Question

Assuming a normal distribution, families spends an average of $500 on their monthly cell phone bills with a standard deviation of $75. what is the probability a random families' bill is between $350 and $400?

Answer 1

The probability is 0.69.

We find the z-scores that correspond with each end of the range we're looking for.  To find a z-score, we use the formula:
$z=\frac{X-\mu}{\sigma} \\ \\=\frac{350-500}{75}=\frac{-150}{75}=-2$

$z=\frac{400-500}{75}=\frac{-100}{75}=-1.33$

Using a z-table (http://www.z-table.com) we see that the area under the curve left of the score 350 would be 0.228.  The area under the curve left of the score 400 would be 0.918.  To find just the range between 350 and 400, we subtract the area left of 350 from the area left of 400; this will give us just from 350 to 400:

0.918-0.228 = 0.69.