7 Probability intervals - standard normal distribution

We use probability models to make predictions in situations where there is not sufficient data available to make a definite statement. Any statement based on these models carries with it a risk of being proved incorrect by events. Notice that the normal probability curve extends to infinity in both directions. Theoretically any value of the normal random variable is possible, although, of course, values far from the mean position (zero) are very unlikely.

Consider the diagram in Figure 13:

Figure 13

No alt text was set. Please request alt text from the person who provided you with this resource.

The shaded area is 95 % of the total area. If we look at the entry in Table 1 (at the end of the Workbook) corresponding to Z = 1.96 we see the value 4750. This means that the probability of Z taking a value between 0 and 1.96 is 0.475. By symmetry, the probability that Z takes a value between − 1.96 and 0 is also 0.475. Combining these results we see that

P ( − 1.96 < Z < 1.96 ) = 0.95 or 95 %

We say that the 95 % probability interval for Z (about its mean of 0) is ( − 1.96 , 1.96 ) . It follows that there is a 5 % chance that Z lies outside this interval.

Task!

Find the 99 % probability interval for Z about its mean, i.e. the value of z 1 in the diagram:

No alt text was set. Please request alt text from the person who provided you with this resource.

The shaded area is 99 % of the total area

First, note that 99 % corresponds to a probability of 0.99. Find z 1 such that

P ( 0 < Z < z 1 ) = 1 2 × 0.99 = 0.495 :

We look for a table value of 4950. The nearest we get is 4949 and 4951 corresponding to Z = 2.57 and Z = 2.58 respectively. We choose Z = 2 . 58 .

Now quote the 99 % probability interval:

( − 2.58 , 2.58 ) or − 2.58 < Z < 2.58 .

Notice that the risk of Z lying outside this wider interval is reduced to 1 % .

Task!

Find the value of Z

  1. which is exceeded on 5 % of occasions
  2. which is exceeded on 99 % of occasions.
  1. The value is z 1 , where P ( Z > z 1 ) = 0.05 . Hence P ( 0 < Z < z 1 ) = 0.5 − 0.05 = 0.45 This corresponds to a table entry of 4500. The nearest values are 4495 ( Z = 1.64 ) and 4505 ( Z = 1.65 ).

    Hence the required value is Z 1 = 1.65 .

  2. Values less than z 1 occur on 1 % of occasions. By symmetry values greater than ( − z 1 ) occur on 1 % of occasions so that P ( 0 < z < − z 1 ) = 0.49 . The nearest table corresponding to 4900 is 4901 ( Z = 2.33 ) .

    Hence the required value is z 1 = − 2.33 .