JEE Mathematics 2023 Question with Solution

Given: Number of observations is $n = 10$, mean is $\bar{x} = 20$, and standard deviation is $\sigma = 8$ for the wrongly recorded data. One observation $50$ should be $40$.

Find: The correct variance.

First, the variance of the wrong data is

$$\sigma^2 = 8^2 = 64$$

Using

$$\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$$

we get

$$64 = \frac{\sum x^2}{10} - 20^2$$ $$64 = \frac{\sum x^2}{10} - 400$$ $$\frac{\sum x^2}{10} = 464 \Rightarrow \sum x^2 = 4640$$

Also, the sum of observations for the wrong data is

$$\sum x = n\bar{x} = 10 \times 20 = 200$$

Now correct the error.

Correct sum:

$$\sum x_{\text{correct}} = 200 - 50 + 40 = 190$$

Correct sum of squares:

$$\sum x^2_{\text{correct}} = 4640 - 50^2 + 40^2$$ $$= 4640 - 2500 + 1600 = 3740$$

So the correct mean is

$$\bar{x}_{\text{correct}} = \frac{190}{10} = 19$$

Now the correct variance is

$$\sigma_{\text{correct}}^2 = \frac{\sum x^2_{\text{correct}}}{n} - \bar{x}_{\text{correct}}^2$$ $$= \frac{3740}{10} - 19^2$$ $$= 374 - 361 = 13$$

Therefore, the correct variance is $13$ and the correct option is C.

Given: One value is corrected from $50$ to $40$.

Find: The corrected variance.

The key idea is that variance depends on both sum and sum of squares, so both must be updated.

Wrong-data sum is

$$\sum x = 10 \times 20 = 200$$

and wrong-data sum of squares is found from

$$64 = \frac{\sum x^2}{10} - 400$$

which gives

$$\sum x^2 = 4640$$

After correction,

$$\sum x = 200 - 50 + 40 = 190$$ $$\sum x^2 = 4640 - 2500 + 1600 = 3740$$

Hence,

$$\text{Variance} = \frac{3740}{10} - \left(\frac{190}{10}\right)^2 = 374 - 361 = 13$$

Therefore, the correct option is C.