In a large section of a statistics class the points for the

In a large section of a statistics class, the points for the final exam are normally distributed, with a mean of 72 and a standard deviation of 9. Grades are to be assigned according to the following rule:

The top 10% receive A\'s.

The next 20% receive B\'s.

The middle 40% receive C\'s.

The next 20% received D\'s.

The bottom 10% receive F\'s.

Find the lowest score on the final exam that would qualify a student for an A, a B, a C, and a D.

Solution

a)

For an A:

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.1 =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    72      
z = the critical z score =    1.281551566      
s = standard deviation =    9      
          
Then          
          
x = critical value =    83.53396409   [ANSWER]

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b)

For a B:

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.1 - 0.2 =    0.7      
          
Then, using table or technology,          
          
z =    0.524400513      
          
As x = u + z * s,          
          
where          
          
u = mean =    72      
z = the critical z score =    0.524400513      
s = standard deviation =    9      
          
Then          
          
x = critical value =    76.71960461   [ANSWER]  

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C)

For a C:

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.1 - 0.2 - 0.4 =   0.3      
          
Then, using table or technology,          
          
z =    -0.524400513      
          
As x = u + z * s,          
          
where          
          
u = mean =    72      
z = the critical z score =    -0.524400513      
s = standard deviation =    9      
          
Then          
          
x = critical value =    67.28039539   [ANSWER]

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d)

For a D:

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.1 = 0.2 - 0.4 - 0.2 =   0.1      
          
Then, using table or technology,          
          
z =    -1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    72      
z = the critical z score =    -1.281551566      
s = standard deviation =    9      
          
Then          
          
x = critical value =    60.46603591   [ANSWER]