1 Answer

Nitesh Tosniwal · Answer 1 · 2020-03-02T19:44:48+0000

Answer:

A. 6.5

Explanation:

First we find the average
$\bar{x}$ of the 9 data:

$\bar{x} =(\sum_(x=1)^(n)x_(i))/(n)$

Where n is the data number, that in this case is 9.

$\bar{x} =(1+ 2+ 11+ 8+ 16+ 16+20+ 16+ 18)/(9)=12\\$

The formula of the standard deviation
$\sigma$ is:

$\sigma=\sqrt{\frac{\sum_(x=1)^(n)(x_(i)-\bar{x})^(2)}{n}}$

We replace the data and find the value of the standard deviation:

$\sigma=\sqrt{((1-12)^(2)+(2-12)^(2)+(11-12)^(2)+(8-12)^(2)+(16-12)^(2)+(16-12)^(2)+(20-12)^(2)+(16-12)^(2)+(18-12)^(2))/(9)}$

$\sigma=\sqrt{((-11)^(2)+(-10)^(2)+(-1)^(2)+(-4)^(2)+(4)^(2)+(4)^(2)+(8)^(2)+(4)^(2)+(6)^(2))/(9)}=6,54$

We approximate the number and the solution is 6,5

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