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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