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?

Respuesta :

Answer:

[tex]P(X>4.5)=P(\frac{X-\mu}{\sigma}>\frac{4.5-\mu}{\sigma})=P(Z>\frac{4.5-3.2}{1.32})=P(z>0.985)[/tex]

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

[tex]P(z>0.985)=1-P(z<0.985)[/tex]

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

[tex]P(z>0.985)=1-P(z<0.985)=1-0.838=0.162[/tex]

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".  

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:

[tex]X \sim N(3.2,1.32)[/tex]  

Where [tex]\mu=3.2[/tex] and [tex]\sigma=1.32[/tex]

We are interested on this probability

[tex]P(X>4.5)[/tex]

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X>4.5)=P(\frac{X-\mu}{\sigma}>\frac{4.5-\mu}{\sigma})=P(Z>\frac{4.5-3.2}{1.32})=P(z>0.985)[/tex]

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

[tex]P(z>0.985)=1-P(z<0.985)[/tex]

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

[tex]P(z>0.985)=1-P(z<0.985)=1-0.838=0.162[/tex]