Chapter 12: Statistics

To calculate the arithmetic mean of observations given in the form of a frequency table, we use

$$\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}.$$

Here $$x_i$$ are the marks and $$f_i$$ the corresponding number of students.

Add the frequencies and the products:

$$\sum f_i = 1+1+3+4+3+2+4+4+1+1+2+3+1 = 30,$$

$$\sum f_i x_i = 10+20+108+160+150+112+240+280+72+80+176+276+95 = 1779.$$

Therefore,

$$\bar{x} = \dfrac{1779}{30} = 59.3.$$

The mean mark obtained by the students is $$59.3$$.

Given grouped data

Total number of observations (states/UT) $$N = 6+11+7+4+4+2+1 = 35$$.

1. Direct (actual-mean) method

Mean $$\bar x = \dfrac{\sum f_i x_i}{N}=\dfrac{1390}{35}=39.714285\ldots \approx 39.7\%$$.

2. Assumed-mean method

Choose an assumed mean $$A = 45$$.

Mean $$\bar x = A + \dfrac{\sum f_i d_i}{N}=45 + \dfrac{-185}{35}=39.714285\ldots \approx 39.7\%$$.

3. Step-deviation method

Take $$A = 40$$ and class width $$h = 10$$.

Mean $$\bar x = A + h\left( \dfrac{\sum f_i u_i}{N} \right)=40 + 10\left(\dfrac{-1}{35}\right)=39.714285\ldots \approx 39.7\%$$.

Mean percentage of female teachers in rural primary schools is therefore about $$39.7\%$$ by all three methods.

Given grouped data

To find the mean we use the assumed-mean method.

1. Compute each class mark (mid-value) $$x_i$$:
   

   Class	$$x_i = \dfrac{l + u}{2}$$
   20 − 60	$$40$$
   60 − 100	$$80$$
   100 − 150	$$125$$
   150 − 250	$$200$$
   250 − 350	$$300$$
   350 − 450	$$400$$

2. Choose an assumed mean $$A$$.
   Because $$125$$ lies near the centre, take $$A = 125$$.

3. Find deviations $$d_i = x_i - A$$ and products $$f_i d_i$$:

   $$x_i$$	$$d_i$$	$$f_i$$	$$f_i d_i$$
   40	$$40-125=-85$$	7	$$7(-85)=-595$$
   80	$$80-125=-45$$	5	$$5(-45)=-225$$
   125	$$0$$	16	$$16\times0=0$$
   200	$$200-125=75$$	12	$$12\times75=900$$
   300	$$300-125=175$$	2	$$2\times175=350$$
   400	$$400-125=275$$	3	$$3\times275=825$$

4. Add the columns:

   Total number of bowlers $$N = \sum f_i = 7+5+16+12+2+3 = 45$$

   $$\sum f_i d_i = -595 -225 +0 +900 +350 +825 = 1\,255$$

5. Calculate the mean:

   $$\bar{x} = A + \dfrac{\sum f_i d_i}{N} = 125 + \dfrac{1\,255}{45}$$

   $$\dfrac{1\,255}{45} = 27.888\ldots$$

   $$\therefore \; \bar{x} \approx 125 + 27.888 = 152.888$$

Interpretation

The mean number of wickets is about $$153$$. It represents the average (typical) number of wickets taken by a bowler in the given list of one-day international matches.

Step 1 : Write the observations in ascending order
$$0, 1, 2, 2, 2, 3, 3, 4, 5, 6$$

Step 2 : Prepare a frequency table

Step 3 : Locate the highest frequency
The highest frequency is $$3$$, which corresponds to the observation $$2$$.

Step 4 : State the mode
The mode of a data set is the observation that occurs most frequently. Hence,
$$\text{Mode}=2\text{ wickets}.$$

The information supplied is grouped, so we look for the class interval that occurs most often (the modal class).

The greatest frequency is 8, so the modal class is $$3-5$$.

For a grouped (continuous) distribution the mode is obtained from

$$\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h$$

• $$l = 3$$ (lower limit of the modal class)
• $$h = 2$$ (class width)
• $$f_1 = 8$$ (frequency of the modal class)
• $$f_0 = 7$$ (frequency of the class just before it)
• $$f_2 = 2$$ (frequency of the class just after it)

Substituting these values,

$$\text{Mode}=3+\frac{8-7}{2\times8-7-2}\times2$$

$$\text{Mode}=3+\frac{1}{7}\times2$$

$$\text{Mode}=3+\frac{2}{7}$$

$$\text{Mode}\approx3.29$$

Hence, the modal family size is about $$3.3$$ members.

