Step 1: State the variance scaling property.
When every observation in a data set is multiplied by a constant $k$, the new variance equals $k^2$ times the original variance. In symbols, $\operatorname{Var}(kX) = k^2 \operatorname{Var}(X)$. This happens because variance measures squared spread, and scaling stretches distances by $k$, so squared distances grow by $k^2$.
Step 2: Identify the scaling factor.
Here each of $w, x, y, z$ is multiplied by $5$, so $k = 5$.
Step 3: Square the scaling factor.
\[ k^2 = 5^2 = 25 \]
Step 4: Apply the formula.
Given $\operatorname{Var}(w, x, y, z) = 9$, the new variance is \[ \operatorname{Var}(5w, 5x, 5y, 5z) = 25 \times 9 \]
Step 5: Compute the product.
\[ 25 \times 9 = 225 \]
Step 6: Verify intuition.
If the mean of $w, x, y, z$ is $\mu$, the mean of the scaled data is $5\mu$. Each deviation $(5w_i - 5\mu)^2 = 25(w_i - \mu)^2$, so the average of squared deviations is $25$ times larger. This confirms the formula.
Step 7: State the final answer.
The variance of $5w, 5x, 5y, 5z$ is \[ \boxed{225} \]