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(B) Given: Mean $\bar{x} = 16$ and Standard Deviation $\sigma = 16$. Let $y_i = x_i - 5$. The mean of the new observations $y_i$ is $\bar{y} = \bar{x}

Vedclass · Accepted Answer

$377$

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(B) Given: Mean $\bar{x} = 16$ and Standard Deviation $\sigma = 16$. Let $y_i = x_i - 5$. The mean of the new observations $y_i$ is $\bar{y} = \bar{x} - 5 = 16 - 5 = 11$. The standard deviation remains unchanged when a constant is subtracted from each observation,so $\sigma_y = 16$. We know that $\sigma_y^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2$. Substituting the values: $16^2 = \frac{\sum (x_i-5)^2}{50} - 11^2$. $256 = \frac{\sum (x_i-5)^2}{50} - 121$. $\frac{\sum (x_i-5)^2}{50} = 256 + 121 = 377$. Thus,the mean of $(x_i-5)^2$ is $377$.

Expenditure in $Rs. (x)$	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$
No. of families $(f)$	$24$	$33$	$37$	$b$	$25$

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 $(x_1-5)^2, (x_2-5)^2, \ldots, (x_{50}-5)^2$ is

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