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 mean 38 and std deviation 7.9 (a) If the number of spares, 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 probability of needing more than 40 spares, given the demand is at least 38?

Explanation / Answer

a) probability of  demand exceeds the supply =P(X>40)=1-P(X<40)=1-P(Z<(40-38)/7.9)=1-P(Z<0.2532)

=1-0.5999 =0.4001

b) for 89th percentile ; value of z score =1.2265

therefore number of spares required =mean+z*Std deviation=38+1.2265*7.9 =47.69~ 48

c) P(X>40|X>38 )=P(Z>0.2532)/P(Z>0)=0.4001/0.5 =0.8001

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