Question:
If both the mean and the standard deviation of 50 observations \(x_1,x_2,\ldots,x_{50}\) are equal to 16, then the mean of \((x_1-4)^2,(x_2-4)^2,\ldots,(x_{50}-4)^2\) is :
If both the mean and the standard deviation of 50 observations \(x_1,x_2,\ldots,x_{50}\) are equal to 16, then the mean of \((x_1-4)^2,(x_2-4)^2,\ldots,(x_{50}-4)^2\) is :
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The mean of \((x_i-c)^2\) equals variance plus \((\text{mean}-c)^2\).
Updated On: Nov 5, 2026
- \(400\)
- \(480\)
- \(380\)
- \(525\)
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The Correct Option is A
Solution and Explanation
Mean \(\bar x=16\), standard deviation \(=16\), so variance \(=256\).
Mean of \((x_i-4)^2=\frac{1}{50}\sum(x_i-\bar x+\bar x-4)^2=\sigma^2+(\bar x-4)^2\) (the cross term vanishes since \(\sum(x_i-\bar x)=0\)).
\(=256+(16-4)^2=256+144=400\).
Answer: (1) \(400\)
Mean of \((x_i-4)^2=\frac{1}{50}\sum(x_i-\bar x+\bar x-4)^2=\sigma^2+(\bar x-4)^2\) (the cross term vanishes since \(\sum(x_i-\bar x)=0\)).
\(=256+(16-4)^2=256+144=400\).
Answer: (1) \(400\)
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