Question

The time a song plays on the radio varies from song to song. The time songs play varies according to a normal distribution with mean 3.2 minutes and standard deviation 1.32. What is the probability that a randomly selected song will play longer than 4.5 minutes?

Answer 1

`P(X>4.5)=P((X-\mu)/(\sigma)>(4.5-\mu)/(\sigma))=P(Z>(4.5-3.2)/(1.32))=P(z>0.985)`

And we can find this probability using the complement rule and we got:

`P(z>0.985)=1-P(z<0.985)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(z>0.985)=1-P(z<0.985)=1-0.838=0.162`

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".

Solution to the problem

Let X the random variable that represent the time a song plays of a population, and for this case we know the distribution for X is given by:

`X \sim N(3.2,1.32)`

Where
`\mu=3.2` and
`\sigma=1.32`

We are interested on this probability

`P(X>4.5)`

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>4.5)=P((X-\mu)/(\sigma)>(4.5-\mu)/(\sigma))=P(Z>(4.5-3.2)/(1.32))=P(z>0.985)`

And we can find this probability using the complement rule and we got:

`P(z>0.985)=1-P(z<0.985)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(z>0.985)=1-P(z<0.985)=1-0.838=0.162`