We know that
z score =(X-mean)/standard deviation =(100-86)/12 =1.17
Part a.
Corresponding score =mean+z*standard deviation =100+1.17*15 =117.55
Part b.
Corresponding score =mean+z*standard deviation =500+1.17*100 =617
Part c.
since z score of 1.17 corresponds to 88th percentile (Using Excel formula ''=NORMSDIST(1.17)''), this tells us that her aptitude is better than 88% of her peers.
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