EX 3.13 flash drives

A store carries flash drives with either $1\mathrm{GB}$, $2\mathrm{GB}$, $4\mathrm{GB}$, $8\mathrm{GB}$, or $16\mathrm{GB}$ of memory. The accompanying table gives the distribution of $Y=$ the amount of memory in a purchased drive:

$y$	1	2	4	8	16
$p(y)$	.05	.10	.35	.40	.10

Let’s first determine $F\left(y\right)$ for each of the five possible values of $Y$ :

$$\begin{array}{rl}F(1)& =P(Y\le 1)=P(Y=1)=p(1)=0.05\\ F(2)& =P(Y\le 2)=P(Y=1\text{ or }2)=p(1)+p(2)=0.15\\ F(4)& =P(Y\le 4)=P(Y=1\text{ or }2\text{ or }4)=p(1)+p(2)+p(4)=0.50\\ F(8)& =P(Y\le 8)=p(1)+p(2)+p(4)+p(8)=0.90\\ F(16)& =P(Y\le 16)=1\end{array}$$

Now for any other number $y,F\left(y\right)$ will equal the value of $F$ at the closest possible value of $Y$ to the left of $y$ . For example, $F\left(2.7\right)=P\left(Y\le 2.7\right)=P\left(Y\le 2\right)=F\left(2\right)=.15$ $F\left(7.999\right)=P\left(Y\le 7.999\right)=P\left(Y\le 4\right)=F\left(4\right)=.50$

If $y$ is less than $1,F\left(y\right)=0$ [e.g. $F\left(.58\right)=0$ ], and if $y$ is at least $16,F\left(y\right)=1$ [e.g., $F\left(25\right)=1]$. The cdf is thus $F\left(y\right)=\{\begin{array}{cc}0& y<1\\ .05& 1\le y<2\\ .15& 2\le y<4\\ .50& 4\le y<8\\ .90& 8\le y<16\\ 1& 16\le y\end{array}$ A graph of this cdf is shown in Figure 3.5.

Figure 3.5 A graph of the cdf of Example 3.13