MHT CET · Maths · Statistics
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
- A 378
- B 377
- C 357
- D 397
Answer & Solution
Correct Answer
(B) 377
Step-by-step Solution
Detailed explanation
\(\vec{x}=16 \text { and S.D. }=16\)
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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