If both the means and the standard deviation | vedclass

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Question

Mean $\left( \mu  \right) = \frac{{\sum {{x_i}} }}{{50}} = 16$

Standard deviation $\left( \sigma  \right) = \sqrt {\frac{{\sum

Vedclass · Accepted Answer

$400$

Vedclass · Answer

Mean $\left( \mu  ight) = \frac{{\sum {{x_i}} }}{{50}} = 16$

Standard deviation $\left( \sigma  ight) = \sqrt {\frac{{\sum {x_i^2} }}{{50}} - {{\left( \mu  ight)}^2}}  = 16$

$ \Rightarrow \left( {256} ight) 	imes 2 = \frac{{\sum {x_i^2} }}{{50}}$

$\Rightarrow$ New mean

$ = \frac{{\sum {{{\left( {{x_i} - 4} ight)}^2}} }}{{50}} = \frac{{\sum {x_i^2 + 16 	imes 50 - 8\sum {{x_i}} } }}{{50}}$

$ = \left( {256} ight) 	imes 2 + 16 - 8 	imes 16 = 400$

If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, …., (x_{50} - 4)^2$ is

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