Question

The amounts of electricity bills for all households in a city have a normal probability distribution with a mean of $140 and a standard deviation of $30. Find the probability that the mean amount of electric bills for a random sample of 75 households selected from this city will be:

a. between $132 and $136
b. within $6 of the population mean
c. more than the population mean by at least $4

Answer 1

Answer)

As the population is normally distributed, we can use standard normal z table to estimate the probability here.

Given mean = 140

S.d = 30

N = sample size = 75

A)

P(132<x<136) = p(x<136) - p(x<132)

Z = (x-mean)/(s.d/√n)

P(x<136)

Z = (136-140)/(30/√75)

Z = -1.15

From z table, p(z<-1.15) = 0.1251

P(x<132)

Z = (132-140)/(30/√75)

Z = -2.31

From z table, p(z<-2.31) = 0.0104

Required probability is 0.1251 - 0.0104

= 0.1147

B)

Within $6 means

P(134<x<146)

P(x<146)

Z = (146-140)/(30/√75)

Z = 1.73

From z table, p(z<1.73) = 0.9582

P(x<134)

Z = -1.73

P(z<-1.73) = 0.0418

Required probability is 0.9582 - 0.0418

= 0.9164

C)

P(x>144)

Z = (144-140)/(30/√75)

Z = 1.15

P(z>1.15)

= 0.1251