This matches Option C.

Hence, the correct answer is Option C.

We have eight numbers: $$3,\,7,\,9,\,12,\,13,\,20,\,x \text{ and } y.$$

The mean (average) is stated to be $$10.$$ By definition,

$$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Number of observations}}.$$

There are $$8$$ observations, so

$$\dfrac{3+7+9+12+13+20+x+y}{8}=10.$$

Multiplying both sides by $$8$$ we obtain

$$3+7+9+12+13+20+x+y=80.$$

Adding the known numbers first,

$$3+7+9+12+13+20=64,$$

and substituting this into the previous equation gives

$$64+x+y=80.$$

So

$$x+y=80-64=16.$$

Next we use the information about variance. For a data set of size $$n,$$ the variance $$\sigma^{2}$$ is given by the formula

$$\sigma^{2}=\dfrac{\sum x_{i}^{2}}{n}-\mu^{2},$$

where $$\sum x_{i}^{2}$$ is the sum of the squares of the observations and $$\mu$$ is the mean.

Here $$\sigma^{2}=25,\; n=8 \text{ and } \mu=10.$$ Substituting these values, we get

$$25=\dfrac{\sum x_{i}^{2}}{8}-10^{2}.$$

Since $$10^{2}=100,$$ we have

$$25=\dfrac{\sum x_{i}^{2}}{8}-100.$$

Adding $$100$$ to both sides,

$$25+100=\dfrac{\sum x_{i}^{2}}{8},$$

so

$$\dfrac{\sum x_{i}^{2}}{8}=125.$$

Multiplying by $$8,$$

$$\sum x_{i}^{2}=125 \times 8=1000.$$

We now separate the squares of the known numbers:

$$3^{2}=9,\;7^{2}=49,\;9^{2}=81,\;12^{2}=144,\;13^{2}=169,\;20^{2}=400.$$

Adding these squares,

$$9+49+81+144+169+400=852.$$

Hence

$$x^{2}+y^{2}=1000-852=148.$$

We already found $$x+y=16.$$ To connect $$x+y$$ and $$x^{2}+y^{2},$$ we recall the algebraic identity

$$\bigl(x+y\bigr)^{2}=x^{2}+2xy+y^{2}.$$

Writing this identity with our known values,

$$16^{2}=x^{2}+2xy+y^{2}.$$

Since $$x^{2}+y^{2}=148,$$ the equation becomes

$$256=148+2xy.$$

Subtracting $$148$$ from both sides,

$$256-148=2xy,$$

so

$$2xy=108.$$

Finally, dividing by $$2,$$ we get

$$xy=\dfrac{108}{2}=54.$$

So, the answer is $$54$$.

We have a grouped data set where the variable $$x$$ takes the values $$0,2,4,8,\ldots ,2^n$$ and the corresponding frequencies are $$^nC_0,\,^nC_1,\,^nC_2,\ldots ,^nC_n$$.

The mean (arithmetic average) of any discrete distribution is given by the well-known formula

$$\text{Mean}\;=\;\frac{\displaystyle \sum (\text{value})\times(\text{frequency})}{\displaystyle \sum (\text{frequency})}.$$

First we evaluate the denominator. By the Binomial Theorem,

$$\sum_{k=0}^{n} {^nC_k}=2^{\,n},$$

because it is the expansion of $$(1+1)^n.$$ Hence the total of all frequencies equals $$2^{\,n}.$$

The question itself tells us that the mean is

$$\frac{728}{2^{\,n}}.$$

Comparing this with the definition of the mean, we see that the numerator of the fraction, namely the sum of the products of each value with its frequency, must therefore be $$728$$. Symbolically,

$$\sum_{k=0}^{n} (\text{value})\times(\text{frequency}) \;=\;728.$$

Now let us compute this numerator explicitly. Observe that except for the first value $$0$$, every other value is a power of $$2$$ that matches its index in the combination:

$$\begin{aligned} \text{Numerator} &=\;0\cdot {^nC_0} \;+\; 2\cdot{^nC_1}\;+\;4\cdot{^nC_2}\;+\;8\cdot{^nC_3}\;+\;\ldots+\;2^{\,n}\cdot{^nC_n}\\[4pt] &=\;\sum_{k=1}^{n} 2^{\,k}\,{^nC_k}. \end{aligned}$$

