Answer:
0.1827
Step-by-step explanation:
Given mean of exponential distribution = 100
==> 1/χ = 100  ==> χ = 1/100 ==> χ = 0.01
PDF of χ , f(x) = χe^(-χx), x ≥ 0
===> f(x) = 0.01e^(-0.01x), x ≥ 0
Now we find the probability that the demand will exceed 170 cfs during the early afternoon on a randomly selected day
P(X>170) = ∞∫170 f(x)dx
P(X>170) = ∞∫170 0.01 e^(0.01x) dx
P(X>170) = [e^(-0.01x) / -0.01]^∞  base 170
P(X>170) = -1 [e^-∞  - e^-0.01*170]
P(X>170) = e^-1.7
P(X>170) = 0.1827
The probability that the demand will exceed 170 cfs during the early afternoon on a randomly selected day is 0.1827
