Answer
$\begin{aligned} \mathbb{E}(X) &=4.808 \\ \operatorname{Var}(X) &=6.15\end{aligned}$
Work Step by Step
$\begin{aligned} \mathbb{E}(X) &=\sum_{x=1}^{15} x f(x)=1\times0.038+2\times0.102+3\times0.172+4\times0.204+5\times0.174+6\times0.124+7\times0.080+8\times0.036+9\times0.028+10\times0.022+15\times0.020=4.808 \\ \mathbb{E}\left(X^{2}\right) &=\sum_{x=1}^{15} x^{2} f(x)=29.264 \\ \operatorname{Var}(X) &=\mathbb{E}\left(X^{2}\right)-\mathbb{E}(X)^{2}={6 .15} \end{aligned}$
