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
$$ \begin{aligned} \Var\bigg(\frac{1}{3}Y\bigg) &= \bigg(\frac{1}{3}\bigg)^2 \cdot \Var(Y) \\ & = \frac{1}{9}\Var(Y) \\ & = \frac{1}{9}\cdot9 \\ & = 1 \end{aligned} $$
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.
