Question

According to a report by Scarborough Research, the average monthly household cellular phone bill is $73. Suppose local monthly household cell phone bills are normally distributed with a standard deviation of $11.

a. What is the probability that a randomly selected monthly cell phone bill is more than $100?
b. What is the probability that a randomly selected monthly cell phone bill is between $60 and $83?
c. What is the probability that a randomly selected monthly cell phone bill is between $80 and $90?
d. What is the probability that a randomly selected monthly cell phone bill is no more than $55?

Answer 1

a)
`P(X>100)=P((X-\mu)/(\sigma)>(100-\mu)/(\sigma))=P(Z>(100-73)/(11))=P(Z>2.45)`

And we can find this probability using the complement rule:

`P(z>2.45)=1-P(z<2.45)=1-0.993=0.007`

b)
`P(60<X<83)=P((60-\mu)/(\sigma)<(X-\mu)/(\sigma)<(83-\mu)/(\sigma))=P((60-73)/(11)<Z<(83-73)/(11))=P(-1.182<z<0.909)`

And we can find this probability with this difference:

`P(-1.182<z<0.909)=P(z<0.909)-P(z<-1.182)=0.818-0.119=0.699`

c)
`P(80<X<90)=P((80-\mu)/(\sigma)<(X-\mu)/(\sigma)<(90-\mu)/(\sigma))=P((80-73)/(11)<Z<(90-73)/(11))=P(0.636<z<1.55)`

And we can find this probability with this difference:

`P(0.636<z<1.55)=P(z<1.55)-P(z<0.636)=0.939-0.738=0.201`

d)
`P(X<55)=P((X-\mu)/(\sigma)<(55-\mu)/(\sigma))=P(Z<(55-73)/(11))=P(Z<-1.636)`

And we can find this probability using the normal standard table:

`P(z<-1.636)=0.051`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the average monthly household cellular phone bill of a population, and for this case we know the distribution for X is given by:

`X \sim N(73,11)`

Where
`\mu=73` and
`\sigma=11`

We are interested on this probability

`P(X>100)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X>100)=P((X-\mu)/(\sigma)>(100-\mu)/(\sigma))=P(Z>(100-73)/(11))=P(Z>2.45)`

And we can find this probability using the complement rule:

`P(z>2.45)=1-P(z<2.45)=1-0.993=0.007`

Part b

`P(60<X<83)=P((60-\mu)/(\sigma)<(X-\mu)/(\sigma)<(83-\mu)/(\sigma))=P((60-73)/(11)<Z<(83-73)/(11))=P(-1.182<z<0.909)`

And we can find this probability with this difference:

`P(-1.182<z<0.909)=P(z<0.909)-P(z<-1.182)=0.818-0.119=0.699`

Part c

`P(80<X<90)=P((80-\mu)/(\sigma)<(X-\mu)/(\sigma)<(90-\mu)/(\sigma))=P((80-73)/(11)<Z<(90-73)/(11))=P(0.636<z<1.55)`

And we can find this probability with this difference:

`P(0.636<z<1.55)=P(z<1.55)-P(z<0.636)=0.939-0.738=0.201`

Part d

`P(X<55)=P((X-\mu)/(\sigma)<(55-\mu)/(\sigma))=P(Z<(55-73)/(11))=P(Z<-1.636)`

And we can find this probability using the normal standard table:

`P(z<-1.636)=0.051`