Step 1  Locate the modal (highest–frequency) class
From the frequency column we see that the maximum frequency is
$$f_1 = 7$$ which occurs for the class $$40\;\text{–}\;55$$.
Hence $$40\;\text{–}\;55$$ is the modal class.

Step 2  Note the symbols needed in the mode-formula

• Lower class boundary of the modal class: $$l = 40$$
• Class width: $$h = 55 - 40 = 15$$
• Frequency of the modal class: $$f_1 = 7$$
• Frequency of the class just before it: $$f_0 = 3$$ (class $$25\;\text{–}\;40$$)
• Frequency of the class just after it: $$f_2 = 6$$ (class $$55\;\text{–}\;70$$)

Step 3  Apply the grouped-data mode formula

$$\text{Mode} = l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\,\times h$$

Substituting the values:

$$\begin{aligned} \text{Mode} &= 40 + \dfrac{7-3}{2\times7-3-6}\times15\\[4pt] &= 40 + \dfrac{4}{14-3-6}\times15\\[4pt] &= 40 + \dfrac{4}{5}\times15\\[4pt] &= 40 + 0.8\times15\\[4pt] &= 40 + 12\\[4pt] &= 52. \end{aligned}$$

Therefore the mode of the marks = 52.

Step 4  Mean of the distribution (for comparison)
The totals of $$f_i x_i$$ have already been obtained in the given table:

$$\sum f_i x_i = 1860,\;\; \sum f_i = 30$$

Hence the mean

$$\bar{x}=\dfrac{\sum f_i x_i}{\sum f_i}=\dfrac{1860}{30}=62.$$

Step 5  Interpretation

• The most frequently obtained mark (mode) is $$52$$.
• The average mark (mean) is $$62$$, which is higher than the mode.
• Since $$\bar{x} > \text{mode}$$, the distribution is positively skewed; a few students who scored high marks have pulled the mean upward, while the bulk of the class scored around $$52$$ marks.

Step 1 : Convert the ‘less than’ data into a continuous grouped-frequency table

Total number of girls $$n=51$$.

Step 2 : Locate the median class

Half of the observations $$=\dfrac{n}{2}=\dfrac{51}{2}=25.5$$.
The cumulative frequency just greater than $$25.5$$ is $$29$$, which corresponds to the class $$145\text{–}150\text{ cm}$$.

Hence,

• Median class $$=145\text{–}150$$
• Lower class boundary $$l = 144.5\text{ cm}$$
• Class width $$h = 5\text{ cm}$$
• Frequency of the class $$f = 18$$
• Cumulative frequency preceding the class $$c_f = 11$$

Step 3 : Apply the grouped-data median formula

$$\text{Median}=l+\left(\dfrac{\dfrac{n}{2}-c_f}{f}\right)h$$

Substituting the values,

$$\text{Median}=144.5+\left(\dfrac{25.5-11}{18}\right)\times5$$

$$\phantom{\text{Median}}=144.5+\left(\dfrac{14.5}{18}\right)\times5$$

$$\phantom{\text{Median}}=144.5+4.028\approx148.53$$

Step 4 : State the result

The median height of the 51 girls is approximately $$148.5\text{ cm}$$.

Given data

Total frequency is $$N = 100$$, and the median is $$525$$.

Step 1: Equation from the total frequency

Sum of the known frequencies is

$$2 + 5 + 12 + 17 + 20 + 9 + 7 + 4 = 76.$$

Therefore,

$$76 + x + y = 100 \quad\Rightarrow\quad x + y = 24. \qquad \text{...(1)}$$

Step 2: Locate the median class

Half of the total frequency is

$$\dfrac{N}{2} = \dfrac{100}{2} = 50.$$

The given median value $$525$$ lies in the class $$500\text{ \– }600$$, so the median class is $$500\text{ \– }600$$.

