The SAT scores of students are normally distributed with a m

The SAT scores of students are normally distributed with a mean of 950 and a standard deviation of 200.

a. Maha’s SAT score was 1390. What percentage of students has obtained scores more than Maha?

b. What percentage of students scored between 1100 and 1200?

c. There were 165 students who scored below 1432. How many students took the SAT?

d. What’s the score above it 80% of students got?

Solution

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1390      
u = mean =    950      
          
s = standard deviation =    200      
          
Thus,          
          
z = (x - u) / s =    2.2      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.2   ) =    0.013903448 = 1.3903448% [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    1100      
x2 = upper bound =    1200      
u = mean =    950      
          
s = standard deviation =    200      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.75      
z2 = upper z score = (x2 - u) / s =    1.25      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.773372648      
P(z < z2) =    0.894350226      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.120977579 = 12.0977% [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1432      
u = mean =    950      
          
s = standard deviation =    200      
          
Thus,          
          
z = (x - u) / s =    2.41      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   2.41   ) =    0.99202374

Thus, there are 1432/0.99202374 = 1443.51 = 1444 who took the test. [ANSWER]

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D)

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 =    950      
z = the critical z score =    -0.841621234      
s = standard deviation =    200      
          
Then          
          
x = critical value =    781.6757533   [ANSWER]