Statistics – Mean and Variance – JEE Main 28 Jan 2026 Shift 1

Solution:

Let the two unknown observations be $a$ and $b$.
Given Mean $\bar{x} = 9$ and $n=10$:
$$\sum x_i = 90$$
Sum of known 8 observations $= 2+3+5+10+11+13+15+21 = 80$.
$$\therefore a + b = 90 – 80 = 10 \quad \dots(1)$$

Given Variance $\sigma^2 = 34.2$:
$$\frac{\sum x_i^2}{10} – (\bar{x})^2 = 34.2$$
$$\sum x_i^2 = 10(34.2 + 81) = 1152$$

Sum of squares of known observations:
$$4 + 9 + 25 + 100 + 121 + 169 + 225 + 441 = 1094$$
$$\therefore a^2 + b^2 = 1152 – 1094 = 58 \quad \dots(2)$$

Using $(a+b)^2 = a^2 + b^2 + 2ab$:
$$100 = 58 + 2ab \Rightarrow 2ab = 42 \Rightarrow ab = 21$$
Since $a+b=10$ and $ab=21$, the roots are 3 and 7.
So the observations are $\{2, 3, 3, 5, 7, 10, 11, 13, 15, 21\}$.

Arranging in ascending order: $2, 3, 3, 5, 7, 10, 11, 13, 15, 21$.
Median $M$ is the average of the 5th and 6th terms:
$$M = \frac{7 + 10}{2} = 8.5$$

Calculate Mean Deviation about Median ($M.D.(M) = \frac{1}{n}\sum |x_i – M|$):
$$|2-8.5| = 6.5$$
$$|3-8.5| = 5.5$$
$$|3-8.5| = 5.5$$
$$|5-8.5| = 3.5$$
$$|7-8.5| = 1.5$$
$$|10-8.5| = 1.5$$
$$|11-8.5| = 2.5$$
$$|13-8.5| = 4.5$$
$$|15-8.5| = 6.5$$
$$|21-8.5| = 12.5$$

Sum $= 6.5+5.5+5.5+3.5+1.5+1.5+2.5+4.5+6.5+12.5 = 50$.
$$M.D. = \frac{50}{10} = 5$$

Ans. (1)