The mean and the variance of five observation | vedclass

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(A) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given $n = 5$,$\bar{x} = 4$,and $\sigma^2 = 5.2$.Sum of observations: $\sum x_i = n 	imes

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given $n = 5$,$\bar{x} = 4$,and $\sigma^2 = 5.2$.Sum of observations: $\sum x_i = n 	imes \bar{x} = 5 	imes 4 = 20$.Given $x_1 = 3, x_2 = 4, x_3 = 4$,we have $3 + 4 + 4 + x_4 + x_5 = 20$,so $x_4 + x_5 = 9$ $(i)$.Variance formula: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.$5.2 = \frac{\sum x_i^2}{5} - 4^2$ $\Rightarrow 5.2 = \frac{\sum x_i^2}{5} - 16$ $\Rightarrow \sum x_i^2 = 5 	imes 21.2 = 106$.Sum of squares: $3^2 + 4^2 + 4^2 + x_4^2 + x_5^2 = 106$ $\Rightarrow 9 + 16 + 16 + x_4^2 + x_5^2 = 106$ $\Rightarrow x_4^2 + x_5^2 = 65$ $(ii)$.We know $(x_4 - x_5)^2 = 2(x_4^2 + x_5^2) - (x_4 + x_5)^2$.$(x_4 - x_5)^2 = 2(65) - (9)^2 = 130 - 81 = 49$.Therefore,$|x_4 - x_5| = \sqrt{49} = 7$.

The mean and the variance of five observations are $4$ and $5.20,$ respectively. If three of the observations are $3, 4,$ and $4,$ then the absolute value of the difference of the other two observations is

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