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:

$P(29$

And we can find this probability with this difference:

$P(-3$

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

$P(-3$

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 %

$P(29$

The z score is given by:

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

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

$P(29$

And we can find this probability with this difference:

$P(-3$

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

$P(-3$

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