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Varvara68 [4.7K]
3 years ago
5

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)
Mathematics
1 answer:
erma4kov [3.2K]3 years ago
5 0
Given that a<span> 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}&#10;      cy, & \text{if} \ \ 0\ \textless \ y\ \textless \ 1 \\&#10;      (2-y), & \text{if} \ \ 1\leq y\ \textless \ 2 \\&#10;      0, & \text{elsewhere}&#10;    \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.
</span>
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