Scores on a management aptitude examination are normally dis

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.

Solution

a)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.99      
          
Then, using table or technology,          
          
z =    2.326347874      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    72      
z = the critical z score =    2.326347874      
s = standard deviation =    8      
          
Then          
          
x = critical value =    90.61078299 = 91 [answer]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.2      
          
Then, using table or technology,          
          
z =    -0.841621234      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    72      
z = the critical z score =    -0.841621234      
s = standard deviation =    8      
          
Then          
          
x = critical value =    65.26703013 = 65 [answer]