To evaluate this sum, we again use the Binomial Theorem. For any real number $$a,$$

$$\sum_{k=0}^{n} {^nC_k}a^{\,k} = (1+a)^{n}.$$

Choosing $$a=2,$$ we get

$$\sum_{k=0}^{n} {^nC_k}2^{\,k} = (1+2)^{n}=3^{\,n}.$$

Notice that the left-hand side includes the term with $$k=0,$$ which is $$2^{\,0}{^nC_0}=1\cdot1=1.$$ Therefore,

$$\sum_{k=1}^{n} 2^{\,k}{^nC_k}=3^{\,n}-1.$$

But this is exactly the numerator we need, and we already equated that numerator to $$728$$. Hence

$$3^{\,n}-1 = 728.$$

Adding $$1$$ to both sides gives

$$3^{\,n}=729.$$

We recognize $$729$$ as a power of $$3$$ because $$3^{6}=729.$$ Therefore,

$$n=6.$$

So, the answer is $$6$$.

We have an increasing arithmetic progression whose eleven terms are written as $$b_1,\,b_2,\,b_3,\ldots ,b_{11}$$.

Let the first term be denoted by $$a$$ and the common difference by $$d$$.

Thus the general term is $$b_k = a + (k-1)d$$ for $$k = 1,2,\ldots ,11$$.

For an A.P. with $$n$$ terms, the mean (average) of all the terms is given by the formula $$\mu = \dfrac{\text{first term} + \text{last term}}{2}.$$ Here the last term is $$a + 10d$$ because the eleventh term has index $$k = 11$$. So we obtain $$\mu = \dfrac{a + \bigl(a + 10d\bigr)}{2} = \dfrac{2a + 10d}{2} = a + 5d.$$

We now recall the general formula for the variance of the $$n$$ terms of an arithmetic progression. If the terms are $$a,\,a+d,\,a+2d,\ldots ,a+(n-1)d$$, the variance (taking the divisor as $$n$$, i.e. population variance) is $$\sigma^2 = \dfrac{d^{2}\,(n^{2}-1)}{12}.$$ This formula is obtained by expanding each squared deviation $$\bigl(a+(k-1)d-\mu\bigr)^2$$, adding them for all $$k$$ from $$1$$ to $$n$$, and finally dividing by $$n$$.

In our problem $$n = 11$$, so we substitute $$n = 11$$ into the formula: $$\sigma^2 = \dfrac{d^{2}\,\bigl(11^{2}-1\bigr)}{12}.$$

We compute the square and the subtraction inside the brackets: $$11^{2} = 121,\quad 121 - 1 = 120.$$ Hence $$\sigma^2 = \dfrac{d^{2}\,\times 120}{12}.$$

Simplifying the fraction $$\dfrac{120}{12}$$ gives $$10$$, so $$\sigma^2 = 10\,d^{2}.$$

The question states that the variance of the eleven terms is $$90$$. We therefore set $$10\,d^{2} = 90.$$

Dividing both sides by $$10$$ gives $$d^{2} = 9.$$

Taking the positive square root (because the progression is increasing and thus $$d > 0$$) yields $$d = 3.$$

So, the answer is $$3$$.

We have a grouped frequency distribution with three classes: 10‒20, 20‒30 and 30‒40. Their respective frequencies are 2, $$x$$ and 2, while the variance is given as $$50$$. Our goal is to determine the unknown frequency $$x$$.

For every class we first find the class mark (mid-point). Using the formula “mid-point = (lower limit + upper limit)/2”, we get

$$m_1 = \frac{10+20}{2}=15, \qquad m_2 = \frac{20+30}{2}=25, \qquad m_3 = \frac{30+40}{2}=35.$$

Let the total frequency be $$N$$ and the arithmetic mean be $$\bar{x}$$. Clearly

$$N = 2 + x + 2 = x + 4.$$

Using “mean $$\bar{x} = \dfrac{\sum f m}{\sum f}$$”, we compute the numerator

$$\sum f m = 2(15) + x(25) + 2(35) = 30 + 25x + 70 = 25x + 100.$$

Hence the mean is

$$\bar{x} = \frac{25x + 100}{x + 4}.$$

Next, for variance we use the shortcut formula for grouped data:

$$\sigma^{2} = \frac{\sum f m^{2}}{N} - \bar{x}^{2}.$$

We therefore need $$\sum f m^{2}$$:

$$\sum f m^{2} = 2(15^{2}) + x(25^{2}) + 2(35^{2}) = 2(225) + x(625) + 2(1225) = 450 + 625x + 2450 = 625x + 2900.$$

Because the variance is given as $$\sigma^{2} = 50$$, we write

$$50 = \frac{625x + 2900}{x + 4} - \left( \frac{25x + 100}{x + 4} \right)^{2}.$$

To remove denominators we multiply every term by $$(x + 4)^{2}$$:

$$50(x + 4)^{2} = (625x + 2900)(x + 4) - (25x + 100)^{2}.$$

Now we simplify each expression separately.

