the manager of an electrical supply store measured

First sort the numbers into ascending order (from least to greatest):
$0.118, 0.154, 0.287, 0.412, 0.56$

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

$\mathrm{Mean}=\frac{0.118+ 0.154+ 0.287+ 0.412+ 0.56}{5}=\frac{1.531}{5}=0.3062$

So the mean is $0.3062$

Use this formula to find the Standard Deviation:
$\mathrm{Standard} \mathrm{Deviation}: \mathrm{\sigma }=\sqrt{\frac{1}{\mathrm{N}-1}\sum _{\mathrm{i}=0}^{\mathrm{N}}{({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}})}^2}$

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 $5$

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

Subtract $5-1$ to get $4$

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

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

Expand the summation (replace each $x_i$ with the respective number)

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

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

Raise ${\displaystyle -0.1882}$ to the power of ${\displaystyle 2}$.

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

Subtract ${\displaystyle 0.3062}$ from ${\displaystyle 0.154}$.

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

Raise ${\displaystyle -0.1522}$ to the power of ${\displaystyle 2}$.

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

Subtract ${\displaystyle 0.3062}$ from ${\displaystyle 0.287}$.

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

Raise ${\displaystyle -0.0192}$ to the power of ${\displaystyle 2}$.

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

Subtract ${\displaystyle 0.3062}$ from ${\displaystyle 0.412}$.

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

Raise ${\displaystyle 0.1058}$ to the power of ${\displaystyle 2}$.

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

Subtract ${\displaystyle 0.3062}$ from ${\displaystyle 0.56}$.

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

Raise ${\displaystyle 0.2538}$ to the power of ${\displaystyle 2}$.

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

Add ${\displaystyle 0.03541924}$ and ${\displaystyle 0.02316484}$.

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

Add ${\displaystyle 0.05858408}$ and ${\displaystyle 0.00036864}$.

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

Add ${\displaystyle 0.05895272}$ and ${\displaystyle 0.01119364}$.

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

Add ${\displaystyle 0.07014636}$ and 0.06441444${\displaystyle 0.06441444}$.

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

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

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

Divide ${\displaystyle 0.1345608}$ by ${\displaystyle 4}$.

${\displaystyle \sqrt{0.0336402}}$

The result can be shown in multiple forms.

Exact Form:

${\displaystyle \sqrt{0.0336402}}$

Decimal Form:

$0.18341264\dots$