MHT CET · Maths · Statistics
The mean of 100 obseryations is 50 and their standard deviation is 5 , then the sum of all squares of all the observations is
- A 252500
- B 250500
- C 250000
- D 255000
Answer & Solution
Correct Answer
(A) 252500
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \bar{x}=50, \text { Standard deviation }=5 \\
& \mathrm{n}=100 \\
& \text { Mean } \bar{x}=50 \\
& \Rightarrow \frac{\sum x_{\mathrm{i}}}{100}=50 \\
& \Rightarrow \sum x_{\mathrm{i}}=5000 \\
& \text { Now, standard deviation }=\sqrt{\frac{\sum x_i^2}{\mathrm{n}}-\left(\frac{\sum x_{\mathrm{i}}}{\mathrm{n}}\right)^2} \\
& \Rightarrow 5=\sqrt{\frac{\sum x_{\mathrm{i}}^2}{100}-\left(\frac{5000}{100}\right)^2} \\
& \Rightarrow 25=\frac{\sum x_i^2}{100}-(50)^2 \\
& \Rightarrow \frac{\sum x_i^2}{100}=25+2500 \\
& \Rightarrow \sum x_i^2=252500
\end{aligned}\)
& \bar{x}=50, \text { Standard deviation }=5 \\
& \mathrm{n}=100 \\
& \text { Mean } \bar{x}=50 \\
& \Rightarrow \frac{\sum x_{\mathrm{i}}}{100}=50 \\
& \Rightarrow \sum x_{\mathrm{i}}=5000 \\
& \text { Now, standard deviation }=\sqrt{\frac{\sum x_i^2}{\mathrm{n}}-\left(\frac{\sum x_{\mathrm{i}}}{\mathrm{n}}\right)^2} \\
& \Rightarrow 5=\sqrt{\frac{\sum x_{\mathrm{i}}^2}{100}-\left(\frac{5000}{100}\right)^2} \\
& \Rightarrow 25=\frac{\sum x_i^2}{100}-(50)^2 \\
& \Rightarrow \frac{\sum x_i^2}{100}=25+2500 \\
& \Rightarrow \sum x_i^2=252500
\end{aligned}\)
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