Construct the cumulative-frequency column up to the median class:

Step 3: Apply the median formula

For the median class $$500\text{ \– }600$$:

• Lower class boundary $$l = 500$$
• Class width $$h = 100$$
• Cumulative frequency just before the median class $$cf = 36 + x$$
• Frequency of the median class $$f = 20$$

The median formula is

$$\text{Median} = l + \left(\dfrac{\tfrac{N}{2} - cf}{f}\right) h.$$

Substituting the values,

$$525 = 500 + \left(\dfrac{50 - (36 + x)}{20}\right) \times 100$$

$$525 = 500 + \left(\dfrac{14 - x}{20}\right) \times 100$$

$$525 = 500 + 5(14 - x)$$

$$525 = 570 - 5x$$

$$5x = 45 \quad\Rightarrow\quad x = 9.$$

Step 4: Find $$y$$ from equation (1)

$$x + y = 24 \quad\Rightarrow\quad 9 + y = 24 \quad\Rightarrow\quad y = 15.$$

Conclusion

The required values are $$x = 9$$ and $$y = 15$$.

Step 1: Prepare a working table

Step 2: Calculate the mean

The mean for grouped data by the direct method is

$$\overline{x} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{162}{20} = 8.1$$

Method chosen and reason

The direct method was used because:

• there are only seven classes, and
• the class marks and products $$f_i x_i$$ are small integers, so calculation is straightforward without the extra steps of the assumed-mean or step-deviation methods.

Hence, the mean number of plants per house is 8.1.

Step 1 : Compute the class-marks

The class mark (mid-point) of each class is
$$x_i = \dfrac{\text{upper limit} + \text{lower limit}}{2}$$

Step 2 : Choose an assumed mean and the step size

• Assumed mean  A = 550 (the middle class mark)
• Class width  h = 20

Define the deviations $$d_i = x_i - A$$ and step-deviations $$u_i = \dfrac{d_i}{h}$$.

Step 3 : Prepare the working table

Step 4 : Apply the step-deviation mean formula

$$\bar{x}= A + h\,\frac{\sum f_i u_i}{\sum f_i}$$

Substituting the values:
$$\bar{x}= 550 + 20\,\frac{-12}{50}= 550 + 20(-0.24)= 550 - 4.8= 545.2$$

Hence, the mean daily wage of the workers is ₹ 545.20.

For grouped data the mean is found by
$$\bar{x}=\dfrac{\sum f_i x_i}{\sum f_i},$$
where $$x_i$$ is the class-mark (mid-value) of each class and $$f_i$$ its frequency.

Total frequency

$$\sum f_i = 7+6+9+13+f+5+4 = 44+f$$

Total of $$f_i x_i$$

$$\sum f_i x_i = 84+84+144+234+20f+110+96 = 752+20f$$

The mean pocket allowance is given to be Rs 18, therefore

$$18 = \dfrac{752+20f}{44+f}$$

Solve for $$f$$:

$$752 + 20f = 18(44 + f)$$
$$752 + 20f = 792 + 18f$$
$$20f - 18f = 792 - 752$$
$$2f = 40$$
$$f = 20$$

Hence, the missing frequency is 20.

Given data

The class width is the same everywhere:

$$h = 68-65 = 3$$

Step 1: Find class-marks $$x_i$$

Step 2: Choose an assumed mean

Take the class-mark of the central class as the assumed mean:
$$A = 75.5$$

Step 3: Compute deviations in convenient units

Let $$u_i = \dfrac{x_i - A}{h}$$. With $$h = 3$$ we get:

Step 4: Apply the step-deviation mean formula

$$\bar{x} = A + h\,\dfrac{\sum f_i u_i}{\sum f_i}$$

$$\bar{x} = 75.5 + 3\times\dfrac{4}{30} = 75.5 + 3\times 0.133\overline{3} = 75.5 + 0.4 = 75.9$$

Mean number of heartbeats per minute

Therefore, the mean heart-beat rate for the 30 women is $$75.9$$ beats per minute.

Given grouped data

The class-size is the same for every class, $$h = 3$$. We apply the step–deviation (assumed-mean) method.

1. Find the class marks $$x_i$$ (mid-points):
   50–52 → $$x_1 = 51$$,  53–55 → $$x_2 = 54$$,
   56–58 → $$x_3 = 57$$,  59–61 → $$x_4 = 60$$,
   62–64 → $$x_5 = 63$$.
2. Choose the assumed mean $$a$$.
   Take $$a = 57$$ (the class mark of the class with the largest frequency).
3. Compute the step-deviations $$d_i = \dfrac{x_i - a}{h}$$ and $$f_i d_i$$.