First expand the left side:

$$(x + 4)^{2} = x^{2} + 8x + 16,$$

so

$$50(x + 4)^{2} = 50x^{2} + 400x + 800.$$

Next expand the first product on the right:

$$(625x + 2900)(x + 4) = 625x^{2} + 2500x + 2900x + 11600 = 625x^{2} + 5400x + 11600.$$

Then expand the square:

$$(25x + 100)^{2} = (25x)^{2} + 2(25x)(100) + 100^{2} = 625x^{2} + 5000x + 10000.$$

Subtracting that square from the previous expansion gives

$$(625x^{2} + 5400x + 11600) - (625x^{2} + 5000x + 10000) = 400x + 1600.$$

So the equation becomes

$$50x^{2} + 400x + 800 = 400x + 1600.$$

Subtract $$400x + 1600$$ from both sides:

$$50x^{2} + 400x + 800 - 400x - 1600 = 0,$$

which simplifies to

$$50x^{2} - 800 = 0.$$

Dividing every term by $$50$$ yields

$$x^{2} - 16 = 0.$$

This factors as

$$(x - 4)(x + 4) = 0,$$

leading to

$$x = 4 \quad \text{or} \quad x = -4.$$

Since a frequency cannot be negative, we discard $$x = -4$$ and keep

$$x = 4.$$

So, the answer is $$4$$.

We begin by recalling that the variance of a data set having $$N$$ observations $$x_1,\;x_2,\;\dots ,x_N$$ is given by the formula

$$\text{Variance}= \frac{\displaystyle\sum_{i=1}^{N}x_i^{2}}{N}-\left(\frac{\displaystyle\sum_{i=1}^{N}x_i}{N}\right)^{2}.$$

First we deal with the variance of the first $$n$$ natural numbers, i.e. $$1,2,3,\dots ,n.$$

The number of terms is clearly $$N=n.$$

The sum of the first $$n$$ natural numbers is the well-known result

$$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}.$$

The sum of the squares of the first $$n$$ natural numbers is

$$\sum_{k=1}^{n}k^{2}= \frac{n(n+1)(2n+1)}{6}.$$

Substituting these expressions into the variance formula, we obtain

$$\text{Var}_{\text{first }n}= \frac{\dfrac{n(n+1)(2n+1)}{6}}{n}-\left(\frac{\dfrac{n(n+1)}{2}}{n}\right)^{2}.$$

Simplifying the fractions one at a time, we have

$$\text{Var}_{\text{first }n}= \frac{(n+1)(2n+1)}{6}-\left(\frac{n+1}{2}\right)^{2}.$$

Taking the common factor $$(n+1)$$ out, the expression becomes

$$\text{Var}_{\text{first }n}= (n+1)\left[\frac{2n+1}{6}-\frac{n+1}{4}\right].$$

We now combine the two fractions inside the square brackets by using the common denominator 12:

$$\frac{2n+1}{6}=\frac{2(2n+1)}{12}=\frac{4n+2}{12}, \quad \frac{n+1}{4}=\frac{3(n+1)}{12}=\frac{3n+3}{12}.$$

Subtracting gives

$$\frac{4n+2}{12}-\frac{3n+3}{12}= \frac{(4n+2)-(3n+3)}{12}= \frac{n-1}{12}.$$

Hence,

$$\text{Var}_{\text{first }n}= (n+1)\left(\frac{n-1}{12}\right)=\frac{n^{2}-1}{12}.$$

According to the question, this variance equals 10, so

$$\frac{n^{2}-1}{12}=10 \;\Longrightarrow\; n^{2}-1=120 \;\Longrightarrow\; n^{2}=121 \;\Longrightarrow\; n=11.$$

