# Showing that a certain random variable has variance $1$

I am calculating this example at the moment. Are my calculations right?

The random variable $Y$ has a mean of 15 and a variance of 9. $Z = 1/3 (Y − 15)$. Show that $Z$ has variance $1$.

$Z = \frac13 (Y − 15)$ $, \newcommand{\stdev}{\mathrm{stdev}}\stdev(Y)=3$

$$\stdev(Z)=\frac13 (Y − 15) = \frac13 (Y)$$ $$\stdev(Y)=3$$ $$\stdev(Z)=\frac13 Y = \frac13 3 = 1$$ $$\mathrm{var}(Z) = 1^2 = 1$$

Is my argumentation right?

You're correct, though I'm not exactly sure why you converted variances to standard deviations and then back, since you can derive your answer from two properties of variance, namely $$\newcommand{\Var}{\mathrm{Var}} \Var(X-c) = \Var(X)$$ $$\Var(cX) = c^2 \cdot \Var(X)$$
where $$X$$ is a random variable and $$c$$ is a constant.
The first property tells us that $$\Var(Y-15) = \Var(Y) = 9$$. Using the second, we find that
\begin{aligned} \Var\bigg(\frac{1}{3}Y\bigg) &= \bigg(\frac{1}{3}\bigg)^2 \cdot \Var(Y) \\ & = \frac{1}{9}\Var(Y) \\ & = \frac{1}{9}\cdot9 \\ & = 1 \end{aligned}