Totals:  $$\sum f_i = 400$$,  $$\sum f_i d_i = 25$$.

Mean

$$\bar x = a + h\,\dfrac{\sum f_i d_i}{\sum f_i} = 57 + 3\left(\dfrac{25}{400}\right) = 57 + 3\times 0.0625 = 57 + 0.1875 = 57.1875$$

Therefore, the mean number of mangoes in a packing box is approximately $$57.19$$.

The step–deviation (assumed-mean) method was used because the class intervals are of equal width and the calculations become shorter.

Step 1 : Write the class marks x_i

Step 2 : Choose an assumed mean A
Take the class mark of the central class, $$A = 225$$.

Step 3 : Find deviations d_i = x_i − A
The common class width is $$h = 50$$. Use the step–deviation variable $$u_i = \dfrac{d_i}{h}$$.

Step 4 : Compute the mean

$$\bar{x} = A + h\,\dfrac{\sum f_i u_i}{\sum f_i} = 225 + 50\times\dfrac{-7}{25}$$
$$\bar{x} = 225 - 50\times0.28 = 225 - 14 = 211$$

The mean daily expenditure on food is ₹211.

Given grouped data

Total number of observations: $$N = 4 + 9 + 9 + 2 + 4 + 2 = 30$$

Step 1 – Find the class marks
The class mark $$x_i$$ for each class is

• $$x = \dfrac{0.00 + 0.04}{2} = 0.02$$
• $$x = \dfrac{0.04 + 0.08}{2} = 0.06$$
• $$x = \dfrac{0.08 + 0.12}{2} = 0.10$$
• $$x = \dfrac{0.12 + 0.16}{2} = 0.14$$
• $$x = \dfrac{0.16 + 0.20}{2} = 0.18$$
• $$x = \dfrac{0.20 + 0.24}{2} = 0.22$$

Step 2 – Form $$f_i x_i$$ and add

Step 3 – Calculate the mean

$$\bar x = \dfrac{\sum f_i x_i}{N} = \dfrac{2.96}{30} = 0.098\,666\ldots$$

Step 4 – Write the answer
Rounding to three decimal places, $$\bar x \approx 0.099$$ ppm.

Step 1 : Prepare a working table

Step 2 : Apply the assumed-mean formula

Take the assumed mean $$A=17$$ (the class mark of 14 – 20).

The mean $$\bar{x}$$ is given by $$\bar{x}=A+\dfrac{\sum f_i d_i}{\sum f_i}$$.

Substituting the values:

$$\bar{x}=17+\dfrac{-181}{40}=17-4.525=12.475$$.

Step 3 : Conclusion

The mean number of days a student was absent is about $$12.5$$ days (correct to one decimal place).

Step 1 : Write the class marks

Step 2 : Compute $$f_i x_i$$

Step 3 : Find the mean

The mean literacy rate $$\overline{x}$$ for grouped data is

$$\overline{x}=\dfrac{\sum f_i x_i}{\sum f_i}=\dfrac{2430}{35}\approx69.43\%.$$

Thus, the average (mean) literacy rate of the 35 cities is about 69.4 % (to one decimal place).

Step 1 : Identify the modal class

The highest frequency is 23, which occurs in the class 35 – 45.
Hence, the modal class is 35 – 45.

Step 2 : Convert to class boundaries

As the classes are stated in the inclusive form (5 – 15, 15 – 25, …), we make them continuous by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit:

The quantities required for the mode formula are therefore:

• lower class boundary of the modal class: $$l = 34.5$$
• class width: $$h = 10$$
• frequency of the modal class: $$f_1 = 23$$
• frequency of the class just before: $$f_0 = 21$$
• frequency of the class just after: $$f_2 = 14$$

Step 3 : Calculate the mode

For grouped data, $$\text{Mode} = l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\,h$$

Substituting the values:

$$\text{Mode} = 34.5 + \dfrac{23-21}{2\times23 - 21 - 14}\times10$$
$$= 34.5 + \dfrac{2}{46 - 35}\times10$$
$$= 34.5 + \dfrac{2}{11}\times10$$
$$= 34.5 + 1.818\text{ (approximately)}$$
$$\boxed{\text{Mode} \approx 36.3\,\text{years}}$$

Step 4 : Compute the mean

First find the class mark $$x_i$$ (mid-point) of each class and form $$f_i x_i$$.

