JEE Main 2026 Apr 8 Shift 2, Mathematics Q7: Measure of Dispersion

JEE Main2026Apr 8, Shift 2Mathematics

A set of four observations has mean $1$ and variance $13$ . Another set of six observations has mean $2$ and variance $1$ . Then, the variance of all these $10$ observations is equal to :

A
$5.96$
B
$6.14$
C
$6.04$
D
$6.24$

Solution

Combined-mean formula: $\overset{z}{ˉ} = \frac{n _{1} x ˉ + n _{2} y ˉ}{n _{1} + n _{2}} = \frac{4 ( 1 ) + 6 ( 2 )}{10} = 1.6$ .

Combined-variance formula:

$σ^{2} = \frac{n _{1} σ _{1}^{2} + n _{2} σ _{2}^{2}}{n _{1} + n _{2}} + \frac{n _{1} n _{2}}{( n _{1} + n _{2} ) ^{2}} (\overset{x}{ˉ} - \overset{y}{ˉ})^{2}$

First term: $\frac{4 ( 13 ) + 6 ( 1 )}{10} = \frac{58}{10} = 5.8$ .

Second term: $\frac{4 \cdot 6}{100} (1 - 2)^{2} = \frac{24}{100} = 0.24$ .

$∴$ $σ^{2} = 5.8 + 0.24 = 6.04$ .

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q7: Measure of Dispersion

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