JEE Main 2026 Apr 4 Shift 2, Mathematics Q8: Measure of Dispersion

JEE Main2026Apr 4, Shift 2Mathematics

For 10 observations $x_{1}, x_{2}, \dots, x_{10}$ , if $i = 1 \sum 10 (x_{i} + 2)^{2} = 180$ and $i = 1 \sum 10 (x_{i} - 1)^{2} = 90$ , then their standard deviation is:

A
2
B
$3$
C
$22$
D
3

Solution

Let $S_{1} = \sum x_{i}$ and $S_{2} = \sum x_{i}^{2}$ . Expand both sums:

$\sum (x_{i} + 2)^{2} = S_{2} + 4 S_{1} + 40 = 180 \Rightarrow S_{2} + 4 S_{1} = 140$ (I)

$\sum (x_{i} - 1)^{2} = S_{2} - 2 S_{1} + 10 = 90 \Rightarrow S_{2} - 2 S_{1} = 80$ (II)

Subtract (II) from (I): $6 S_{1} = 60 \Rightarrow S_{1} = 10$ . Then $S_{2} = 100$ .

Mean $= S_{1} /10 = 1$ . Variance $= \frac{S _{2}}{10} - (mean)^{2} = 10 - 1 = 9$ .

$σ = 3$ .

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JEE Main 2026 Apr 4 Shift 2, Mathematics Q8: Measure of Dispersion

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