JEE Mathematics 2025 Question with Solution

Given: Mean of the data is $5.5$, number of values is $10$, and $\sum_{i=1}^{10} x_i^2 = 371$.

Find: The variance of the corrected data after replacing $4, 5$ by $6, 8$.

From the mean,

$$\sum_{i=1}^{10} x_i = 10 \times 5.5 = 55$$

The incorrect values contribute

$$4 + 5 = 9$$

and the correct values contribute

$$6 + 8 = 14$$

So the corrected sum is

$$55 - 9 + 14 = 60$$

Hence the corrected mean is

$$\frac{60}{10} = 6$$

Now adjust the sum of squares. The squares of the incorrect values are

$$4^2 + 5^2 = 16 + 25 = 41$$

The squares of the correct values are

$$6^2 + 8^2 = 36 + 64 = 100$$

Therefore, the corrected sum of squares is

$$371 - 41 + 100 = 430$$

Using the variance formula,

$$\text{Variance} = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$$

Substituting the corrected values,

$$\text{Variance} = \frac{430}{10} - \left(\frac{60}{10}\right)^2 = 43 - 36 = 7$$

Therefore, the variance of the corrected data is $7$. The correct option is A.

Given: Original mean is $5.5$, so original sum is $55$. Also, $\sum x_i^2 = 371$.

Find: Corrected variance.

Instead of rebuilding the whole data set, directly correct the totals:

$$\sum x_i' = 55 + (6+8) - (4+5) = 60$$

and

$$\sum (x_i')^2 = 371 + (6^2+8^2) - (4^2+5^2) = 430$$

Now apply

$$\text{Variance} = \frac{\sum (x_i')^2}{10} - \left(\frac{\sum x_i'}{10}\right)^2$$

So,

$$\text{Variance} = \frac{430}{10} - \left(\frac{60}{10}\right)^2 = 43 - 36 = 7$$

This works because variance depends only on the corrected sum and corrected sum of squares. Therefore, the correct option is A.