Next, we consider the first $$m$$ even natural numbers, namely $$2,4,6,\dots ,2m.$$ Observe that each term is exactly twice the corresponding term in the set $$1,2,3,\dots ,m.$$

If a data set is multiplied by a constant factor $$c,$$ its variance is multiplied by $$c^{2}.$$ Here the constant factor is $$c=2,$$ so the variance of $$2,4,6,\dots ,2m$$ is

$$\text{Var}_{\text{even}} = 2^{2}\times\text{Var}_{\text{first }m}=4\times\frac{m^{2}-1}{12}=\frac{m^{2}-1}{3}.$$

We are told that this variance equals 16, hence

$$\frac{m^{2}-1}{3}=16 \;\Longrightarrow\; m^{2}-1=48 \;\Longrightarrow\; m^{2}=49 \;\Longrightarrow\; m=7.$$

Finally, the problem asks for $$m+n,$$ and we have found $$m=7$$ and $$n=11,$$ so

$$m+n = 7+11 = 18.$$

So, the answer is $$18$$.

We are told that for the first five observations $$x_1,\,x_2,\,x_3,\,x_4,\,x_5$$ the mean is $$\bar x = 10$$ and the standard deviation is $$\sigma = 3$$. From the definition, the variance of these five observations is the square of the standard deviation, so

$$\sigma^2 = 3^2 = 9.$$

For any data set, the variance is related to the sum of the squares of the observations by the formula

$$\sigma^2 \;=\; \frac{\displaystyle\sum_{i=1}^{n} x_i^{\,2}}{n}\;-\;\bigl(\bar x\bigr)^{2},$$

where $$n$$ is the number of observations and $$\bar x$$ is their mean. Here $$n = 5$$, $$\bar x = 10$$ and $$\sigma^2 = 9$$, so substituting these values we get

$$9 \;=\; \frac{\displaystyle\sum_{i=1}^{5} x_i^{\,2}}{5}\;-\;10^{2}.$$

First move $$10^{2} = 100$$ to the left side:

$$9 + 100 \;=\; \frac{\displaystyle\sum_{i=1}^{5} x_i^{\,2}}{5}.$$

So

$$109 \;=\; \frac{\displaystyle\sum_{i=1}^{5} x_i^{\,2}}{5}.$$

Multiplying both sides by 5 gives the total of the squares of the first five observations:

$$\sum_{i=1}^{5} x_i^{\,2} \;=\; 5 \times 109 \;=\; 545.$$

Now we introduce a sixth observation, $$x_6 = -50$$. The problem asks for the variance of the six numbers $$x_1,\,x_2,\,x_3,\,x_4,\,x_5,\,x_6$$.

First find the new mean. The sum of the original five observations is $$5 \times 10 = 50$$. Adding the new value -50 gives

$$\text{new sum} = 50 + (-50) = 0.$$

With six observations the new mean is

$$\mu = \frac{0}{6} = 0.$$

Next we need the sum of the squares of all six observations. We already have the sum of squares of the first five, namely 545. Add the square of the sixth value:

$$(-50)^{2} = 2500,$$

so

$$\sum_{i=1}^{6} x_i^{\,2} \;=\; 545 + 2500 \;=\; 3045.$$

Using the variance formula again, but now with $$n = 6$$ and mean $$\mu = 0$$, we have

$$\sigma_{\text{new}}^{2} \;=\; \frac{\displaystyle\sum_{i=1}^{6} x_i^{\,2}}{6}\;-\;\mu^{2}.$$

Because $$\mu = 0$$, the term $$\mu^{2}$$ vanishes, leaving

$$\sigma_{\text{new}}^{2} = \frac{3045}{6}.$$

Dividing, we obtain

$$\sigma_{\text{new}}^{2} = 507.5.$$

Hence, the correct answer is Option B.

