Question

A major television manufacturer has determined that its 44 inch screens have a mean service life that can be modeled by a normal distribution with a mean of 6 years and a standard deviation of one-half year (6 months). What is the probability that the service life of that product is between 5 and 7 years

Answer 1

`P(5<X<7)=P((5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(7-\mu)/(\sigma))=P((5-6)/(0.5)<Z<(7-6)/(0.5))=P(-2<z<2)`

And we can find this probability with thie difference:

`P(-2<z<2)=P(z<2)-P(z<-2)`

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

`P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545`

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

`X \sim N(6,0.5)`

Where
`\mu=6` and
`\sigma=0.5`

We are interested on this probability

`P(5<X<7)`

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(5<X<7)=P((5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(7-\mu)/(\sigma))=P((5-6)/(0.5)<Z<(7-6)/(0.5))=P(-2<z<2)`

And we can find this probability with thie difference:

`P(-2<z<2)=P(z<2)-P(z<-2)`

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

`P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545`