SOLUTION: Scores on a management aptitude examination are normally distributed with a mean of 72 and a standard deviation of 8. a. What is the lowest score that will place a manager in t

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Question 841820: Scores on a management aptitude examination are normally distributed with a mean of 72 and a standard deviation of 8.
a. What is the lowest score that will place a manager in the top 1% (99th percentile) of the distribution? Your answer is Please round to an integer number.
b. What is the highest score that will place a manager in the bottom 20% (20th percentile) of the distribution? Your answer is Please round to an integer number.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Using TI:

*Note: z+=+blue%28x+-+mu%29%2Fblue%28sigma%29

z = invNorm(area to left of desired z).

a)z =  invNorm(.99) = 2.326

  2.362 = (x-72)/8

 8(2.326) + 72 = x = 91

b)z = invNorm(.20) = -.842

   8(-.842) + 72 = x = 65