We are given ten numbers $$x_1, x_2, \ldots , x_{10}$$. Among them, the first four have an arithmetic mean of $$11$$, while the last six have an arithmetic mean of $$16$$. From the definition of mean, the sum of the first four observations is obtained by multiplying their mean by the count:

$$\text{Sum of first four}=4\times 11=44.$$

In exactly the same way, the sum of the remaining six observations equals their mean times their number:

$$\text{Sum of last six}=6\times 16=96.$$

Adding the two partial sums gives the total sum of all ten observations:

$$\sum_{i=1}^{10}x_i=44+96=140.$$

The overall mean (denoted by $$\bar{x}$$) is therefore the total sum divided by the total number of observations:

$$\bar{x}=\frac{\sum_{i=1}^{10}x_i}{10}=\frac{140}{10}=14.$$

Next we are told that the sum of squares of all the observations equals $$2000$$, that is,

$$\sum_{i=1}^{10}x_i^{\,2}=2000.$$

To compute the variance, we recall the identity

$$\sigma^2=\frac{1}{n}\sum_{i=1}^{n}x_i^{\,2}-\bar{x}^{\,2},$$

where $$n=10$$ is the number of observations, $$\sum x_i^{\,2}$$ is the sum of squares, and $$\bar{x}$$ is the mean. We substitute the known values one by one.

First, the average of the squares is

$$\frac{1}{n}\sum_{i=1}^{10}x_i^{\,2}=\frac{2000}{10}=200.$$

Next, we square the mean already found:

$$\bar{x}^{\,2}=14^{2}=196.$$

Putting these two numbers into the variance formula we obtain

$$\sigma^2 = 200 - 196 = 4.$$

Finally, the standard deviation $$\sigma$$ is the positive square root of the variance:

$$\sigma=\sqrt{\sigma^2}=\sqrt{4}=2.$$

Hence, the correct answer is Option C.

The formula for the standard deviation of $$n$$ observations $$x_1,x_2,\dots ,x_n$$ is

$$\sigma \;=\;\sqrt{\dfrac{1}{n}\,\sum_{i=1}^{n}(x_i-\bar x)^2},$$

where $$\bar x$$ is the arithmetic mean of the observations.

Here the four numbers are $$-1,\,0,\,1,\,k$$ with $$k \gt 0$$, so $$n = 4$$ and the given standard deviation is $$\sigma = \sqrt{5}$$.

First, we calculate the mean:

$$\bar x\;=\;\dfrac{(-1)+0+1+k}{4}\;=\;\dfrac{0+k}{4}\;=\;\dfrac{k}{4}.$$

Next, we write each squared deviation from the mean:

$$$ \begin{aligned} (-1-\bar x)^2 &= \Bigl(-1-\dfrac{k}{4}\Bigr)^2,\\[4pt] (0-\bar x)^2 &= \Bigl(0-\dfrac{k}{4}\Bigr)^2,\\[4pt] (1-\bar x)^2 &= \Bigl(1-\dfrac{k}{4}\Bigr)^2,\\[4pt] (k-\bar x)^2 &= \Bigl(k-\dfrac{k}{4}\Bigr)^2. \end{aligned} $$$

We expand each expression completely:

$$$ \begin{aligned} \Bigl(-1-\dfrac{k}{4}\Bigr)^2 &= (-1)^2+2(-1)\Bigl(-\dfrac{k}{4}\Bigr)+\Bigl(-\dfrac{k}{4}\Bigr)^2 \\ &= 1+\dfrac{k}{2}+\dfrac{k^{2}}{16},\\[6pt] \Bigl(0-\dfrac{k}{4}\Bigr)^2 &= \dfrac{k^{2}}{16},\\[6pt] \Bigl(1-\dfrac{k}{4}\Bigr)^2 &= 1-\,\dfrac{k}{2}+\dfrac{k^{2}}{16},\\[6pt] \Bigl(k-\dfrac{k}{4}\Bigr)^2 &= \Bigl(\dfrac{3k}{4}\Bigr)^2 = \dfrac{9k^{2}}{16}. \end{aligned} $$$

Now we add these four squared deviations:

$$$ \begin{aligned} S &= \left(1+\dfrac{k}{2}+\dfrac{k^{2}}{16}\right) +\left(\dfrac{k^{2}}{16}\right) +\left(1-\dfrac{k}{2}+\dfrac{k^{2}}{16}\right) +\left(\dfrac{9k^{2}}{16}\right)\\[6pt] &= \bigl(1+1\bigr)\;+\; \Bigl(\dfrac{k}{2}-\dfrac{k}{2}\Bigr)\;+\; \Bigl(\dfrac{k^{2}}{16}+\dfrac{k^{2}}{16}+\dfrac{k^{2}}{16}+\dfrac{9k^{2}}{16}\Bigr)\\[6pt] &= 2 + 0 + \dfrac{12k^{2}}{16}\\[6pt] &= 2 + \dfrac{3k^{2}}{4}. \end{aligned} $$$

