The mean and standard deviation of 15… - JEE Main 2024 Mathematics PYQ

JEE Main 2024 — Mathematics Probability & Statistics

2024integermedium

The mean and standard deviation of $15$ observations were found to be $12$ and $3$ respectively. On rechecking it was found that an observation was read as $10$ in place of $12.$ If $\mu$ and ${\sigma }^{2}$ denote the mean and variance of the correct observations respectively, then $15(\mu +{\mu }^{2}+{\sigma }^{2})$ is equal to _________.

Official previous-year question

Held on 27 Jan 2024 · Verified 6 Jul 2026.

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Solution

Given: Incorrect mean, $x=12$ and incorrect standard deviation, ${\sigma }^{'}=3$ .

$\Rightarrow \frac{{\Sigma x}_{i}}{15}=12$

$\Rightarrow {\Sigma x}_{i}=180$

Correct mean is given by,

$\mu =\frac{{\Sigma x}_{i}-10+12}{15}$

$\Rightarrow \mu =\frac{182}{15}$

Also, $\displaystyle {\sigma }^{'}=\sqrt{\frac{\sum {x}_{i}^{2}}{15}-144}=3$

$\Rightarrow \frac{\sum {x}_{i}^{2}}{15}=9+144$

$\Rightarrow \sum {x}_{i}^{2}=2295$

Correct ${\Sigma x}_{i}^{2}=2295-100+144=2339$

${\sigma }^{2}$ (correct variance) $=\frac{2339}{15}-\frac{182\times 182}{15\times 15}$

$\Rightarrow 15(\mu +{\mu }^{2}+{\sigma }^{2})=15(\frac{182}{15}+\frac{182\times 182}{15\times 15}+\frac{2339}{15}-\frac{182\times 182}{15\times 15})$

$\Rightarrow 15(\mu +{\mu }^{2}+{\sigma }^{2})=15(\frac{182}{15}+\frac{2339}{15})$

$\Rightarrow 15(\mu +{\mu }^{2}+{\sigma }^{2})=2521$

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