Please explain 6 The scores on a college aptitude test foll

Please explain :)

6. The scores on a college aptitude test follow N(490, 70). Use the “NORM.INV” function in Excel and
properties of areas to answer the following questions. Round your answers to the nearest whole
number score.
a) The bottom 10% of test takers score at or below what score?
b) The top 15% of test takers score at or above what score?
c) What scores delineate the middle 60% of test takers?

Solution

a)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.1      
          
Then, using table or technology,          
          
z =    -1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    490      
z = the critical z score =    -1.281551566      
s = standard deviation =    70      
          
Then          
          
x = critical value =    400.2913904   [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.85      
          
Then, using table or technology,          
          
z =    1.036433389      
          
As x = u + z * s,          
          
where          
          
u = mean =    490      
z = the critical z score =    1.036433389      
s = standard deviation =    70      
          
Then          
          
x = critical value =    562.5503373   [ANSWER]

**********************

c)

The lower end must have a left tailed area of 0.2, then the right end has an area of 0.8.

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,          
          
where          
          
u = mean =    490      
z = the critical z score =    -0.841621234      
s = standard deviation =    70      
          
Then          
          
x = critical value =    431.0865136   [LOWER END, ANSWER]  
  
First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.8      
          
Then, using table or technology,          
          
z =    0.841621234      
          
As x = u + z * s,          
          
where          
          
u = mean =    490      
z = the critical z score =    0.841621234      
s = standard deviation =    70      
          
Then          
          
x = critical value =    548.9134864   [UPPER END, ANSWER]

Thus, it is between 431.0865136 AND 548.9134864.