Question

Based on the Normal model N(100, 16) describing IQ scores from Exercise 22, what percent of applicants would you expect to have scores

a) over 80?

b) under 90?

c) between 112 and 132?

d) over 125?

Answer 1

solution:

the given informatio aas follows:

X follows the normal distribution with mean = = 100

standard dev iation = = 16

a)

percentage of applicant have score over 80 = P(X > 80)

corresponding z score =

P(X > 80) = 1 - value of z to the left of -1.25 = 1 - 0.1056 = 0.8944

percentage of applicant have score over 80 = 89.44%

b)

percentage of applicant have score under 90 = P(X < 90)

corresponding z score =

P(X < 90) = value of z to the left of -0.63 = 0.2643

percentage of applicant have score under 90 = 26.43%

c)

between 112 and 132 = P(112 < X < 132)

for X = 112,

for X = 132 ,.

P(112 < X < 132) = (value of z to the left of 2) - (value of z to the left of -0.75)

P(112 < X < 132) = 0.9772 - 0.7734 = 0.2038

percentage of applicant have score between 112 and 132 = 20.38%

d)

over 125 = P(X > 125)

P(X > 125) = 1 - value of z to the left of 1.56 = 1 - 0.9406 = 0.0594

percentage of applicant have score over 125 = 5.94%