By the standard deviation formula, we must have

$$\sigma = \sqrt{\dfrac{S}{n}} \;\; \Longrightarrow \;\; \sqrt{5} = \sqrt{\dfrac{2+\dfrac{3k^{2}}{4}}{4}}.$$

Squaring both sides removes the square roots:

$$$ 5 \;=\; \dfrac{2+\dfrac{3k^{2}}{4}}{4}. $$$

Multiplying by $$4$$ on both sides gives

$$$ 20 = 2+\dfrac{3k^{2}}{4}. $$$

Subtracting $$2$$ from each side, we get

$$$ 18 = \dfrac{3k^{2}}{4}. $$$

Multiplying by $$4$$ gives

$$$ 72 = 3k^{2}. $$$

Dividing by $$3$$, we obtain

$$$ k^{2} = 24. $$$

Finally, taking the positive square root (because $$k \gt 0$$) yields

$$$ k = \sqrt{24} = 2\sqrt{6}. $$$

Hence, the correct answer is Option D.

First, we list the marks already known: $$45,\;54,\;41,\;57,\;43$$. Let the unknown sixth mark be $$x$$.

We are told that the mean (average) score of all six tests is $$48$$. The definition of the mean for $$n$$ observations is

$$\text{Mean} \;=\; \dfrac{\text{Sum of all observations}}{n}.$$

Applying this with $$n=6$$, we have

$$\dfrac{45+54+41+57+43+x}{6}=48.$$

First we add the five known marks:

$$45+54=99,$$ $$99+41=140,$$ $$140+57=197,$$ $$197+43=240.$$

So the equation becomes

$$\dfrac{240+x}{6}=48.$$

Multiplying both sides by $$6$$ gives

$$240+x = 48\times6.$$

Since $$48\times6 = 288,$$ we obtain

$$240+x = 288.$$

Subtracting $$240$$ from both sides, we get

$$x = 288-240 = 48.$$

Hence the sixth mark is $$48$$, and the complete list of marks is $$45,\;54,\;41,\;57,\;43,\;48.$$

Now we calculate the standard deviation. For a set of $$n$$ observations $$x_1,x_2,\ldots,x_n$$ with mean $$\bar{x}$$, the population variance is defined as

$$\sigma^2 = \dfrac{\displaystyle\sum_{i=1}^{n}(x_i-\bar{x})^2}{n}.$$

The standard deviation is the positive square root of the variance:

$$\sigma = \sqrt{\sigma^2}.$$

Our mean is $$\bar{x}=48$$. We form each deviation $$d_i = x_i - 48$$:

$$$ \begin{aligned} 45-48 &= -3,\\ 54-48 &= +6,\\ 41-48 &= -7,\\ 57-48 &= +9,\\ 43-48 &= -5,\\ 48-48 &= 0. \end{aligned} $$$

Next we square each deviation and list the squares:

$$$ (-3)^2=9,\; (+6)^2=36,\; (-7)^2=49,\; (+9)^2=81,\; (-5)^2=25,\; 0^2=0. $$$

Adding these squared deviations, we have

$$$ 9+36=45,\; 45+49=94,\; 94+81=175,\; 175+25=200,\; 200+0=200. $$$

So the sum of squared deviations is $$\sum (x_i-\bar{x})^2 = 200.$$

The variance is therefore

$$\sigma^2 = \dfrac{200}{6} = \dfrac{100}{3}.$$

Taking the square root, we obtain the standard deviation:

$$\sigma = \sqrt{\dfrac{100}{3}} = \dfrac{10}{\sqrt{3}}.$$

Hence, the correct answer is Option 3.

We are given four distinct marks $$2,\;3,\;5,\;7$$ with corresponding frequencies $$f_1=(x+1)^2,\;f_2=(2x-5),\;f_3=(x^2-3x),\;f_4=x$$. The total number of students is stated to be $$20$$, so by definition of a frequency distribution we must have

$$f_1+f_2+f_3+f_4 \;=\;20.$$

Substituting the four expressions we get

$$ (x+1)^2 \;+\;(2x-5)\;+\;(x^2-3x)\;+\;x \;=\;20. $$

Now we expand and collect like terms step by step. First, expand the square: $$ (x+1)^2 = x^2 + 2x + 1.$$ Placing this into the equation,

