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

Answer:

$P(29<X<89)=P(\frac{29-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{89-\mu}{\sigma})=P(\frac{29-59}{10}<Z<\frac{89-59}{10})=P(-3<z<3)$

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

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

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

$P(29<X<89)=P(\frac{29-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{89-\mu}{\sigma})=P(\frac{29-59}{10}<Z<\frac{89-59}{10})=P(-3<z<3)$

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