SOLUTION: 8. The average starting salary of this year's MBA students is $35,000 with a standard deviation of $5,000. Furthermore, it is known that the starting salaries are normally distrib

Question 1083845: 8. The average starting salary of this year's MBA students is $35,000 with a standard deviation of $5,000. Furthermore, it is known that the starting salaries are normally distributed. What are the minimum and the maximum starting salaries of the middle 95% of MBA graduates?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
average is 35,000
standard deviation is 5,000

middle 95% of the normal distribution curve leave 2.5% on each end.

2.5% on each end is your alpha.

z-score is 100% - 2.5% = 97.5% of the normal distribution curve to the left of it.

look for a z-score with .975 to the left of it.

that z-score will be approximately 1.96

since the normal distribution curve is symmetric about the mean, that means the your critical z-score is plus or minus 1.96.

the formula for z-score is z = (x-m)/s

z is the z-score.
x is the raw-score.
m is the raw-mean.
s is the standard deviation.

for your data, to find the high side value of x, you would do:

1.96 = (x-35,000) / 5,000

solve for x to get x = 1.96 * 5,000 + 35,000 = 44,800.

to find the low side value of x, you would do:

-1.96 = (x-35,000) / 5,000

solve for x to get x = -1.96 * 5000 + 35,000 = 25,200.

your minimum starting salary is 25,200.

your maximum starting salary is 44,800.

this is at 95% confidence interval.

this is what it looks like on the normal distribution curve.

$$$

there are slight differences in my answer and the calculator's answer that are due to rounding.

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