The distribution of SAT exam scores matches the Normal distr

The distribution of SAT exam scores matches the Normal distribution with a mean =1509 and standard deviation =321.

a. What is the probability that a student scores 1700 or less?

b. What is the probability that a student scores 1900 or more?

c. What is the maximum score for the bottom 20% of SAT scores?

d. What is the minimum score for the top 10% of SAT scores?

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1700      
u = mean =    1509      
          
s = standard deviation =    321      
          
Thus,          
          
z = (x - u) / s =    0.595015576      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0.595015576   ) =    0.72408347 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1900      
u = mean =    1509      
          
s = standard deviation =    321      
          
Thus,          
          
z = (x - u) / s =    1.218068536      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.218068536   ) =    0.111598965 [ANSWER]

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

c)

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 =    1509      
z = the critical z score =    -0.841621234      
s = standard deviation =    321      
          
Then          
          
x = critical value =    1238.839584   [ANSWER]

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

d)

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,          
          
where          
          
u = mean =    1509      
z = the critical z score =    1.281551566      
s = standard deviation =    321      
          
Then          
          
x = critical value =    1920.378053   [ANSWER]  
  


          

The distribution of SAT exam scores matches the Normal distribution with a mean =1509 and standard deviation =321. a. What is the probability that a student sco
The distribution of SAT exam scores matches the Normal distribution with a mean =1509 and standard deviation =321. a. What is the probability that a student sco

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