Assume that the GRE test scores are normally distributed wit

Assume that the GRE test scores are normally distributed with a mean score of = 620 and a standard deviation = 40. Then:

a.) What is the 90th percentile score value?

b.) What percent of the test takers scored 620 or higher?

c.) What percent of the test takers scored between 580 and 660?

Solution

a)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    620      
z = the critical z score =    1.281551566      
s = standard deviation =    40      
          
Then          
          
x = critical value =    671.2620626   [answer]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    620      
u = mean =    620      
          
s = standard deviation =    40      
          
Thus,          
          
z = (x - u) / s =    0      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0   ) =    0.5 = 50% [answer]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    580      
x2 = upper bound =    660      
u = mean =    620      
          
s = standard deviation =    40      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.682689492 = 68.2689492% [answer]