Converting nonstandard normal into standard normal

The key to handling nonstandard normal problems is converting the problem into one involving the standard normal distribution. Recall the following conversion.

For a population with mean of and standard deviation of the z score or standard score corresponding to x is given by

Example 1

Assume a population has a normal distribution with mean of 23 and standard deviation of 3. Find the area between x = 17 and x = 29.

Solution .

First make a sketch of the given distribution and the region whose area we need to find.

While the caption on the computer generated graph contains the answer to this question, we will go over how to do this problem using the tables.

In order to find P(17<x<29) we must convert to the standard z-scale. To do this convert each endpoint in the original problem to a z score. Using

with x = 17, , and , we get z = (17 - 23)/3 = (-6)/3 = -2.

Repeating the process for the other endpoint of x = 29, we get

z = (29 - 23)/3 = (6)/3 = 2.

The orginal problem of finding P(17<x<29) can now be done by finding P(-2<z<2). Because of symmetry, P(-2<z<2) = 2P(0<z<2) = 2(.4772) = .9544

The graph below shows the corresponding area under the standard normal distribution. Note that, as we've seen before, the computer generated graphs may have captions indicating solutions that are a little different from the values obtained by using the tables.

Example 2

If a normal distribution has a mean of 5 and a standard deviation of 1.6, find P(x>2.6).

Solution .

Here is the graph showing the area we need to compute.

We use the conversion formula

to convert 2.6 to a z score. We get, z = (2.6 - 5)/1.6 = (-2.4)/1.6 = -1.5.

The corresponding region under the standard normal is pictured below.

Thus, P(x>2.6) = P(z>-1.5) = .5 + .4332 = .9332

Example 3

If a normal distribution has a mean of 100 with a standard deviation of 12, find P(103<x<115).

Solution .

Sketching the region we get

We must convert to the z-scale using

As a result, 103 corresponds to z = (103 - 100)/12 = 3/12 = .25

and 115 corresponds to z = (115 - 100)/12 =12 = 15/12 = 1.25.

The corresponding region on the z-scale is pictured below.

P(103<x<115) = P(.25<z<1.25) = P(0<z<1.25) - P(0<z<.25) = .3944 - .0987 = .2957