Given a population of 5000 scores with a mean of 86 and a standard deviation of 10, how many scores are above 116?

Answer 1

first compute probability of being above 116:

for x = score, z-score is

z = (x - mean)/standard deviation.

compute z score for mean 86, standard deviation 86 and score of 110:

z = (116 - 86)/10 = 3

compute probability of z-score exceeding 3:

p(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013

use probability to determine number of scores:

0.0013 * 5000 = 6.5, so 6.