EX 4.12 EX 4.10 continued

Variance of EX 4.10 expected value of EX 4.9

For $X=$ weekly gravel sales, we computed $E\left(X\right)=\frac{3}{8}.$ Since

$$\begin{array}{rl}E\left(X^2\right)& ={\int }_{-\infty }^{\infty }{x}^2\cdot f\left(x\right)dx\\ & ={\int }_0^1{x}^2\cdot \frac{3}{2}\left(1-x^2\right)dx\\ & ={\int }_0^1\frac{3}{2}\left(x^2-x^4\right)dx\\ & =\frac{1}{5}\end{array}$$ $$\begin{array}{rl}V\left(X\right)& =\frac{1}{5}-{\left(\frac{3}{8}\right)}^2\\ & =\frac{19}{320}\approx 0.059\end{array}$$

and ${\sigma }_X=\mathrm{0.244.}$