$X_1,X_3$ and $X_5$ are from a population with average = $\mu$ and stdev = $\sigma^2$.
My professor did:
$$V(\Theta_1) = V(\frac{X_1+X_3+X_5}{3}) = \frac{1}{9}(V(X_1)+V(X_3)+V(X_5)) = 1/9 * 3\sigma^2 = \sigma^2/3$$
In another exercise he calculated the expected value of this population this way:
$$E(\Theta_1) = \frac{1}{3}*(E(X_1)+E(X_3)+E(X_5))=\frac{1}{3}(3*\mu) = \mu$$
Why is the 1/3 squared on the variance?
Let $a$ be a real-valued constant. Let $X$ be a random variable with mean $\operatorname{E}[X] = \mu$ and variance $\operatorname{Var}[X] = \sigma^2$. Then consider the variance $\operatorname{Var}[aX]$. Recalling the definition of variance $$\operatorname{Var}[X] = \operatorname{E}[(X - \operatorname{E}[X])^2],$$ we have $$\begin{align*} \operatorname{Var}[aX] &= \operatorname{E}[(aX - \operatorname{E}[aX])^2] \\ &= \operatorname{E}[(aX - a\operatorname{E}[X])^2] \\ &= \operatorname{E}[a^2 (X - \operatorname{E}[X])^2] \\ &= a^2 \operatorname{E}[(X - \operatorname{E}[X])^2] \\ &= a^2 \operatorname{Var}[X], \end{align*}$$ as claimed.