Solution Found!
The weekly gravel demand X (in tons) follows the pdf
Chapter 3, Problem 10E(choose chapter or problem)
The weekly gravel demand \(X\) (in tons) follows the pdf
\(f(x)=\left(\frac{1}{5}\right) e^{-x / 5}, \quad 0<x<\infty\) .
However, the owner of the gravel pit can produce at most only 4 tons of gravel per week. Compute the expected value of the tons sold per week by the owner.
Equation Transcription:
Text Transcription:
X
f(x)=(⅕)e^-x/5, 0 < x < infinity
Questions & Answers
QUESTION:
The weekly gravel demand \(X\) (in tons) follows the pdf
\(f(x)=\left(\frac{1}{5}\right) e^{-x / 5}, \quad 0<x<\infty\) .
However, the owner of the gravel pit can produce at most only 4 tons of gravel per week. Compute the expected value of the tons sold per week by the owner.
Equation Transcription:
Text Transcription:
X
f(x)=(⅕)e^-x/5, 0 < x < infinity
ANSWER:Solution :
Step 1 of 2:
The weekly gravel demand X (in tons) ~exponential distribution with pdf
f(x) = (1/5)
The owner of the gravel pit can produce at most 4 tons of gravel per week.we have to compute the expected value of the tons sold per week by the owner.