Total frequency: $$\sum f_i = 6+11+21+23+14+5 = 80$$
Sum of $$f_i x_i$$: $$\sum f_i x_i = 60+220+630+920+700+300 = 2830$$

Therefore, the mean age is
$$\text{Mean} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{2830}{80} = 35.375$$

$$\boxed{\text{Mean} \approx 35.4\,\text{years}}$$

Step 5 : Comparison and interpretation

• The mean age is about 35.4 years, while the mode is about 36.3 years.
• Both measures lie in the same class interval (35 – 45 years), showing that most patients are in their mid-thirties to early forties.
• The small difference (≈ 0.9 year) indicates that the distribution of ages is fairly symmetrical around this interval.

The class–interval with the greatest frequency is the modal class.

Here the maximum frequency is $$61$$, so

• Modal class  = $$60\,\text{–}\,80$$
• Lower limit $$l = 60$$
• Class width $$h = 80 - 60 = 20$$
• Frequency of the modal class $$f_1 = 61$$
• Frequency of the class just before it $$f_0 = 52$$
• Frequency of the class just after it $$f_2 = 38$$

For grouped data,

$$ \text{Mode} = l + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h $$

Substitute the values:

$$ f_1 - f_0 = 61 - 52 = 9 $$
$$ 2f_1 - f_0 - f_2 = 2\times61 - 52 - 38 = 122 - 90 = 32 $$

$$ \text{Mode} = 60 + \frac{9}{32}\times20 $$

$$ \quad\;\, = 60 + \frac{9 \times 20}{32} = 60 + \frac{180}{32} $$

$$ \quad\;\, = 60 + 5.625 = 65.625 $$

Therefore, the modal lifetime of the electrical components is approximately $$65.6\,\text{hours}$$.

Given data

Total number of families $$N = 200$$ and class width $$h = 500\,\text{rupees}$$.

1. Mode

The highest frequency is $$40$$, so the modal class is $$1500 - 2000$$.

For the modal class:

• Lower limit $$l = 1500$$
• $$f_1 = 40$$ (frequency of the modal class)
• $$f_0 = 24$$ (frequency of the class just before)
• $$f_2 = 33$$ (frequency of the class just after)

Formula: $$\text{Mode} = l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\,\times h$$

Substituting:

$$\text{Mode} = 1500 + \dfrac{40 - 24}{2\times40 - 24 - 33}\times500$$
$$ = 1500 + \dfrac{16}{80 - 57}\times500$$
$$ = 1500 + \dfrac{16}{23}\times500$$
$$ = 1500 + 347.826 \approx 1847.8$$

Hence, modal monthly expenditure $$\approx \text{Rs }1848$$.

2. Mean (Step-deviation method)

Take assumed mean $$A = 2750$$ (the mid value of the fourth class).

$$\sum f_i d_i = -72 - 80 - 33 + 0 + 30 + 44 + 48 + 28 = -35$$

Mean:

$$\overline{x} = A + \dfrac{\sum f_i d_i}{N}\,h$$
$$ = 2750 + \dfrac{-35}{200}\times500$$
$$ = 2750 - 87.5$$
$$ = 2662.5$$

Therefore, mean monthly expenditure $$= \text{Rs }2662.5$$.

Step 1 : Prepare the frequency table with class marks

Total number of states $$N=\sum f_i=35$$.

Step 2 : Mean by the assumed-mean method

Take assumed mean $$A=32.5$$ (class 30–35), class width $$h=5$$.

$$\sum f_i d_i=-23$$

Mean:
$$\bar{x}=A+h\dfrac{\sum f_i d_i}{\sum f_i}=32.5+5\times\dfrac{-23}{35}=32.5-3.2857\approx29.2$$

Step 3 : Mode of the grouped data

The highest frequency is 10 (class 30–35), so this is the modal class.

