According to a Salary Survey Study, the average base salary for a brand manager

Question

According to a Salary Survey Study, the average base salary for a brand manager in Houston, Texas, is $89,500 and the average base salary for a brand manager in Los Angeles, California, is $99,750. Assume that salaries are normally distributed, the standard deviation for brand managers in Houston is $21,500, and the standard deviation for brand managers in Los Angeles is $19,725.

a) What is the probability that a brand manager in Houston has a base salary in excess of $110,000?

b) What is the probability that a brand manager in Los Angeles has a base salary in excess of $95,000 but less than $105,500?

c) What is the probability that a brand manager in Houston has a base salary of less than $80,500?

d) How much would a brand manager Mr. Jones in Los Angeles have to make in order to have a salary higher than 98% of the brand managers in Houston?

e) What percentages of brand managers in LA earn more salary than Mr. Jones?

I need help putting this in excel please.

Explanation / Answer

Answer to the question is as follows:

In excel use NORM.DIST to find the cumulative probability.

Houston, Mu = 89000, Srdev = 21500
California, Mu = 99750, Stdev = 19725

a. P(houston > 110000) = P(Zh> (110000-89000)/21500 ) =.1644
b. P(95000<la<105500) = P(95000-89000/21500 <Zla< 105500-89000/21500) = .7786-.3901 = .3885
c. P(houston<80500) = P(Zh<80500-89000/21500) = .3463
d. P(houston<=salaryla) = .98 , So, Zhouston = 2.055

2.055 = (salaryla-89000)/21500
salaryla = 2.055*21500+89000 = $133182.5
This must be brand manager Mr. Jones salary in LA to have salary higher than 98% of the brand managers in Houston

e. P(lasalart>jones) = P( Z> (133182.5-99750) /19725)) = 0.0450
So, 4.5% of LA managers have more salary than Mr. Jones

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