Finding Probabilities from z-scores

When we worked with the uniform distribution, we could obtain probabilities easily since they corresponded to rectangular regions. Since the curve corresponding to the standard normal distribution is difficult to work with, tables and technology have been developed to do this. In the examples we will proceed as if we are using tables and use computed generated results to check our work.

We will be using the standard normal distribution through out this section. Before we do let's summarize some properties that we will be using.

- Mean is zero.

- Standard deviation is one.

- Total area under the curve is 1.

Because of symmetry we have

- Area to right of mean is 0.5.

- Area to left of mean is 0.5.

When doing problems like these, it is a good idea to make a rough sketch of the curve and mark off the region you need to work with.

Example 1

Find the area between z = 0 and z = 1.

Solution .

The table in the text is set up to tell us the area under the curve from z = 0 to z = some positive value. Using the table (A-2) in the text, we get .3413. This is pictured below.

Note that each unit on the z-scale corresponds to the lenght of one standard deviation which is one for the standard normal distribution.

Example 2

Find the area between z = -1 and z = 0.

Solution .

You'll note that the table does not give negative values for z. However, if we note the symmetry of the standard normal, we can just use the area from z = 0 to z = 1. The answer again is .3413.

Remember that these areas correspond to probabilities. As a result, we can summarize the last two examples by saying that the probability that z is between 0 and 1 is .3413 and that the probability that z is between -1 and 0 is .3413. Notation for this is P(0<z<1) = .4313 and P(-1<z<0) =.3413.

Example 3

Find P(-1<z<2).

Solution .

We do this by adding the areas of two regions together.

P(-1<z<2) = P(-1<z<0) + P(0<z<2) = .3413 + .4772 = .8185

When you compare the caption on the computer generated figure above with the answer we have just obtained, you'll note a slight difference in the results. This is because we have added entries in a table that have been rounded to four places. On the other hand the software used to generate the graph and caption on it has worked with ten digit values and then rounded off. From time to time you may note other slight differences that are due to rounding.

Example 4

Find the probability that z is between 1.1 and 2.5.

Solution .

Since these z values have the same sign, we will need to find the area from 0 to 2.5 and subtract the area from 0 to 1.1.

P(1.1<z<2.5) = P(0<z<2.5) - P(0<z<1.1) = .4938 - .3643 = .1295

Example 5

Find P(-2.1<z<-1.54).

Solution .

P(-2.1<z<-1.54) = P(-2.1<z<0) - P(-1.54<z<0) = .4821 - .4382 = .0439

(Note that to find P(-1.54<z<0) you need to go down the z column in the table until you reach 1.5 and then look in the column labeled .04 to find the .4382)

Note that the result is given in the caption on the picture in scientific notation. It agrees with what we got using the table, since .4392e-1 means .4392 times ten to the negative one which is the same as .04392

Example 6

Find the area to the right of z = 2.

Solution .

We need to find P(z>2). The table tells us that the area from 0 to 2 is .4772. If we subtract this from .5 (which is the area from 0 to infinity), we'll get the area we need. Thus,

P(z>2) = .5 - P(0<z<2) = .5 - .4772 = .0228.

Example 7

Find the area to the left of z = 2.

Solution .

P(z<2) = P(z<0) + P(0<z<2) = .5 + .4772 = .9772

Example 8

Find the area to the right of z = -1.

Solution .

P(z>-1) = P(-1<z<0) + P(z>0) = .3413 + .5 = .8413

Example 9

Find P(z<-1).

Solution .

This is the area to the left of z = -1.

P(z<-1) = .5 - P(-1<z<0) = .5 - .3413 = .1587