$$ x^2 + 2x + 1 \;+\;2x - 5 \;+\;x^2 - 3x \;+\;x \;=\;20. $$

Combine the $$x^2$$ terms: $$x^2 + x^2 = 2x^2.$$

Combine the $$x$$ terms: $$2x + 2x - 3x + x = 2x.$$

Combine the constant terms: $$1 - 5 = -4.$$

So the left-hand side simplifies to

$$ 2x^2 + 2x - 4. $$

Equating this to $$20$$ and bringing every term to one side gives

$$ 2x^2 + 2x - 4 - 20 = 0, $$

which is

$$ 2x^2 + 2x - 24 = 0. $$

Dividing every term by $$2$$ for convenience,

$$ x^2 + x - 12 = 0. $$

This is a quadratic equation. The quadratic formula $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a = 1,\,b = 1,\,c = -12$$ gives

$$x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-12)}}{2(1)} = \dfrac{-1 \pm \sqrt{1 + 48}}{2} = \dfrac{-1 \pm 7}{2}.$$

Thus

$$x = \dfrac{-1 + 7}{2} = 3$$ or $$x = \dfrac{-1 - 7}{2} = -4.$$

Because a frequency cannot be negative, $$x=-4$$ is inadmissible. Therefore we take

$$ x = 3. $$

We now compute each actual frequency:

$$\begin{aligned} f_1 &= (x+1)^2 = (3+1)^2 = 16,\\ f_2 &= 2x-5 = 2(3)-5 = 1,\\ f_3 &= x^2 - 3x = 3^2 - 3(3) = 9 - 9 = 0,\\ f_4 &= x = 3. \end{aligned}$$

A quick check: $$16 + 1 + 0 + 3 = 20,$$ so the frequencies are consistent.

The mean (average) mark is defined as

$$\bar{M} = \dfrac{\sum (\text{mark} \times \text{frequency})}{\sum (\text{frequency})}.$$

Substituting the calculated values,

$$\begin{aligned} \sum (\text{mark} \times \text{frequency}) &= 2 \times 16 \;+\; 3 \times 1 \;+\; 5 \times 0 \;+\; 7 \times 3\\ &= 32 + 3 + 0 + 21\\ &= 56. \end{aligned}$$

Therefore

$$ \bar{M} = \dfrac{56}{20} = 2.8. $$

Hence, the correct answer is Option D.

We have a total of five observations whose arithmetic mean is given to be $$\mu = 4$$. By definition of the mean,

$$\mu \;=\;\dfrac{\text{Sum of all observations}}{n},\qquad n = 5.$$

Substituting the known values,

$$4 \;=\;\dfrac{\text{Sum of all observations}}{5}\;\;\Longrightarrow\;\;\text{Sum of all observations}=4\times 5=20.$$

Out of the five observations, three are already known: $$3,\;4,\;4.$$ Let the remaining two unknown observations be denoted by $$x$$ and $$y.$$ Then, from the sum obtained above,

$$x + y + 3 + 4 + 4 = 20$$

$$\Longrightarrow\;x + y = 20 - (3 + 4 + 4) = 20 - 11 = 9.$$

Next, we use the information about the variance. For a set of $$n$$ observations, the (population) variance is defined as

$$\sigma^{2} = \dfrac{\displaystyle\sum_{i=1}^{n}(x_{i} - \mu)^{2}}{n}.$$

Here $$\sigma^{2}=5.20$$ and $$n = 5,$$ so

$$\sum_{i=1}^{5}(x_{i}-4)^{2}=n\sigma^{2}=5\times 5.20=26.$$

We now separate this total into the contributions from the known observations and the unknown ones.

The squared deviations of the known observations are

$$\begin{aligned} (3-4)^{2} &= (-1)^{2}=1,\\ (4-4)^{2} &= 0,\\ (4-4)^{2} &= 0. \end{aligned}$$

