Question

The amount of Jen's monthly phone bill is normally distributed with a mean of $59 and a standard deviation of $10. What percentage of her phone bills are between $29 and $89?

Answer 1

`P(29<X<89)=P((29-\mu)/(\sigma)<(X-\mu)/(\sigma)<(89-\mu)/(\sigma))=P((29-59)/(10)<Z<(89-59)/(10))=P(-3<z<3)`

And we can find this probability with this difference:

`P(-3<z<3)=P(z<3)-P(z<-3)`

And in order to find these probabilities using the normal standard distribution or excel and we got.

`P(-3<z<3)=P(z<3)-P(z<-3)=0.9987-0.00135=0.99735`

So we expect about 99.735% of values between $29 and $89

Explanation:

Let X the random variable that represent the amount of Jens monthly phone of a population, and for this case we know the distribution for X is given by:

`X \sim N(59,10)`

Where
`\mu=59` and
`\sigma=10`

We are interested on this probability first in order to find a %

`P(29<X<89)`

The z score is given by:

`z=(x-\mu)/(\sigma)`

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

`P(29<X<89)=P((29-\mu)/(\sigma)<(X-\mu)/(\sigma)<(89-\mu)/(\sigma))=P((29-59)/(10)<Z<(89-59)/(10))=P(-3<z<3)`

And we can find this probability with this difference:

`P(-3<z<3)=P(z<3)-P(z<-3)`

And in order to find these probabilities using the normal standard distribution or excel and we got.

`P(-3<z<3)=P(z<3)-P(z<-3)=0.9987-0.00135=0.99735`

So we expect about 99.735% of values between $29 and $89