Mathematics · Statistics

JEE Main 2026 — 8 April, Evening Shift — Question 26

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 :

Option A:
$5.96$
Option B:
$6.14$
Option C:
$6.04$
Correct
Option D:
$6.24$

Answer: C

Step-by-step solution

Given $\overline{\mathrm{x}}=1, \sigma_{1}^{2}=13$ $\overline{\mathrm{y}}=2, \sigma_{2}^{2}=1$ Combined variance $=\frac{\mathrm{n}_{1} \sigma_{1}^{2}+\mathrm{n}_{2} \sigma_{2}^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)^{2}}(\overline{\mathrm{x}}-\overline{\mathrm{y}})^{2}$ $=\frac{4 \times 13+6 \times 1}{10}+\frac{4 \times 6}{10^{2}}(2-1)^{2}$ $\frac{58}{10}+\frac{24}{100}=\frac{580+24}{100}=\frac{604}{100}=6.04$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Mathematics
Chapter: Statistics
Topic: Measures of Dispersion