The serum cholesterol levels for men in one age group are normally distributed with a mean of 178.2 and a standard deviation of 40.8. All units are in mg/100 mL. Find the 9th percentile.
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Standard Value = Z = (X-178.2)/40.8
Percentile rank = 100 - (value of Z*100)= 9
value of Z = 0.91
Value of Z = 0.91-0.5 = 0.41
So standard Value of Z = 1.34
so 9 th percentile score value is = (1.34*40.8)+178.2 = 232.872
Percentile rank = 100 - (value of Z*100)= 9
value of Z = 0.91
Value of Z = 0.91-0.5 = 0.41
So standard Value of Z = 1.34
so 9 th percentile score value is = (1.34*40.8)+178.2 = 232.872