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# Normal Distribution

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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 Summary

Probability values have been calculated using Normal Distribution

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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 ...

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