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If both mean and the standard deviation of 50 observations $x_1, x_2, \ldots ., x_{50}$ are equal to 16 , then mean of $\left(x_1-5\right)^2,\left(x_2-5\right)^2, \ldots \ldots .,\left(x_{50}-5\right)^2$ is
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Verified Answer
The correct answer is:
377
$\vec{x}=16 \text { and S.D. }=16$
if we subtract 5 from each observations mean will become $16-5=11$ and S.D. remains unchanged i.e., 16
$\begin{aligned}
& \Rightarrow \sqrt{\frac{\sum\left(x_i-5\right)^2}{50}-11^2}=16 \\
& \Rightarrow \frac{\sum\left(x_i-5\right)^2}{50}=256+121=377
\end{aligned}$
hence, the required mean is 377
if we subtract 5 from each observations mean will become $16-5=11$ and S.D. remains unchanged i.e., 16
$\begin{aligned}
& \Rightarrow \sqrt{\frac{\sum\left(x_i-5\right)^2}{50}-11^2}=16 \\
& \Rightarrow \frac{\sum\left(x_i-5\right)^2}{50}=256+121=377
\end{aligned}$
hence, the required mean is 377
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