Question

The lifetime of a certain type of TV tube has a normal distribution with a mean of 61 and a standard deviation of 6 months. What portion of the tubes lasts between 57 and 59 months?

Answer 1

`P(57<X<59)=P((57-\mu)/(\sigma)<(X-\mu)/(\sigma)<(59-\mu)/(\sigma))=P((57-61)/(6)<Z<(59-61)/(6))=P(-0.67<Z<-0.33)`

`P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)`

`P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)=0.371-0.251=0.119`

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

`X \sim N(61,6)`

Where
`\mu=61` and
`\sigma=6`

We are interested on this probability

`P(57<X<59)`

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(57<X<59)=P((57-\mu)/(\sigma)<(X-\mu)/(\sigma)<(59-\mu)/(\sigma))=P((57-61)/(6)<Z<(59-61)/(6))=P(-0.67<Z<-0.33)`

And we can find this probability like this:

`P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)`

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

`P(-0.67<z<-0.33)=P(z<-0.33)-P(z<-0.67)=0.371-0.251=0.119`