Mathematics · Statistics

JEE Main 2025 — 3 April, Evening Shift — Question 22

Let the Mean and Variance of five observations $x_{1}=$ $1, x_{2}=3, x_{3}=a, x_{4}=7$ and $x_{5}=b, a>b$ , be

5 and 10 respectively. Then the Variance of the observations $n+x_{n}, n=1.2, \ldots, 5$ is

Answer: B

$5=\frac{1+3+a+7+b}{5}$

$\Rightarrow a+b=14$

$\frac{1+9+a^{2}+49-b^{2}}{5}-(5)^{2}=10$

$a^{2}+b^{2}=116$

$\Rightarrow a=10, b=4$

New digits: $2,5,13,11,9$

Var $=\frac{4+26+169+121+81}{5}-64$

Var $=16$

Answer key and solution verified before publishing.

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