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

Answer:

a) $P(X>100)=P(\frac{X-\mu}{\sigma}>\frac{100-\mu}{\sigma})=P(Z>\frac{100-73}{11})=P(Z>2.45)$

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) $P(60<X<83)=P(\frac{60-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{83-\mu}{\sigma})=P(\frac{60-73}{11}<Z<\frac{83-73}{11})=P(-1.182<z<0.909)$

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) $P(80<X<90)=P(\frac{80-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{90-\mu}{\sigma})=P(\frac{80-73}{11}<Z<\frac{90-73}{11})=P(0.636<z<1.55)$

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) $P(X<55)=P(\frac{X-\mu}{\sigma}<\frac{55-\mu}{\sigma})=P(Z<\frac{55-73}{11})=P(Z<-1.636)$

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

$P(z<-1.636)=0.051$

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

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

$P(X>100)=P(\frac{X-\mu}{\sigma}>\frac{100-\mu}{\sigma})=P(Z>\frac{100-73}{11})=P(Z>2.45)$

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

$P(60<X<83)=P(\frac{60-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{83-\mu}{\sigma})=P(\frac{60-73}{11}<Z<\frac{83-73}{11})=P(-1.182<z<0.909)$

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

$P(80<X<90)=P(\frac{80-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{90-\mu}{\sigma})=P(\frac{80-73}{11}<Z<\frac{90-73}{11})=P(0.636<z<1.55)$

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

$P(X<55)=P(\frac{X-\mu}{\sigma}<\frac{55-\mu}{\sigma})=P(Z<\frac{55-73}{11})=P(Z<-1.636)$

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

$P(z<-1.636)=0.051$