# Can I get some help please? thanks!

Sep 20, 2017

$\overline{x} = 16.1 \text{ (nearest tenth)}$

sigma=3.6" (nearest tenth")

#### Explanation:

The mean is the sum of the numbers divided by how many there are:

$\overline{x} = \frac{1}{n} \sum {x}_{i}$

$\overline{x} = \frac{20 + 16 + 18 + 14 + 9 + 20 + 16}{7}$

$\overline{x} = \frac{113}{7} = 16.1428 \ldots . = 16.1 \text{ (nearest tenth)}$

For the standard deviation, there are a several formulae one can use. Whichever one you use, there will be a lot of arithmetic to do!

We will use:

sigma=sqrt(1/nsum(x_i-barx)^2

We have to subtract the mean from each ${x}_{i}$ value, square these differences, and then add:

${\left(20 - \frac{113}{7}\right)}^{2} = \frac{729}{49}$

${\left(16 - \frac{113}{7}\right)}^{2} = \frac{1}{49}$

${\left(18 - \frac{113}{7}\right)}^{2} = \frac{169}{49}$

${\left(14 - \frac{113}{7}\right)}^{2} = \frac{225}{49}$

${\left(9 - \frac{113}{7}\right)}^{2} = \frac{2500}{49}$

${\left(20 - \frac{113}{7}\right)}^{2} = \frac{729}{49}$

${\left(16 - \frac{113}{7}\right)}^{2} = \frac{1}{49}$

The sum is $\frac{729 + 1 + 169 + 225 + 2500 + 729 + 1}{49} = \frac{4354}{49}$

So

$\sigma = \sqrt{\frac{1}{7} \times \frac{4354}{49}}$

$\sigma = \frac{\sqrt{622}}{7} = 3.56284 \ldots \ldots .$

sigma=3.6" (nearest tenth")