Question

Question

Answer 1

b) \[\large \begin{align} E(X)&= \sum_{x}x\cdot p(x)\\
\, \\

&=0(0.21)+1(0.39)+2(0.18)+\ldots+6(0.01)\end{align}\]

c) \[\large Var(X)=E(X^2)-[E(X)]^2\]
\[\large \begin{align} E(X^2)&= \sum_{x}x^2\cdot p(x)\\


&=0^2(0.21)+1^2(0.39)+2^2(0.18)+\ldots+6^2(0.01)\end{align}\]
And then for \([E(X)]^2\), you just find the result in b), and square it.

Answer 2

For the cdf in a):
\[ F(X)=\begin{cases}0 && x<0\\
0.21 &&0\le x <1 \\
0.6 &&1\le x <2 \\
0.78 &&2\le x <3 \\
0.94 &&3\le x <4 \\
0.97 &&4\le x <5 \\
0.99 &&5\le x <6 \\
1 && x \ge1 \\\end{cases}\]