Question

WNAE, an all-news AM station, finds that the distribution of the lengths of time listeners are tuned to the station follows the normal distribution. The mean of the distribution is 15.0 minutes and the standard deviation is 3.5 minutes.
What is the probability that a particular listener will tune in:
a. More than 20 minutes?
b. For 20 minutes or less?
c. Between 10 and 12 minutes?

Answer 1

Answer:

a)  $P(X>20)=P(\frac{X-\mu}{\sigma}>\frac{20-\mu}{\sigma})=P(Z>\frac{20-15}{3.5})=P(z>1.43)$

And we can find this probability using the complement rule, the normal standard distribution or excel or a calculator

$P(z>1.43)=1-P(z<1.43)=1-0.9236=0.076$

b) $P(X<20)=P(\frac{X-\mu}{\sigma}<\frac{20-\mu}{\sigma})=P(Z<\frac{20-15}{3.5})=P(z<1.43)$

And we can find this probability using the normal standard distribution or excel or a calculator

$P(z<1.43)=0.9236$

c) $P(<10X<12)=P(\frac{10-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{12-\mu}{\sigma})=P(\frac{10-15}{3.5}<Z<\frac{12-15}{3.5})=P(-1.43<z<-0.86)$

And we can find this probability using the normal standard distribution or excel or a calculator  with the following difference:

$P(-1.43<z<-0.86)=P(z<-0.86) -P(z<-1.43) =0.195-0.076=0.199$

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 lenghts of a population, and for this case we know the distribution for X is given by:

$X \sim N(15,3.5)$  

Where $\mu=15$ and $\sigma=3.5$

We are interested on this probability

$P(X>20)$

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:

$P(X>20)=P(\frac{X-\mu}{\sigma}>\frac{20-\mu}{\sigma})=P(Z>\frac{20-15}{3.5})=P(z>1.43)$

And we can find this probability using the complement rule, the normal standard distribution or excel or a calculator

$P(z>1.43)=1-P(z<1.43)=1-0.9236=0.076$

Part b

$P(X<20)=P(\frac{X-\mu}{\sigma}<\frac{20-\mu}{\sigma})=P(Z<\frac{20-15}{3.5})=P(z<1.43)$

And we can find this probability using the normal standard distribution or excel or a calculator

$P(z<1.43)=0.9236$

Part c

$P(<10X<12)=P(\frac{10-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{12-\mu}{\sigma})=P(\frac{10-15}{3.5}<Z<\frac{12-15}{3.5})=P(-1.43<z<-0.86)$

And we can find this probability using the normal standard distribution or excel or a calculator  with the following difference:

$P(-1.43<z<-0.86)=P(z<-0.86) -P(z<-1.43) =0.195-0.076=0.199$