Statistics – Mean and Variance – JEE Main 24 Jan 2026 Shift 1

Solution:

Let the original observations be $x_1, x_2, \dots, x_{10}$.
Given $n=10$, Old Mean $\bar{x} = 10$, Old Variance $\sigma^2 = 2$.

Sum of observations:
$$\sum x_i = n \bar{x} = 10 \times 10 = 100$$
Sum of squares of observations (using variance formula $\sigma^2 = \frac{\sum x_i^2}{n} – (\bar{x})^2$):
$$2 = \frac{\sum x_i^2}{10} – (10)^2$$
$$\frac{\sum x_i^2}{10} = 102 \Rightarrow \sum x_i^2 = 1020$$

Now, observation $\alpha$ is replaced by $\beta$.
New Mean = 10.1.
$$100 – \alpha + \beta = 10 \times 10.1 = 101$$
$$\beta – \alpha = 1 \quad \dots(1)$$

New Variance = 1.99.
$$\frac{\sum y_i^2}{10} – (10.1)^2 = 1.99$$
$$\frac{\sum y_i^2}{10} – 102.01 = 1.99$$
$$\frac{\sum y_i^2}{10} = 104 \Rightarrow \sum y_i^2 = 1040$$

Relation between sum of squares:
$$\sum x_i^2 – \alpha^2 + \beta^2 = 1040$$
$$1020 – \alpha^2 + \beta^2 = 1040$$
$$\beta^2 – \alpha^2 = 20$$
$$(\beta – \alpha)(\beta + \alpha) = 20$$

Substitute value from (1):
$$1 \cdot (\beta + \alpha) = 20$$
$$\alpha + \beta = 20$$

Ans. (4)