For the modal class:

• lower boundary $$l=30$$
• $$f_1=10$$ (modal class)
• $$f_0=9$$ (preceding class)
• $$f_2=3$$ (succeeding class)

Mode:
$$\text{Mode}=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h=30+\dfrac{10-9}{2\times10-9-3}\times5=30+\dfrac{1}{8}\times5=30+0.625=30.625\approx30.6$$

Step 4 : Interpretation

The mean student-teacher ratio among the 35 states/UTs is about 29 students per teacher, while the most common (modal) ratio is about 31 students per teacher. Both measures are close, indicating that most states have roughly 30 students per teacher in higher-secondary schools.

To locate the mode in a grouped frequency distribution we first identify the modal class – the class with the greatest frequency.

The highest frequency is 18, so the modal class is $$4000 - 5000$$.

For this class

• Lower limit, $$l = 4000$$
• Class size, $$h = 1000$$
• Frequency of modal class, $$f_1 = 18$$
• Frequency of class just before, $$f_0 = 4$$
• Frequency of class just after, $$f_2 = 9$$

The mode for grouped data is calculated by

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2}\,h$$

Substituting the values,

$$\text{Mode} = 4000 + \frac{18 - 4}{2(18) - 4 - 9}\times1000$$

$$= 4000 + \frac{14}{36 - 4 - 9}\times1000$$

$$= 4000 + \frac{14}{23}\times1000$$

$$= 4000 + 608.6956$$

$$\approx 4608.7$$

Hence, the mode of the given data is approximately $$4609$$ runs.

Step 1  Make the class intervals continuous

Step 2  Identify the modal class

The highest frequency is 20, so the modal class is 40 – 50 (continuous limits 39.5 – 50.5).

Step 3  Collect the required values

• Lower class boundary of modal class: $$l = 39.5$$
• Class size: $$h = 10$$
• Frequency of modal class: $$f_1 = 20$$
• Frequency of class just before modal class: $$f_0 = 12$$
• Frequency of class just after modal class: $$f_2 = 11$$

Step 4  Apply the mode formula for grouped data

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2}\times h$$

Substituting the values,

$$\text{Mode} = 39.5 + \frac{20 - 12}{2(20) - 12 - 11}\times 10$$

Calculate the numerator and denominator:

$$f_1 - f_0 = 8, \quad 2f_1 - f_0 - f_2 = 40 - 12 - 11 = 17$$

Hence,

$$\text{Mode} = 39.5 + \frac{8}{17}\times 10$$

$$= 39.5 + 4.70588\ldots$$

$$\approx 44.2$$

Therefore, the mode of the given distribution is approximately 44.2 cars per 3-minute interval.

Step 1 : Prepare the continuous grouped table

Total number of consumers $$N = \sum f_i = 68$$ and $$\sum f_i x_i = 9320$$.

Step 2 : Mean

$$\text{Mean } \bar{x} = \frac{\sum f_i x_i}{\sum f_i}=\frac{9320}{68}=137.06\text{ units (to 2 d.p.)}$$

Step 3 : Median

• $$\frac{N}{2}=\frac{68}{2}=34$$
• The cumulative frequency just exceeding 34 is 42, which lies in the class 125 – 145. Hence this is the median class.

For the median class 124.5 – 144.5:
$$l = 124.5,\; h = 20,\; f = 20,\; cf = 22$$ (cumulative frequency of the class just before).

$$\text{Median}=l+\Big(\tfrac{\frac{N}{2}-cf}{f}\Big)h = 124.5+\Big(\tfrac{34-22}{20}\Big)\times20 = 124.5+12 = 136.5\text{ units}$$

Step 4 : Mode

The highest frequency 20 also occurs in the class 125 – 145, so this is the modal class.

For this class:
$$l = 124.5,\; h = 20,\; f_1 = 20,\; f_0 = 13,\; f_2 = 14$$

$$\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)h = 124.5+\left(\frac{20-13}{2\times20-13-14}\right)\times20 = 124.5+\frac{7}{13}\times20\approx124.5+10.77 = 135.27\text{ units}$$

Step 5 : Comparison

$$\text{Mean}\; (137.06) \;>\; \text{Median}\; (136.5) \;>\; \text{Mode}\; (135.27)$$

Since $$\bar{x} > \text{Median} > \text{Mode}$$ the distribution is slightly positively skewed, although all three values are close, indicating near symmetry.

