The demand for the number of spares, X, to support a squadron of V22 helicopters

Question

The demand for the number of spares, X, to support a squadron of V22 helicopters for a month of deployed operations can be described by a normal distribution with mean = 38 and standard deviation = 7.9 (a) If the number of spaces, for the deployment is the standard number of 40, what is the probability of not having enough, i.e., the demand exceeds the supply? (b) How many spares are required for the deployment if the probability of having enough is 89

The demand for the number of spares, X, to support a squadron of V22 helicopters for a month of deployed operations can be described by a normal distribution with mean = 38 and standard deviation = 7.9 (a) If the number of spaces, for the deployment is the standard number of 40, what is the probability of not having enough, i.e., the demand exceeds the supply? (b) How many spares are required for the deployment if the probability of having enough is 89

(c) what is the probabilityof needing more than 40 spares, given the demand is at least 38

Explanation / Answer

mean = 38
std. dev. = 7.9

(a)
P(X < 40) = P(z < (40 - 38)/7.9) = P(z < 0.2532)
= 0.5999

(b)
P = 0.89, z-value = 1.2265
xbar = z*sigma + mu
xbar = 1.2265*7.9 + 38
xbar = 47.689

i.e. 48 spares

c)
A indicates an event with demand at least 38
and B indicates an event with demand more than 40

P(A) = 0.5
P(B) = 1 - 0.5999 = 0.4001
P(A and B) = 0.4001

Required probability, P(B|A) = P(A and B)/P(A) = 0.4001/0.5 = 0.8002

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