Incorrect Observation Correction

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:

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

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

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