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# Standard Normal Distribution for Beginning Statistics

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By — McGraw-Hill Professional
Updated on Aug 12, 2011

Practice problems for these concepts can be found at:

Continuous Random Variables and Their Probability Distributions Solved Problems for Beginning Statistics

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:

Continuous Random Variables and Their Probability Distributions Solved Problems for Beginning Statistics

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