Step 1 Form an equation from the total frequency

The sum of all frequencies is 60, so

$$5 + x + 20 + 15 + y + 5 = 60 \[3pt] x + y + 45 = 60 \[3pt] \boxed{x + y = 15}$$

Step 2 Locate the median class

Total number of observations $$N = 60 \;\Rightarrow\; \dfrac N2 = 30$$.

Compute cumulative frequencies up to each class (using the unknown $$x$$ where necessary):

Because $$x + 5 \lt 30 \le x + 25$$ for the unknown $$x$$, the class $$20\text{–}30$$ contains the 30^th item and is therefore the median class.

Step 3 Apply the median formula

For the median class $$20\text{–}30$$:

• lower class boundary $$l = 20$$ (the classes already touch),
• class width $$h = 10$$,
• frequency of the median class $$f = 20$$,
• cumulative frequency preceding the median class $$cf = x + 5$$.

The formula is

$$\text{Median} = l + \frac{\dfrac N2 - cf}{f}\,h$$

Substituting the given median $$28.5$$ and the above values:

$$28.5 = 20 + \frac{30 - (x + 5)}{20}\times 10$$

Simplify step by step:

$$28.5 - 20 = \frac{25 - x}{20}\times 10$$
$$8.5 = \frac{25 - x}{2}$$
$$17 = 25 - x$$
$$\boxed{x = 8}$$

Step 4 Find $$y$$ using the relation from Step 1

$$x + y = 15 \;\Rightarrow\; 8 + y = 15 \;\Rightarrow\; \boxed{y = 7}$$

Verification (optional)

All frequencies are now 5, 8, 20, 15, 7, 5, whose sum is
$$5 + 8 + 20 + 15 + 7 + 5 = 60$$, as required, and the computed median indeed equals 28.5.

Step 1 : Rewrite the ‘less–than’ data as a regular grouped distribution

Total number of policy holders $$N = 100$$.

Step 2 : Locate the median class

Position of the median term  $$=\dfrac{N}{2}=\dfrac{100}{2}=50$$.

The cumulative frequency just greater than (or equal to) 50 is $$78$$, which corresponds to the class $$35\;\text{–}\;40$$.
Hence the median class is $$35\;\text{–}\;40$$.

Step 3 : Apply the median formula

• Lower limit of the median class $$l = 35$$
• Cumulative frequency of the class before it $$cf = 45$$
• Frequency of the median class $$f = 33$$
• Class width $$h = 40 - 35 = 5$$

Formula :  $$\text{Median} = l + \left(\dfrac{\frac{N}{2} - cf}{f}\right) \times h$$

Substitute the values :

$$\text{Median} = 35 + \left(\dfrac{50 - 45}{33}\right) \times 5$$

$$\text{Median} = 35 + \dfrac{5}{33} \times 5 = 35 + \dfrac{25}{33}$$

$$\text{Median} \approx 35 + 0.76 = 35.76$$

Step 4 : Conclusion

The median age of the 100 policy holders is approximately $$35.8\;\text{years}$$.

Step 1 : Convert the class–intervals into continuous form

Each stated class is inclusive. To make the classes continuous subtract $$0.5$$ from every lower limit and add $$0.5$$ to every upper limit.

Step 2 : Compute cumulative frequencies

Total number of leaves $$N = 40$$, so $$\dfrac{N}{2} = 20$$.

Step 3 : Locate the median class

The cumulative frequency just greater than $$20$$ is $$29$$, which belongs to the class $$144.5\,\text{–}\,153.5\,\text{mm}$$. Hence this is the median class.

Step 4 : Apply the median formula

• Lower boundary of median class $$l = 144.5\,\text{mm}$$
• Class width $$h = 153.5 - 144.5 = 9\,\text{mm}$$
• Cumulative frequency preceding the class $$F = 17$$
• Frequency of the class $$f = 12$$

Formula: $$\text{Median} = l + \left(\dfrac{\dfrac{N}{2} - F}{f}\right) h$$

Substituting: $$\text{Median} = 144.5 + \left(\dfrac{20 - 17}{12}\right) \times 9$$

$$\text{Median} = 144.5 + \left(\dfrac{3}{12}\right) \times 9 = 144.5 + 2.25 = 146.75$$

Therefore, the median length of the leaves is $$146.75\,\text{mm}$$.

