A normal population has mean and standard deviation . (a) What proportion of the population is greater than ? (b) What is the probability that a randomly chosen value will be less than .

0.0171

0.89158

Step-by-step explanation:

Given :

μ = 60

Standard deviation , σ = 17

The probability that a randomly chosen score is greater than 96;

P(Z > Zscore)

Zscore = (score, x - μ) / σ

Zscore = (96 - 60) / 17 = 2.118

P(Z > 2.118) = 1 - P(Z < 2.118) = 1 - 0.9829 = 0.0171

The probability that a randomly chosen score is less than 81;

P(Z < Zscore)

Zscore = (score, x - μ) / σ

Zscore = (81 - 60) / 17 = 1.235

P(Z < 1.235) = 0.89158

answer: b

step-by-step explanation: