= Solution
The <random variable> $T$ is the sum $E+(-S)$. The <linearity of expectation> gives
$$
\mathbb E[T]=\mathbb E[E]-\mathbb E[S]=\frac12-\frac12=0.
$$
Multiplication by $-1$ reverses the <mean> but preserves the <variance>. Under the independence assumption used in part (b), the <variance of a sum> gives
$$
\operatorname{Var}(T)=\operatorname{Var}(E)+\operatorname{Var}(-S)
=\frac1{12}+\frac1{12}=\frac16.
$$
Thus both predictions agree with the direct <integrals>. For the actual elapsed time, adding twenty-five changes the <mean> to twenty-five minutes and leaves the <variance> at $1/6$ minute squared.
Without independence, the correct expression is $\operatorname{Var}(T)=1/6-2\operatorname{Cov}(E,S)$ when both marginals are uniform. This explains why independent endpoint errors must be part of the intended model rather than a consequence of the display format.
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