If both mean and variance of 50 observations | vedclass

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(C) Given that the mean $\bar{x} = 16$ and variance $\sigma^2 = 256$ for $n = 50$ observations.$	ext{Mean} = \frac{\sum x_i}{n}$ $\Rightarrow 16 = \f

Vedclass · Accepted Answer

$377$

Vedclass · Answer

(C) Given that the mean $\bar{x} = 16$ and variance $\sigma^2 = 256$ for $n = 50$ observations.$	ext{Mean} = \frac{\sum x_i}{n}$ $\Rightarrow 16 = \frac{\sum x_i}{50}$ $\Rightarrow \sum x_i = 800$.$	ext{Variance} = \frac{\sum x_i^2}{n} - (\bar{x})^2 \Rightarrow 256 = \frac{\sum x_i^2}{50} - (16)^2$.$256 = \frac{\sum x_i^2}{50} - 256$ $\Rightarrow \frac{\sum x_i^2}{50} = 512$ $\Rightarrow \sum x_i^2 = 25600$.We need to find the mean of $(x_i - 5)^2$ for $i = 1, 2, \ldots, 50$.$	ext{New Mean} = \frac{\sum (x_i - 5)^2}{50} = \frac{\sum (x_i^2 - 10x_i + 25)}{50}$.$= \frac{\sum x_i^2 - 10 \sum x_i + \sum 25}{50} = \frac{25600 - 10(800) + 25(50)}{50}$.$= \frac{25600 - 8000 + 1250}{50} = \frac{18850}{50} = 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 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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