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Normal Distribution

Foreign service employees receive housing allowances when posted abroad. These allowances average $30000 annually. Assume that a normal distribution applies and the standard deviation is $5000.
A) What is the probability that a diplomat will receive a housing allowance exceeding $35000?
B) What is the probability that a diplomat will receive a housing allowance between $27800 and $32200?
C) What percentage of employees will receive allowances between $25000 and $35000?

Solution Preview

Mean=M = $30,000
Standard deviation =s = $5,000

A) What is the probability that a diplomat will receive a housing allowance exceeding $35000?

Mean=M = $30,000
Standard deviation =s = $5,000
x= $35,000
z=(x-M )/s =1 =(35000-30000)/5000
Cumulative Probability corresponding to z= 1 is 0.8413
Therefore probability corresponding to x> $35,000 is 1-Prob(Z)= 0.1587 ...

Solution Summary

Probability values have been calculated using Normal Distribution

$2.19