2021-11-26

Martian potatoes begin to sprout very quickly after planting. Suppose X is the number of days after planting until a Martian potato sprouts. Then X has the following probability density function:

$f\left(x\right)=\left\{\begin{array}{ll}\frac{2}{7}{e}^{-x}+\frac{3}{14}{e}^{\frac{-x}{2}}+\frac{1}{14}{e}^{\frac{-x}{4}}& ,0

a) What is the probability that X is within 1 standard deviation of its expected value?
b) What is the probability that $X=.6$?

### Answer & Explanation

Novembruuo

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

Step 2
Consider, $E\left(X\right)={\int }_{0}^{\mathrm{\infty }}xf\left(x\right)dx$
$={\int }_{0}^{\mathrm{\infty }}x\left(\frac{2}{7}{e}^{-x}+\frac{3}{14}{e}^{\frac{-x}{2}}+\frac{1}{14}{e}^{\frac{-x}{4}}\right)dx$
$={\int }_{0}^{\mathrm{\infty }}\left(\frac{2}{7}x{e}^{-x}+\frac{3}{14}x{e}^{\frac{-x}{2}}+\frac{1}{14}x{e}^{\frac{-x}{4}}\right)dx$
$=\frac{16}{7}$
$E\left({X}^{2}\right)={\int }_{0}^{\mathrm{\infty }}{x}^{2}f\left(x\right)dx$
$={\int }_{0}^{\mathrm{\infty }}{x}^{2}\left(\frac{2}{7}{e}^{-x}+\frac{3}{14}{e}^{\frac{-x}{2}}+\frac{1}{14}{e}^{\frac{-x}{4}}\right)dx$
$={\int }_{0}^{\mathrm{\infty }}\left(\frac{2}{7}{x}^{2}{e}^{-x}+\frac{3}{14}{x}^{2}{e}^{\frac{-x}{2}}+\frac{1}{14}{x}^{2}{e}^{\frac{-x}{4}}\right)dx$
$=\frac{4}{7}+\frac{24}{7}+\frac{64}{7}$
$=\frac{92}{7}$
Step 3
Therefore, $SD\left(X\right)=\sqrt{E\left({X}^{2}\right)-{\left[E\left(X\right)\right]}^{2}}$
$=\sqrt{\frac{92}{7}-{\left(\frac{16}{7}\right)}^{2}}$
$=2.8139$
The probability that X is within 1 standard deviation of its expected value is 0.8810 and it is calculated below:

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