Question

A gas station operates two pumps, each of which can pump up to 10,000 gallons of gas in a month. the total of gas pumped at the station in a month is a random variable y (measured in 10,000 gallons) with a probability density function (p.d.f.) given by compute e(y)

Answer 1

Given that a gas station operates two pumps, each of which can pump up to 10,000 gallons of gas in a month and that the total of gas pumped at the station in a month is a random variable y (measured in 10,000 gallons) with a probability density function (p.d.f.) given by

$f(y)=\begin{cases} cy, & \text{if} \ \ 0\ \textless \ y\ \textless \ 1 \\ (2-y), & \text{if} \ \ 1\leq y\ \textless \ 2 \\ 0, & \text{elsewhere} \end{cases}$

Part A:

The value of c that makes f(y) a pdf is obtained as follows:

$F(\infty)= \int\limits^{\infty}_{-\infty} {f(y)} \, dy=1 \\ \\ \Rightarrow \int\limits^1_0 {cy} \, dy +\int\limits^2_1 {(2-y)} \, dy=1 \\ \\ \Rightarrow \left. \frac{cy^2}{2} \right]^1_0+\left[2y- \frac{y^2}{2} \right]^2_1=1 \\ \\ \Rightarrow \frac{c}{2} +4-2-2+ \frac{1}{2} =1 \\ \\ \Rightarrow \frac{c}{2} = \frac{1}{2} \\ \\ \Rightarrow \bold{c=1}$



Part B:

We compute E(y) as follows:

$E(y)=\int\limits^{\infty}_{-\infty} {yf(y)} \, dy \\ \\ =\int\limits^1_0 {y^2} \, dy +\int\limits^2_1 {(2y-y^2)} \, dy \\ \\ =\left. \frac{y^3}{3} \right]^1_0+\left[y^2- \frac{y^3}{3} \right]^2_1 \\ \\ = \frac{1}{3} +4- \frac{8}{3} -1+ \frac{1}{3} \\ \\ =1$

Therefore, E(y) = 1.