Martian potatoes begin to sprout very quickly after planting. Suppose X is the number of...

Step 1
a) From the given information, the probability density function for X is,
$f(x)=\{\begin{array}{ll}\frac{2}{7}{e}^{-x}+\frac{3}{14}{e}^{\frac{-x}{2}}+\frac{1}{14}{e}^{\frac{-x}{4}}& ,0<x<\mathrm{\infty }\\ 0& otherwise\end{array}$

Step 2
Consider, ${\displaystyle E\left(X\right)={\int }_0^{\mathrm{\infty }}xf\left(x\right)dx}$
${\displaystyle ={\int }_0^{\mathrm{\infty }}x(\frac{2}{7}{e}^{-x}+\frac{3}{14}{e}^{\frac{-x}{2}}+\frac{1}{14}{e}^{\frac{-x}{4}})dx}$
${\displaystyle ={\int }_0^{\mathrm{\infty }}(\frac{2}{7}x{e}^{-x}+\frac{3}{14}x{e}^{\frac{-x}{2}}+\frac{1}{14}x{e}^{\frac{-x}{4}})dx}$
${\displaystyle =\frac{16}{7}}$
${\displaystyle E\left(X^2\right)={\int }_0^{\mathrm{\infty }}{x}^{2}f\left(x\right)dx}$
${\displaystyle ={\int }_0^{\mathrm{\infty }}{x}^2(\frac{2}{7}{e}^{-x}+\frac{3}{14}{e}^{\frac{-x}{2}}+\frac{1}{14}{e}^{\frac{-x}{4}})dx}$
$={\int }_0^{\mathrm{\infty }}(\frac{2}{7}{x}^2{e}^{-x}+\frac{3}{14}{x}^2{e}^{\frac{-x}{2}}+\frac{1}{14}{x}^2{e}^{\frac{-x}{4}})dx$
${\displaystyle =\frac{4}{7}+\frac{24}{7}+\frac{64}{7}}$
${\displaystyle =\frac{92}{7}}$
Step 3
Therefore, ${\displaystyle SD\left(X\right)=\sqrt{E\left(X^2\right)-{\left[E\left(X\right)\right]}^2}}$
${\displaystyle =\sqrt{\frac{92}{7}-{\left(\frac{16}{7}\right)}^2}}$
${\displaystyle =2.8139}$
The probability that X is within 1 standard deviation of its expected value is 0.8810 and it is calculated below: