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JEE Main 20268 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2026 · 8 April · Evening Shift · Mathematics · Statistics

A set of four observations has mean 1 and variance 13 . Another set...

A set of four observations has mean 11 and variance 1313. Another set of six observations has mean 22 and variance 11. Then, the variance of all these 1010 observations is equal to:

Options

  1. A5.965.96
  2. B6.146.14
  3. C6.046.04
  4. D6.246.24

Correct answer

C. 6.046.04

Step-by-step solution

Given for the first set: n1=4n_1 = 4, xˉ1=1\bar{x}_1 = 1, σ12=13\sigma_1^2 = 13.

σ12=∑xi2n1−(xˉ1)2\sigma_1^2 = \dfrac{\sum x_i^2}{n_1} - (\bar{x}_1)^2

13=∑xi24−12⇒∑xi2=5613 = \dfrac{\sum x_i^2}{4} - 1^2 \Rightarrow \sum x_i^2 = 56

Also, ∑xi=n1xˉ1=4×1=4\sum x_i = n_1 \bar{x}_1 = 4 \times 1 = 4

Given for the second set: n2=6n_2 = 6, xˉ2=2\bar{x}_2 = 2, σ22=1\sigma_2^2 = 1.

σ22=∑yi2n2−(xˉ2)2\sigma_2^2 = \dfrac{\sum y_i^2}{n_2} - (\bar{x}_2)^2

1=∑yi26−22⇒∑yi2=301 = \dfrac{\sum y_i^2}{6} - 2^2 \Rightarrow \sum y_i^2 = 30

Also, ∑yi=n2xˉ2=6×2=12\sum y_i = n_2 \bar{x}_2 = 6 \times 2 = 12

For the combined set of 1010 observations:

Total sum ∑zi=∑xi+∑yi=4+12=16\sum z_i = \sum x_i + \sum y_i = 4 + 12 = 16

Combined mean zˉ=1610=1.6\bar{z} = \dfrac{16}{10} = 1.6

Total sum of squares ∑zi2=∑xi2+∑yi2=56+30=86\sum z_i^2 = \sum x_i^2 + \sum y_i^2 = 56 + 30 = 86

Combined variance σ2=∑zi210−(zˉ)2\sigma^2 = \dfrac{\sum z_i^2}{10} - (\bar{z})^2

σ2=8610−(1.6)2=8.6−2.56=6.04\sigma^2 = \dfrac{86}{10} - (1.6)^2 = 8.6 - 2.56 = 6.04

Answer: 6.046.04
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