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Mean and standard deviation of \( 100 \) items are \( 50 \) and \( 4 \) respectively. The sum of all squares of
the items is
Options:
the items is
Solution:
2061 Upvotes
Verified Answer
The correct answer is:
\( 251600 \)
(C)
\[
\begin{array}{l}
\sigma=\sqrt{\frac{\sum x i^{2}}{n}-\left(\bar{x}^{2}\right)} \\
4=\sqrt{\frac{\sum x i^{2}}{100}-(50)^{2}} \\
16=\frac{\sum x i^{2}}{100}-2500 \\
\Rightarrow \sum x i^{2}=251600
\end{array}
\]
\[
\begin{array}{l}
\sigma=\sqrt{\frac{\sum x i^{2}}{n}-\left(\bar{x}^{2}\right)} \\
4=\sqrt{\frac{\sum x i^{2}}{100}-(50)^{2}} \\
16=\frac{\sum x i^{2}}{100}-2500 \\
\Rightarrow \sum x i^{2}=251600
\end{array}
\]
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