First form the cumulative frequencies.

Total number of lamps $$N = 400$$, so $$\frac{N}{2}=200$$. The first cumulative frequency equal to or just exceeding 200 is 216, which lies in the class 3000 – 3500. Hence

• Median class: 3000 – 3500
• Lower class boundary $$L = 3000$$
• Frequency of the median class $$f = 86$$
• Cumulative frequency of the class just before the median class $$C = 130$$
• Class size (width) $$h = 500$$

Using the formula for grouped data

$$\text{Median} = L + \left(\frac{\tfrac{N}{2} - C}{f}\right)h$$

$$\text{Median} = 3000 + \left(\frac{200 - 130}{86}\right) \times 500$$

$$\text{Median} = 3000 + \left(\frac{70}{86}\right) \times 500$$

$$\text{Median} = 3000 + 0.813953\times 500$$

$$\text{Median} \approx 3000 + 406.98 = 3406.98$$

Rounded to the nearest hour, the median life time is about $$3407\;\text{hours}$$.

Frequency distribution of the number of letters in 100 surnames

Total number of surnames $$N = 100$$. The class intervals given are already continuous (the upper limit of each class equals the lower limit of the next), so no boundary adjustment is required. The class width is $$h = 3$$.

1. Median

Form the cumulative-frequency column:

$$\dfrac{N}{2} = \dfrac{100}{2} = 50.$$ The first cumulative frequency that exceeds $$50$$ is $$76$$, corresponding to the class $$7\,\text{–}\,10$$. Hence the median class is $$7\,\text{–}\,10$$.

For this class:

• Lower class boundary $$l = 7$$
• Cumulative frequency just before the class $$cf = 36$$
• Frequency of the median class $$f = 40$$
• Class width $$h = 3$$

$$\text{Median} = l + \left(\dfrac{\tfrac{N}{2} - cf}{f}\right) h$$

$$\text{Median} = 7 + \left(\dfrac{50 - 36}{40}\right) \times 3 = 7 + \dfrac{14}{40}\times 3 = 7 + 1.05 = 8.05.$$

2. Mean

$$\sum f_i x_i = 15 + 165 + 340 + 184 + 58 + 70 = 832.$$

$$\text{Mean} = \dfrac{\sum f_i x_i}{N} = \dfrac{832}{100} = 8.32.$$

3. Mode

The highest frequency is $$40$$ in the class $$7\,\text{–}\,10$$, so this is the modal class.

• Lower class boundary $$l = 7$$
• $$f_1 = 40$$ (modal-class frequency)
• $$f_0 = 30$$ (frequency of the preceding class)
• $$f_2 = 16$$ (frequency of the succeeding class)
• Class width $$h = 3$$

$$\text{Mode} = l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\,h = 7 + \dfrac{40 - 30}{2\times 40 - 30 - 16}\times 3$$

$$\text{Mode} = 7 + \dfrac{10}{34}\times 3 = 7 + \dfrac{30}{34} \approx 7 + 0.88 = 7.88.$$

Results

• Median number of letters $$\approx 8.05$$
• Mean number of letters $$= 8.32$$
• Modal size of the surnames $$\approx 7.88$$

First arrange the data in a table containing the cumulative frequencies.

Total number of students $$n = 30$$.

The median position is given by $$\frac{n}{2} = \frac{30}{2} = 15$$.

Locate the cumulative frequency just greater than or equal to 15. It is 19, which corresponds to the class 55 – 60. Hence, 55 – 60 kg is the median class.

For the median class we note:

• Lower class boundary $$l = 55$$
• Cumulative frequency of the class preceding the median class $$c_f = 13$$
• Frequency of the median class $$f = 6$$
• Class width $$h = 5$$

Using the formula for the median of grouped data,

$$\text{Median} = l + \left(\frac{\frac{n}{2} - c_f}{f}\right) h$$

Substitute the values:

$$\text{Median} = 55 + \left(\frac{15 - 13}{6}\right) \times 5$$

$$= 55 + \left(\frac{2}{6}\right) \times 5$$

$$= 55 + \frac{10}{6}$$

$$= 55 + 1.666\ldots$$

$$= 56.666\ldots \text{ kg}$$

Rounded to two decimal places, the median weight is $$56.67\text{ kg}$$.