The mean of 5 observations is 5 and their var | vedclass

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(C) Given $n = 5$,mean $\bar{x} = 5$,and variance $\sigma^2 = 124$.Let the observations be $x_1=1, x_2=2, x_3=6, x_4, x_5$.Since $\bar{x} = \frac{\sum

Vedclass · Accepted Answer

$2.8$

Vedclass · Answer

(C) Given $n = 5$,mean $\bar{x} = 5$,and variance $\sigma^2 = 124$.Let the observations be $x_1=1, x_2=2, x_3=6, x_4, x_5$.Since $\bar{x} = \frac{\sum x_i}{5} = 5$,we have $1+2+6+x_4+x_5 = 25$,so $x_4+x_5 = 16$.Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 124$.$\frac{1^2+2^2+6^2+x_4^2+x_5^2}{5} - 5^2 = 124$ $\Rightarrow \frac{1+4+36+x_4^2+x_5^2}{5} = 149$ $\Rightarrow x_4^2+x_5^2 = 745 - 41 = 704$.We have $(x_4+x_5)^2 = x_4^2+x_5^2 + 2x_4x_5$ $\Rightarrow 16^2 = 704 + 2x_4x_5$ $\Rightarrow 256 = 704 + 2x_4x_5$ $\Rightarrow 2x_4x_5 = -448$ $\Rightarrow x_4x_5 = -224$.Solving $x_4+x_5=16$ and $x_4x_5=-224$,the quadratic equation $t^2 - 16t - 224 = 0$ gives $t = \frac{16 \pm \sqrt{256 + 896}}{2} = \frac{16 \pm \sqrt{1152}}{2} = 8 \pm 12\sqrt{2}$.Mean deviation $M$.$D$. $= \frac{1}{5} \sum |x_i - 5| = \frac{1}{5} (|1-5| + |2-5| + |6-5| + |x_4-5| + |x_5-5|)$.$|x_4-5| + |x_5-5| = |x_4+x_5-10| = |16-10| = 6$ (since $x_4, x_5$ are on opposite sides of $5$).$M$.$D$. $= \frac{4+3+1+6}{5} = \frac{14}{5} = 2.8$.

${x_i}$	$5$	$7$	$9$	$10$	$12$	$15$
${f_i}$	$8$	$6$	$2$	$2$	$2$	$6$

The mean of $5$ observations is $5$ and their variance is $124$. If three of the observations are $1, 2$ and $6$,then the mean deviation from the mean of the data is:

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