Practice problems for these concepts can be found at:

The *standard normal distribution* is the normal distribution having mean equal to 0 and standard deviation equal to 1. The letter Z is used to represent the standard normal random variable. The standard normal curve is shown in Fig. 6-8. The curve is centered at the mean, 0, and the z-axis is labeled in standard deviations above and below the mean.

**EXAMPLE 6.3** Table 6.2 illustrates how to use the standard normal distribution table to find the area under the standard normal curve between z = 0 and z = 1.65.

Figure 6-9 shows the corresponding area as the shaded region under the curve. The value 1.65 may be written as 1.6 + .05, and by locating 1.6 under the column labeled z and then moving to the right of 1.6 until you come under the .05 column you find the area .4505. This is the area shown in Fig. 6-9. We express this area as P(0 < Z < 1.65) = .4505.

**EXAMPLE 6.4** The area under the standard normal curve that lies between z = –1.65 and z = 0 is represented as P(–1.65 < Z < 0) and is shown in Fig. 6-10. By symmetry, the following probabilities are equal.

P(–1.65 < Z < 0) = P(0 < Z < 1.65)

From Example 6.3, we know that P(0 < Z < 1.65) = .4505 and therefore, P(–1.65 < Z < 0) = .4505.

EXAMPLE 6.5 The area under the standard normal curve between z = –1.65 and z = 1.65 is represented as P(–1.65 < Z < 1.65) and is shown in Fig. 6-11. The probability P(–1.65 < Z < 1.65) is expressible as

P(–1.65 < Z < 1.65) = P(–1.65 < Z < 0) + P(0 < Z < 1.65)

The probabilities on the right side of the above equation are given in Examples 6.3 and 6.4, and their sum is equal to 0.9010. Therefore, P(–1.65 < Z < 1.65) = .9010.

EXAMPLE 6.6 The probability of the event Z < 1.96 is represented by P(Z < 1.96) and is shown in Fig. 6-12.

The area shown in Fig. 6-12 is partitioned into two parts as shown in Fig. 6-13. The darker of the two areas is equal to P(Z < 0) = .5, since it is one-half of the total area. The lighter of the two areas is found in the standard normal distribution table to be .4750. The sum of the two areas is .5 + .4750 = .9750. Summarizing,

P(Z < 1.96) = P(Z < 0) + P(0 < Z < 1.96) = .5 + .4750 = .9750

The probability in Example 6.6 can be found by giving the MINITAB pull-down **Calc ⇒ Probability Distributions ⇒ Normal**. In the dialog box check cumulative distribution, mean = 0, standard deviation = 1, and input constant equal to 1.96. The following output is produced.

Cumulative Distribution Function

Normal with mean = 0 and standard deviation = 1

x P(X <= x)

1.96 0.975002

In order to find the probability P(Z > 1.96) shown in Fig. 6-14, we use the complement of the event Z > 1.96 which is Z < 1.96.

P(Z > 1.96) = 1 – P(Z < 1.96) = 1 – 0.9750 = 0.0250

Practice problems for these concepts can be found at:

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