The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second). Find the probability that the demand will exceed 170 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

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

$79500

step-by-step explanation: