EX 3.25 (EX 3.24 continued)

The pmf of the number $X$ of DVDs checked out was given as

• $p\left(1\right)=.30$,
• $p\left(2\right)=.25$,
• $p\left(3\right)=.15$,
• $p\left(4\right)=.05$,
• $p\left(5\right)=.10$
• $p\left(6\right)=.15$ , We have
• $\mu =2.85$

$$\begin{array}{rl}E(X^2)& =\sum _{x=1}^6{x}^2\cdot p(x)\\ & =(1^2)(0.30)+(2^2)(0.25)+\cdots +(6^2)(0.15)\\ & =11.35\end{array}$$

Thus ${\sigma }^2=11.35-{\left(2.85\right)}^2=3.2275$ as obtained previously from the definition.