With 10 integers available, \(X\) has PMF
\(\mathbb P(X=x)=\begin{cases}\dfrac1{10}&\text{for }0\le x\le9,x\in\mathbb Z\\\\0&\text{otherwise}\end{cases}\)
We're interested in the statistics of the new random variable \(Y=5X\). To do this, we need to know the PMF for \(Y\). This isn't too hard to find.
\(\mathbb P(Y=y)=\mathbb P(5X=y)=\mathbb P\left(X=\dfrac y5\right)\)
Since the PMF for \(X\) gives a value of \(\dfrac1{10}\) whenever \(x\) is an integer between 0 and 9, it follows that \(\dfrac y5\) must also be an integer for the PMF to give the identical value of \(\dfrac1{10}\). This means
\(\mathbb P(Y=y)=\begin{cases}\dfrac1{10}&\text{for }y\in\{0,5,\ldots,45\}\\\\0&\text{otherwise}\end{cases}\)
Now the mean (expectation) is
\(\mathbb E(Y)=\displaystyle\sum_yy\mathbb P(Y=y)=\frac1{10}\sum_{y\in\{0,\ldots,45\}}y\)
\(\mathbb E(Y)=\dfrac{225}{10}=22.5\)
The variance would be
\(\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2\)
\(\mathbb V(X)=\displaystyle\sum_yy^2\mathbb P(Y=y)-\mathbb E(Y)^2\)
\(\mathbb V(X)=\displaystyle\frac1{10}\sum_{y\in\{0,\ldots,45\}}y^2-\left(\frac{225}{10}\right)^2\)
\(\mathbb V(X)=\dfrac{7125}{10}-\dfrac{50625}{100}=206.25\)
The standard deviation is the square root of the variance, so you have
\(\sqrt{\mathbb V(X)}=\sqrt{206.25}\approx14.3614\)
