JEE MAIN - Mathematics (2025 - 7th April Morning Shift - No. 4)
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
447
445
449
451
Explanation
$$\begin{aligned}
&\text { Let the observations be } x_1, x_2, \ldots, x_{99}, 50\\
&\begin{aligned}
& \text { Mean }=\frac{x_1+x_2+\ldots+x_9+50}{100}=40 \\
& \Rightarrow x_1+x_2+\ldots+x_{99}=4000-50 \\
& \Rightarrow x_1+x_2+\ldots+x_{99}=3950 \\
& \text { Current Mean }=\frac{3950+40}{100} \\
& \mu=\frac{399}{10}=39.9 \\
& (\text { S.D })^2=\sum_{i=1}^{99} \frac{\left(x_i\right)^2+2500}{100}-(40)^2 \\
& \sum_{i=1}^{99} x_i^2=160101 \\
& (\text { Correct S.D })^2=\frac{160101+1600}{100}-\left(\frac{399}{10}\right)^2 \\
& \sigma=5 \\
& 10(\mu+\sigma)=10(39.9+5)=449
\end{aligned}
\end{aligned}$$
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