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TvStatic · Answer 1 · 2023-12-18T03:35:29+0000

Given:

The data is,

11,8,6,9,12,10,12,12,13,12,8,9,9,9,13,11,11,6,12,11

The mean of given data is,

$\bar{x}=10.2$

The varience is given as,

$\begin{gathered} s^2=\sum ^n_(i=1)\frac{\left(x_i-\bar{x}\right)^2}{n-1} \\ =(1)/(20-1)\lbrack(11-10.2)^2+(8-10.2)^2+(6-10.2)^2+(9-10.2)^2+(12-10.2)^2+(10-10.2)^2+(12-10.2)^2+(12-10.2)^2+(13-10.2)^2+(12-10.2)+(8-10.2)^2+(9-10.2)^2+(9-10.2)^2+(9-10.2)^2+(13-10.2)^2(11-10.2)^2+(11-10.2)^2+(6-10.2)+(12-10.2)^(22)+(11-10.2)^2\rbrack \\ =(85.2)/(19) \\ =4.48 \end{gathered}$

So, the value of varience is ,

The standard deviation is,

$s=\sqrt[]{4.48}=2.12$

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