JEE Main 2026 — Statistics Question with Solution

The given numbers are $21,8,17,a,51,103,b,13,67$. There are $9$ numbers in total.

Given that the mean is $40$, the sum of the numbers is:
$\sum x_i=9\times 40=360$

The sum of the known numbers is:
$21+8+17+51+103+13+67=280$

Therefore, the sum of $a$ and $b$ is:
$a+b=360-280=80$

The median of the $9$ numbers is $21$. Arranging the known numbers in ascending order, we get:
$8,13,17,21,51,67,103$

There are $3$ numbers less than $21$ and $3$ numbers greater than $21$. For $21$ to be the median (the $5^{\text{th}}$ observation), one of the unknown numbers must be $\le 21$ and the other must be $\ge 21$. Since $a>b$, we must have:
$a\ge 21$ and $b\le 21$

The mean deviation about the median is $26$, so the sum of absolute deviations from the median is:
$\sum |x_i-21|=9\times 26=234$

The sum of absolute deviations of the known numbers from $21$ is:
$|8-21|+|13-21|+|17-21|+|21-21|+|51-21|+|67-21|+|103-21|$
$=13+8+4+0+30+46+82=183$

The sum of absolute deviations of $a$ and $b$ from $21$ is:
$|a-21|+|b-21|=234-183=51$

Since $a\ge 21$ and $b\le 21$, we can remove the absolute values:
$(a-21)+(21-b)=51$
$a-b=51$

We now have a system of two equations:
$a+b=80$
$a-b=51$

Adding the two equations, we get:
$2a=131$

Answer: $131$