The storm drain capacity of a city is a normal variate with mean capacity of 1.6

Question

The storm drain capacity of a city is a normal variate with mean capacity of 1.65 mgd (million gallons per day) and a standard deviation of 0.2 mgd. The storm drain serves two independen and 0.65 mgd respectively; both drainage areas have the same standard deviation of 0.35 mgd. Calculate the probability that the storm drain capacity will be exceeded? Show all your work. Answers without documentation of how the results were obtained earn t drainage source areas A and B that are also normally distributed with means of 0.75 zero credit !!!!

Explanation / Answer

Lets assume storm drain on a random is D

its Mean capacity = 1.65 mgd

standard deviation s = 0.2 mgd

drainage source area A capacity (lets say x)on a random day x = 0.75 mgd

Standard deviation of drainage source area A on a random day x = 0.35 mgd

drainage source area A capacity (lets say y)on a random day y = 0.65 mgd

Standard deviation of drainage source area A on a random day y = 0.35 mgd

Here if z (total drainage capacity) = x +y

will have mean E(Z) = 0.65 + 0.75 = 1.40 mgd

Variance Z ; z2 = x2 + y2 = 0.352 + 0.352 = 0.245

Std(Z) = sqrt(0.245) = 0.495

Here we have to define a new variable W which is Z - D means drainage capacity of Z exceeds storm drain capacity D.

where Mean of W, E(W) = E(Z) - E(D) = 1.40 - 1.65 = -0.25 mgd

Variance of W, Var(Z) = 0.245 + 0.22 = 0.285

STD(Z) = 0.534 mgd

So, Here

Pr(W > 0 ; -0.25 ; 0.534) = 1 - NORM (W < 0 ; -0.25; 0.534)

Z = (0.25)/0.534 = 0.47

Pr(W > 0 ; -0.25 ; 0.534) = 1 - NORM (W < 0 ; -0.25; 0.534) = 1 - Pr(Z < 0.47) = 1 - 0.6808 = 0.3122

so there is 31.22% probability.

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