JEE Main 2026 Apr 5 Shift 1, Mathematics Q8: Measure of Dispersion

JEE Main2026Apr 5, Shift 1Mathematics

The mean deviation about the mean for the data

$x_{i}$	5	7	9	10	12	15
$f_{i}$	8	6	2	2	2	6

is equal to :

A
$\frac{40}{13}$
B
$\frac{42}{13}$
C
$\frac{44}{13}$
D
$\frac{46}{13}$

Solution

Total frequency $\sum f_{i} = 8 + 6 + 2 + 2 + 2 + 6 = 26$ .

Sum of $f_{i} x_{i} = 40 + 42 + 18 + 20 + 24 + 90 = 234$ , so the mean $\overset{x}{ˉ} = \frac{234}{26} = 9$ .

Deviations $∣ x_{i} - 9∣$ are $4, 2, 0, 1, 3, 6$ . Then $\sum f_{i} ∣ x_{i} - \overset{x}{ˉ} ∣ = 8 (4) + 6 (2) + 2 (0) + 2 (1) + 2 (3) + 6 (6) = 32 + 12 + 0 + 2 + 6 + 36 = 88$ .

Therefore M.D. $= \frac{88}{26} = \frac{44}{13}$ .

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JEE Main 2026 Apr 5 Shift 1, Mathematics Q8: Measure of Dispersion

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