2022-04-01

the manager of an electrical supply store measured th diameters of the rolls of wire in the inventory. The diameter of the rolls are listed below 0.56,  0.622, 0.154, 0.412, 0.287, 0.118

RizerMix

First sort the numbers into ascending order (from least to greatest):

Now let's find the mean. So add up all of the numbers and divide the sum by the number of numbers

So the mean is $0.3062$

Use this formula to find the Standard Deviation:

where $\overline{)\mathrm{x}}$ is the average, ${x}_{i}$ is the ith number, and $N$ is the number of numbers

So we can replace $N$ with $5$

$\sqrt{\frac{1}{5-1}\sum _{\mathrm{i}=0}^{\mathrm{N}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{)\mathrm{x}}\right)}^{2}}$

Subtract $5-1$ to get $4$

$\sqrt{\frac{1}{4}\sum _{\mathrm{i}=0}^{\mathrm{N}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{)\mathrm{x}}\right)}^{2}}$

Replace $\overline{)\mathrm{x}}$ with $0.3062$
$\sqrt{\frac{1}{4}\sum _{\mathrm{i}=0}^{\mathrm{N}}{\left({\mathrm{x}}_{\mathrm{i}}-0.3062\right)}^{2}}$

Expand the summation (replace each ${x}_{i}$ with the respective number)

$\sqrt{\frac{1}{4}\left({\left(0.118-0.3062\right)}^{2}+{\left(0.154-0.3062\right)}^{2}+{\left(0.287-0.3062\right)}^{2}+{\left(0.412-0.3062\right)}^{2}+{\left(0.56-0.3062\right)}^{2}\right)}$

$\sqrt{\frac{1}{4}\left({\left(-0.1882\right)}^{2}+{\left(0.154-0.3062\right)}^{2}+{\left(0.287-0.3062\right)}^{2}+{\left(0.412-0.3062\right)}^{2}+{\left(0.56-0.3062\right)}^{2}\right)}$

Raise $-0.1882$ to the power of $2$.

$\sqrt{\frac{1}{4}\left(0.03541924+{\left(0.154-0.3062\right)}^{2}+{\left(0.287-0.3062\right)}^{2}+{\left(0.412-0.3062\right)}^{2}+{\left(0.56-0.3062\right)}^{2}\right)}$

Subtract $0.3062$ from $0.154$.

$\sqrt{\frac{1}{4}\left(0.03541924+{\left(-0.1522\right)}^{2}+{\left(0.287-0.3062\right)}^{2}+{\left(0.412-0.3062\right)}^{2}+{\left(0.56-0.3062\right)}^{2}\right)}$

Raise $-0.1522$ to the power of $2$.

$\sqrt{\frac{1}{4}\left(0.03541924+0.02316484+{\left(0.287-0.3062\right)}^{2}+{\left(0.412-0.3062\right)}^{2}+{\left(0.56-0.3062\right)}^{2}\right)}$

Subtract $0.3062$ from $0.287$.

$\sqrt{\frac{1}{4}\left(0.03541924+0.02316484+{\left(-0.0192\right)}^{2}+{\left(0.412-0.3062\right)}^{2}+{\left(0.56-0.3062\right)}^{2}\right)}$

Raise $-0.0192$ to the power of $2$.

$\sqrt{\frac{1}{4}\left(0.03541924+0.02316484+0.00036864+{\left(0.412-0.3062\right)}^{2}+{\left(0.56-0.3062\right)}^{2}\right)}$

Subtract $0.3062$ from $0.412$.

$\sqrt{\frac{1}{4}\left(0.03541924+0.02316484+0.00036864+{0.1058}^{2}+{\left(0.56-0.3062\right)}^{2}\right)}$

Raise $0.1058$ to the power of $2$.

$\sqrt{\frac{1}{4}\left(0.03541924+0.02316484+0.00036864+0.01119364+{\left(0.56-0.3062\right)}^{2}\right)}$

Subtract $0.3062$ from $0.56$.

$\sqrt{\frac{1}{4}\left(0.03541924+0.02316484+0.00036864+0.01119364+{0.2538}^{2}\right)}$

Raise $0.2538$ to the power of $2$.

$\sqrt{\frac{1}{4}\left(0.03541924+0.02316484+0.00036864+0.01119364+0.06441444\right)}$

Add $0.03541924$ and $0.02316484$.

$\sqrt{\frac{1}{4}\left(0.05858408+0.00036864+0.01119364+0.06441444\right)}$

Add $0.05858408$ and $0.00036864$.

$\sqrt{\frac{1}{4}\left(0.05895272+0.01119364+0.06441444\right)}$

Add $0.05895272$ and $0.01119364$.

$\sqrt{\frac{1}{4}\left(0.07014636+0.06441444\right)}$

Add $0.07014636$ and 0.06441444$0.06441444$.

$\sqrt{\frac{1}{4}\cdot 0.1345608}$

Combine $\frac{1}{4}$ and $0.1345608$.

$\sqrt{\frac{0.1345608}{4}}$

Divide $0.1345608$ by $4$.

$\sqrt{0.0336402}$

The result can be shown in multiple forms.

Exact Form:

$\sqrt{0.0336402}$

Decimal Form:

$0.18341264\dots$

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