If both mean and variance of 50 observations | vedclass

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(C) Given $n=50$,$\bar{x}=16$,and $\sigma_x^2=256$. Since $\sigma_x^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$,we have $256 = \frac{1}{50} \sum x_i^2 -

Vedclass · Accepted Answer

$377$

Vedclass · Answer

(C) Given $n=50$,$\bar{x}=16$,and $\sigma_x^2=256$. Since $\sigma_x^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$,we have $256 = \frac{1}{50} \sum x_i^2 - 16^2$. $256 = \frac{1}{50} \sum x_i^2 - 256 \implies \frac{1}{50} \sum x_i^2 = 512 \implies \sum x_i^2 = 25600$. We need the mean of $(x_i-5)^2$,which is $\frac{1}{50} \sum_{i=1}^{50} (x_i-5)^2$. Expanding the sum: $\sum (x_i^2 - 10x_i + 25) = \sum x_i^2 - 10 \sum x_i + \sum 25$. Since $\bar{x} = \frac{1}{50} \sum x_i = 16$,then $\sum x_i = 50 	imes 16 = 800$. Thus,$\sum (x_i-5)^2 = 25600 - 10(800) + 50(25) = 25600 - 8000 + 1250 = 18850$. Required mean $= \frac{18850}{50} = 377$.

If both mean and variance of $50$ observations $x_1, x_2, \ldots, x_{50}$ are equal to $16$ and $256$ respectively,then the mean of $(x_1-5)^2, (x_2-5)^2, \ldots, (x_{50}-5)^2$ is

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