Question 629073
Given that a population of scores is normally distributed with the mean equal 110 and standard deviation equal to 8, determine the following:
a)The percentile rank of a score of 120
z(120) = (120-110)/8 = 1.25 
%ile rank of 120 = %ile rank of z = 1.25: 
Ans: normalcdf(-100,1.25) = 0.8944
Round up to 90%ile
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b)the percentile of scores that are below a score of 99.
z(99) = (99-100)/8 = -1/8
P(x < 99) = P(z < -1/8) = normalcdf(-100,-1/8) = 0.4503
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Cheers,
Stan H.