Question

A normal population has mean u=60 and standard deviation o=17. (a) What proportion of the population is greater than 96? (b) What is the probability that a randomly chosen value will be less than 81. Round answers to four decimal places. Part 1 of 2 The proportion of the population greater than 96 is .....

Answer 1

The proportion of the population greater than 96 is approximately 0.0179.

Step-by-step explanation:

To find the proportion of the population greater than 96, we need to calculate the z-score and then use the standard normal distribution table. The formula for calculating the z-score is z = (x - u) / o, where x is the value we are interested in, u is the mean, and o is the standard deviation. For this question, x = 96, u = 60, and o = 17.

Substituting these values into the formula, we get z = (96 - 60) / 17 = 36 / 17 = 2.12.

Looking up the z-score of 2.12 in the standard normal distribution table, we find that the proportion of the population greater than 96 is approximately 0.0179.