JEE Main 2025 Apr 4 Shift 2, Mathematics Q20: Measure of Dispersion

JEE Main2025Apr 4, Shift 2Mathematics

Let the mean and the standard deviation of the observation $2, 3, 3, 4, 5, 7, a, b$ be 4 and $2$ respectively. Then the mean deviation about the mode of these observations is :

A
1
B
$\frac{3}{4}$
C
2
D
$\frac{1}{2}$

Solution

Sum of all eight observations $= 8 \times 4 = 32$ . The six known values sum to $24$ , so $a + b = 8$ .

Variance $= 2$ , so $\sum (x_{i} - 4)^{2} = 16$ . The six known deviations squared sum to $4 + 1 + 1 + 0 + 1 + 9 = 16$ , so $(a - 4)^{2} + (b - 4)^{2} = 0$ , forcing $a = b = 4$ .

Now the data is $2, 3, 3, 4, 4, 4, 5, 7$ . The mode is $4$ (appearing thrice). Mean deviation about the mode:

$\frac{1}{8} \sum ∣ x_{i} - 4∣ = \frac{2 + 1 + 1 + 0 + 0 + 0 + 1 + 3}{8} = \frac{8}{8} = 1$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q20: Measure of Dispersion

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