Solutions - 5100-06

(a)

$$\begin{array}{c}P[X>2]=1-P[X\le 2]=1-F_X(2)\\ \\ \gg \gg 1-0.48\gg \gg 0.52\end{array}$$

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(b)

(1) Construct PMF from CDF:

$$P_X(k)=\{\begin{array}{cc}0.13& k=1\\ 0.35& k=2\\ 0.33& k=3\\ 0.19& k=4\end{array}$$

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(2) Compute $E[X]$, $E[X^2]$, and $\mathrm{Var}[X]$:

$$\begin{array}{c}E[X]=1(0.13)+2(0.35)+3(0.33)+4(0.19)=2.58\\ \\ E[X^2]=1^2(0.13)+2^2(0.35)+3^2(0.33)+4^2(0.19)=7.54\\ \\ \mathrm{Var}[X]=E[X^2]-{(E[X])}^2\gg \gg 7.54-{(2.58)}^2\gg \gg 0.8836\end{array}$$