A statistics teach told four of his students their test scor

A statistics teach told four of his students their test scores in a somewhat jovial manner. He informed them that the distribution of the entire class\'s scores was approximately normal with a mean 72 and a standard deviation of 11, and he gave them the following information about their own achievements. Jack: Your score divides the class. 60% have higher scores, 40 % have the same or a lower score. Jill: Your raw score is 83 John: Your standard score is -0.9 Jane: Only a fourth of the class got higher scores than you. Now find the raw score, standard score, and percentile for all.

Solution

FOR JACK:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.4 = 40TH PERCETNILE [ANSWER]  
          
Then, using table or technology,          
          
z =    -0.253347103   [ANSWER, STANDARD SCORE]  
          
As x = u + z * s,          
          
where          
          
u = mean =    72      
z = the critical z score =    -0.253347103      
s = standard deviation =    11      
          
Then          
          
x = critical value =    69.21318187   [ANSWER, RAW SCORE]

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FOR JILL:

RAW SCORE = 83

z = (x-u)/sigma = (83-72)/11 = 1 [ANSWER, STANDARD SCORE]

By table/technology, the percentile rank is 84.13th PERCENTILE [ANSWER]

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For John:

RAW SCORE = u + z*sigma = 72+(-0.9)*11 = 62.1 [ANSWER]

STANDARD SCORE = -0.9 [ANSWER]

By table/technology, the percentile rank is 18.41th PERCENTILE [ANSWER]

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Jane:

PERCENTILE: 75TH PERCENTILE

By table/technology, the standard score is

STANDARD SCORE = Z = 0.67448975 [ANSWER]

Thus, the raw score is

x = u + z*sigma = 72 + 0.67448975*11 = 79.41938725 [ANSWER]