Question 363691
A set of data is normally distributed with a mean of 200 and a standard deviation of 50. 
What percentage of the scores is between 200 and 167?
z(200) = 0
z(167) = (167-200)/50 = -0.66
P(167< x < 200) = P(-0.66 < z < 0) = 0.2454
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What would be the percentile rank for a score of 176? 
z(176) = (176-200)/50 = -0.48
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The percentile rank of a score is the proportion of 
population below the score.
%ile of 176 = P(z< -0.48) = 0.3156 = 31.56%ile
Rounding the answer would be 32%ile
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Cheers,
Stan H.