Mathematics · Statistics

JEE Main 2025 — 7 April, Morning Shift — Question 22

The mean and standard deviation of $100$ observations are $40$ and $5.1,$ respectively. By mistake one observation is taken as $50$ instead of $40.$ If the correct mean and the correct standard deviation are $\mu$ and $\sigma$ respectively, then $10(\mu+\sigma)$ is equal to

Option A:
$447$
Option B:
$445$
Option C:
$449$
Correct
Option D:
$451$

Answer: C

Step-by-step solution

Let the observations be $x_{1}, x_{2}, \ldots, x_{99}, 50$

Mean $=\frac{x_{1}+x_{2}+\ldots+x_{9}+50}{100}=40$

$\Rightarrow x_{1}+x_{2}+\ldots+x_{99}=4000-50$

$\Rightarrow x_{1}+x_{2}+\ldots+x_{99}=3950$

Current Mean $=\frac{3950+40}{100}$

$\mu=\frac{399}{10}=39.9$

$(S . D)^{2}=\sum_{i=1}^{99} \frac{\left(x_{i}\right)^{2}+2500}{100}-(40)^{2}$

$\sum_{i=1}^{99} x_{i}^{2}=160101$

$(\text { Correct S.D })^{2}=\frac{160101+1600}{100}-\left(\frac{399}{10}\right)^{2}$

$\sigma=5$

$10(\mu+\sigma)=10(39.9+5)=449$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Mathematics
Chapter: Statistics
Topic: Measures of Dispersion