The mean and variance of 5 observations are 5 | vedclass

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Question

$\frac{1+3+5+a+b}{5}=5$

$a+b=16 \ldots \ldots(1)$

$\sigma^2=\frac{\sum x_1^2}{5}-\left(\frac{\sum x}{5}\right)^2$ $8=\frac{1^2+3^2+5^2+a^2+b^2}{

Vedclass · Accepted Answer

$1072$

Vedclass · Answer

$\frac{1+3+5+a+b}{5}=5$

$a+b=16 \ldots \ldots(1)$

$\sigma^2=\frac{\sum x_1^2}{5}-\left(\frac{\sum x}{5}\right)^2$ $8=\frac{1^2+3^2+5^2+a^2+b^2}{5}-25$

$a^2+b^2=130 \ldots \ldots(2)$

$b y(1),(2)$

$a=7, b=9$

${x_i}$	$92$	$93$	$97$	$98$	$102$	$104$	$109$
${f_i}$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

class	0-10	10-20	20-30	30-40
Freq	1	3	4	2

The mean and variance of $5$ observations are $5$ and $8$ respectively. If $3$ observations are $1,3,5$, then the sum of cubes of the remaining two observations is

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class

0-10

10-20

20-30

30-40

Freq

1

3

4

2