Hence the combined contribution from the known observations is $$1+0+0=1.$$ Therefore, the contribution from the unknown observations must be

$$26 - 1 = 25.$$

That is,

$$(x-4)^{2} + (y-4)^{2}=25.$$

We now expand the left-hand side algebraically. Using the identity $$(a-b)^{2}=a^{2}-2ab+b^{2},$$ we get

$$(x-4)^{2}=x^{2}-8x+16,\qquad (y-4)^{2}=y^{2}-8y+16.$$

Adding these two,

$$(x-4)^{2} + (y-4)^{2}=x^{2}+y^{2} - 8(x+y) + 32.$$

But we have already shown that $$(x-4)^{2} + (y-4)^{2}=25,$$ and we also know $$x+y=9.$$ Substituting $$x+y=9$$ into the expanded expression gives

$$x^{2}+y^{2} - 8(9) + 32 = 25.$$

Simplifying step by step,

$$x^{2}+y^{2} - 72 + 32 = 25,$$

$$x^{2}+y^{2} - 40 = 25,$$

$$x^{2}+y^{2} = 25 + 40 = 65.$$

So far we have

$$x + y = 9, \qquad x^{2} + y^{2} = 65.$$

To find the absolute difference $$|x - y|,$$ we employ the identity

$$(x - y)^{2} = x^{2} + y^{2} - 2xy.$$

Therefore we must first obtain $$xy.$$ Squaring the sum $$x + y = 9$$ gives

$$(x + y)^{2} = 9^{2} \;\;\Longrightarrow\;\;x^{2} + 2xy + y^{2} = 81.$$

We already have $$x^{2} + y^{2} = 65,$$ so substituting this value into the equation above yields

$$65 + 2xy = 81 \;\;\Longrightarrow\;\;2xy = 81 - 65 = 16 \;\;\Longrightarrow\;\;xy = 8.$$

Now we are ready to compute $$(x - y)^{2}:$$

$$(x - y)^{2} = x^{2} + y^{2} - 2xy = 65 - 16 = 49.$$

Taking the positive square root (because we are asked for the absolute value),

$$|x - y| = \sqrt{49} = 7.$$

Hence, the correct answer is Option C.

We are told that there are altogether seven observations whose arithmetic mean (average) is $$8$$ and whose variance is $$16$$. Five of these observations are already known: $$2,\,4,\,10,\,12,\,14$$. We must find the product of the two unknown observations; let us denote them by $$x$$ and $$y$$.

First, we recall the formula for the mean of $$n$$ observations:

$$\text{Mean}=\dfrac{\sum_{i=1}^{n}x_i}{n}.$$

Here the mean is $$8$$ and the number of observations is $$7$$, so

$$8=\dfrac{2+4+10+12+14+x+y}{7}.$$

Multiplying both sides by $$7$$ gives

$$56=2+4+10+12+14+x+y.$$

Adding the five known numbers on the right-hand side, we have

$$2+4+10+12+14=42,$$

so the equation becomes

$$56=42+x+y.$$

Subtracting $$42$$ from both sides, we get the sum of the two unknown observations:

$$x+y=14. \quad -(1)$$

Next, we use the formula for the variance of $$n$$ observations:

$$\sigma^{2}=\dfrac{\sum_{i=1}^{n}x_i^{2}}{n}-\left(\text{Mean}\right)^{2}.$$

The variance is given as $$16$$, so

$$16=\dfrac{2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}}{7}-8^{2}.$$

Because $$8^{2}=64$$, we rewrite the equation as

$$16=\dfrac{2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}}{7}-64.$$

Adding $$64$$ to both sides gives

$$16+64=\dfrac{2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}}{7},$$

so

$$80=\dfrac{2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}}{7}.$$

Multiplying by $$7$$ yields the total of all seven squares:

$$560=2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}. \quad -(2)$$

Now we compute the sum of squares of the five known observations:

$$2^{2}=4,\quad4^{2}=16,\quad10^{2}=100,\quad12^{2}=144,\quad14^{2}=196.$$ Adding these, we get

$$4+16+100+144+196=460.$$

Substituting this result into equation (2), we find the combined squares of the unknowns:

$$560=460+x^{2}+y^{2}\;\Longrightarrow\;x^{2}+y^{2}=560-460=100. \quad -(3)$$

We now possess two key relations: the sum $$x+y=14$$ from (1) and the sum of squares $$x^{2}+y^{2}=100$$ from (3). To obtain the product $$xy$$, we use the algebraic identity

$$(x+y)^{2}=x^{2}+y^{2}+2xy.$$

Substituting the known values, we have

$$14^{2}=100+2xy.$$

Since $$14^{2}=196$$, the equation becomes

$$196=100+2xy.$$

Subtracting $$100$$ from both sides gives

$$96=2xy,$$

and dividing by $$2$$ yields

$$xy=48.$$

Therefore, the product of the remaining two observations is $$48$$.

Hence, the correct answer is Option A.