Question

6.1 Section Exercise 66 (calc) Question 3 of 6 (1 point) View problem in a pop-up Electricity bills: According to the U.S. Energy Information Administration, the mean monthly household electricity bill in the United States in 2011 was $109.54. Assume the amounts are normally distributed with standard deviation S25.00. Round the answers to four decimal places Part 1 out of 3 What proportion of bills are greater than $132? The proportion of bills that are greater than S132 is

Answer 1

Let X be Monthly household electricity bill.

Given,

X ~ N($\mu$ = 109.54, $\sigma$ = 25)

Proportion of bills that are greater than $132 = P(X > 132)

= P[Z > (132 - 109.54) / 25]

= P[Z > 0.8984]

= 0.1845 (By Z distribution table)