Test scores on a musical aptitude test are normally distribu

Test scores on a musical aptitude test are normally distributed with a mean of 90 and a standard deviation of 10. Suppose an examining committee is trying to set threshold values for sorting examinees into categories of performance. What score on the aptitude test would place an examinee in the 80^th percentile? What score on the aptitude test would place an examinee in the 30^th percentile? Over what range of scores, equidistant from the mean, would roughly 85 % of examinees fall?

Solution

a)

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 =    90      
z = the critical z score =    0.841621234      
s = standard deviation =    10      
          
Then          
          
x = critical value =    98.41621234   [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.3      
          
Then, using table or technology,          
          
z =    -0.524400513      
          
As x = u + z * s,          
          
where          
          
u = mean =    90      
z = the critical z score =    -0.524400513      
s = standard deviation =    10      
          
Then          
          
x = critical value =    84.75599487   [ANSWER]

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c)

Note that              
              
Lower Bound = X - z(alpha/2) * s              
Upper Bound = X + z(alpha/2) * s      
              
where              
alpha/2 = (1 - confidence level)/2 =    0.075          
X = sample mean =    90          
z(alpha/2) = critical z for the confidence interval =    1.439531471          
s = sample standard deviation =    10          
              
Thus,              
              
Lower bound =    75.60468529          
Upper bound =    104.3953147          
              
Thus, the confidence interval is              
              
(   75.60468529   ,   104.3953147   ) [ANSWER]