Suppose you have a normal random variable x with mu=50 and s=15. Find the probability that x will fall within the interval 30

Suppose you have a normal random variable x with mu=50
and s=15.  Find the probability that x will fall within the interval
30<x<60.

The mean is given as 50 and the
standard deviation as 15. We convert the raw data to z scores using
`z=(a-mu)/sigma` where a is the data point, `mu` is the mean, and
`sigma` the standard deviation.

Thus 30 converts to
z score of -4/3, and 60 converts to
2/3.

We want to know the area under the standard normal
curve between these two z values, as the area is the probability we
are seeking.

Using a TI-83 I got the area to be .6563.
Using the z-table found in most statistics books and finding the values closest to -4/3
and 2/3 I got the area to the left of -4/3 to be .0918, and the area to the left of 2/3
to be .7486. Taking .7486-.0918 gives .6568.

So the
probability that a given x will fall in the range is approximately .6563 (The calculator
computes more than the standard 4 digits in a table, so I use this
value.)