EX 4.9 amount of gravel

The distribution of the amount of gravel (in tons) sold by a particular construction supply company in a given week is a continuous rv $X$ with pdf

$$f\left(x\right)=\{\begin{array}{cc}\frac{3}{2}\left(1-x^2\right)& 0\le x\le 1\\ 0& \text{ otherwise }\end{array}$$

The cdf of sales for any $x$ between 0 and 1 is

$$\begin{array}{rl}F\left(x\right)& ={\int }_0^x\frac{3}{2}\left(1-y^2\right)dy\\ & ={\frac{3}{2}\left(y-\frac{y^3}{3}\right)|}_{y=0}^{y=x}\\ & =\frac{3}{2}\left(x-\frac{x^3}{3}\right)\end{array}$$

The graphs of both $f\left(x\right)$ and $F\left(x\right)$ appear in Figure 4.11.

Figure 4.11 The pdf and cdf for Example 4.9

The $\left(100p\right)$ th percentile of this distribution satisfies the equation

$$p=F\left(\eta \left(p\right)\right)=\frac{3}{2}\left[\eta \left(p\right)-\frac{{\left(\eta \left(p\right)\right)}^3}{3}\right]$$

that is,

$${\left(\eta \left(p\right)\right)}^3-3\eta \left(p\right)+2p=0$$

For the 50th percentile, $p=.5$ , and the equation to be solved is ${\eta }^3-3\eta +1=0$; the solution is $\eta =\eta \left(.5\right)=.347$. If the distribution remains the same from week to week, then in the long run

• $50\%$ of all weeks will result in sales of less than .347 ton
• $50\%$ in more than .347 ton.