Search any question & find its solution
Question:
Answered & Verified by Expert
If the mean of 100 observations is 50 and their standard deviation is 5 , then the sum of squares of all observations is
Options:
Solution:
1573 Upvotes
Verified Answer
The correct answer is:
252500
Given; \(\bar{x}=50\)
\(n=100\) and \(S . D(\sigma)=5\)
As, \(\bar{x}=\frac{\Sigma x_i}{n} \Rightarrow \Sigma x_i=n . \bar{x}=5000\).
Also, S.D. is given by,
\(\begin{aligned}
\sigma & =\sqrt{\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2} \\
\Rightarrow \quad \sigma^2 & =\frac{\Sigma x_i^2}{n}-(\bar{x})^2 \Rightarrow 25=\frac{\Sigma x_i^2}{100}-(50)^2 \\
\Rightarrow \quad \Sigma x_i^2 & =252500
\end{aligned}\)
\(n=100\) and \(S . D(\sigma)=5\)
As, \(\bar{x}=\frac{\Sigma x_i}{n} \Rightarrow \Sigma x_i=n . \bar{x}=5000\).
Also, S.D. is given by,
\(\begin{aligned}
\sigma & =\sqrt{\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2} \\
\Rightarrow \quad \sigma^2 & =\frac{\Sigma x_i^2}{n}-(\bar{x})^2 \Rightarrow 25=\frac{\Sigma x_i^2}{100}-(50)^2 \\
\Rightarrow \quad \Sigma x_i^2 & =252500
\end{aligned}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.