Mean Using Summation Notation

For $$N = 10$$ observations:

$$\bar{x}_{10} = 5 \implies \sum_{i=1}^{10} x_i = 10 \times 5 = 50$$

$$\sigma_{10}^2 = 7 \implies \frac{1}{10}\sum_{i=1}^{10} x_i^2 - (\bar{x}_{10})^2 = 7 \implies \frac{1}{10}\sum_{i=1}^{10} x_i^2 - 25 = 7 \implies \sum_{i=1}^{10} x_i^2 = 320$$

For the first 8 observations ($$N = 8$$):

$$\bar{x}_{8} = 4 \implies \sum_{i=1}^{8} x_i = 8 \times 4 = 32$$

$$\sigma_{8}^2 = 3.5 \implies \frac{1}{8}\sum_{i=1}^{8} x_i^2 - (\bar{x}_{8})^2 = 3.5 \implies \frac{1}{8}\sum_{i=1}^{8} x_i^2 - 16 = 3.5 \implies \sum_{i=1}^{8} x_i^2 = 156$$

Calculating relations for $$x_9$$ and $$x_{10}$$:

$$\sum_{i=1}^{10} x_i - \sum_{i=1}^{8} x_i = x_9 + x_{10} \implies 50 - 32 = x_9 + x_{10} \implies x_9 + x_{10} = 18 \quad \text{--- (1)}$$

$$\sum_{i=1}^{10} x_i^2 - \sum_{i=1}^{8} x_i^2 = x_9^2 + x_{10}^2 \implies 320 - 156 = x_9^2 + x_{10}^2 \implies x_9^2 + x_{10}^2 = 164 \quad \text{--- (2)}$$

$$18^2 = 164 + 2x_9x_{10} \implies 324 = 164 + 2x_9x_{10} \implies x_9x_{10} = 80 \quad \text{--- (3)}$$

Solving the quadratic equation $$t^2 - (x_9+x_{10})t + x_9x_{10} = 0$$:

$$t^2 - 18t + 80 = 0 \implies (t-8)(t-10) = 0 \implies t = 8, 10$$

Since $$x_9 < x_{10}$$: $$x_9 = 8, \quad x_{10} = 10$$

$$3x_9 + 2x_{10} = 3(8) + 2(10) = 24 + 20 = 44$$

Given mean $$=4$$ for $$10$$ observations,

$$\sum x_i=40$$

Given variance

$$=9,$$

$$9=\frac{\sum x_i^2}{10}-16$$

$$\sum x_i^2=250$$

After adding $$4,$$

new sum becomes $$44,$$

hence new mean remains $$4$$

Now,

$$x_1^2+x_2^2=50$$ is replaced by $$3^2+5^2=34$$

Also include

$$4^2=16$$

Hence new sum of squares:

$$250-50+34+16=250$$

Therefore,

$$\sigma^2=\frac{250}{11}-16=\frac{74}{11}$$

Thus,

$$11\sigma^2=74$$

Let the 8 observations be $$x_1, x_2, \dots, x_7, x_8$$, where the incorrect observation is $$x_8 = 10$$.

We are given the incorrect mean ($$\bar{x}_{\text{inc}}$$) is 9 for $$n = 8$$:

$$\bar{x}_{\text{inc}} = \frac{\sum_{i=1}^{8} x_i}{8} = 9 \implies \sum_{i=1}^{8} x_i = 72$$

We can find the sum of the first 7 correct observations by subtracting the incorrect value:

$$\sum_{i=1}^{7} x_i = 72 - 10 = 62$$

Now, adding the correct value (6) gives the new correct sum of all 8 observations:

$$\sum x_{\text{new}} = 62 + 6 = 68$$

Thus, the new correct mean ($$\bar{x}_{\text{new}}$$) is:

$$\bar{x}_{\text{new}} = \frac{68}{8} = 8.5$$

Next, we process the variance. The given incorrect variance is 9.25:

$$\text{Variance}_{\text{inc}} = \frac{\sum_{i=1}^{8} x_i^2}{8} - (\bar{x}_{\text{inc}})^2 = 9.25$$

$$\frac{\sum_{i=1}^{8} x_i^2}{8} - 9^2 = 9.25 \implies \frac{\sum_{i=1}^{8} x_i^2}{8} - 81 = 9.25$$

$$\frac{\sum_{i=1}^{8} x_i^2}{8} = 90.25 \implies \sum_{i=1}^{8} x_i^2 = 722$$

We find the sum of squares of the first 7 correct observations by removing the square of the incorrect value ($10^2 = 100$):

$$\sum_{i=1}^{7} x_i^2 = 722 - 100 = 622$$

Now, adding the square of the correct value ($6^2 = 36$) gives the new correct sum of squares:

$$\sum x_{\text{new}}^2 = 622 + 36 = 658$$

Finally, we calculate the correct variance using the new parameters:

$$\text{Variance}_{\text{new}} = \frac{\sum x_{\text{new}}^2}{8} - (\bar{x}_{\text{new}})^2$$

$$\text{Variance}_{\text{new}} = \frac{658}{8} - (8.5)^2$$

$$\text{Variance}_{\text{new}} = 82.25 - 72.25 = 10$$