Question

The amount of Jen’s monthly phone bill is normally
distributed with a mean of $55 and a standard deviation of $12. What percentage of her phone bills are between $19 and $91?

Answer 1

Answer:

$P(19<X<91)=P(\frac{19-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{91-\mu}{\sigma})=P(\frac{19-55}{12}<Z<\frac{91-55}{12})=P(-3<z<2)$

And we can find this probability with this difference and using the normal standard distribution

$P(-3<z<3)=P(z<3)-P(z<-3)=0.9987 -0.00135 =0.9974$

Step-by-step explanation:

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

$X \sim N(55,12)$  

Where $\mu=55$ and $\sigma=12$

We are interested on this probability

$P(19<X<91)$

And we can use the z score formula given by:

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

Replacing the info we got:

$P(19<X<91)=P(\frac{19-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{91-\mu}{\sigma})=P(\frac{19-55}{12}<Z<\frac{91-55}{12})=P(-3<z<2)$

And we can find this probability with this difference and using the normal standard distribution

$P(-3<z<3)=P(z<3)-P(z<-3)=0.9987 -0.00135 =0.9974$