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?
 A: 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 
$$
\begin{aligned} 
\Var\bigg(\frac{1}{3}Y\bigg) &=  \bigg(\frac{1}{3}\bigg)^2 \cdot \Var(Y) \\
& = \frac{1}{9}\Var(Y) \\
& = \frac{1}{9}\cdot9 \\
& = 1
\end{aligned}
$$
Combining the two is pretty trivial!
There's a nice intuition behind both of these properties if you think about variance as "spread" or "scatter". Shifting a distribution left or right (i.e., adding or subtracting a constant) has no effect on how spread apart it is. Scaling a distribution with multiplication or division, however, stretches or squishes it (respectively), which changes the spread of the distribution and thus its variance.
