Slide Copyright © 2008 Pearson Education, Inc. Chapter 6 The Normal Distribution

Slide Copyright © 2008 Pearson Education, Inc. Definition 6.1

Slide Copyright © 2008 Pearson Education, Inc. Figure 6.2

Slide Copyright © 2008 Pearson Education, Inc. Example 6.1 A midwestern college has an enrollment of 3264 female students. Records show that the mean height of these students is 64.4 inches and that the standard deviation is 2.4 inches. Here the variable is height, and the population consists of the 3264 female students attending the college. Frequency and relative-frequency distributions for these heights appear in Table 6.1 on the next slide. The table shows, for instance, that 7.35% (0.0735) of the students are between 67 and 68 inches tall.

Slide Copyright © 2008 Pearson Education, Inc. Table 6.1 a.Show that the variable “height” is approximately normally distributed for this population. b.Identify the normal curve associated with the variable “height” for this population. c.Discuss the relationship between the percentage of female students whose heights lie within a specified range and the corresponding area under the associated normal curve.

Slide Copyright © 2008 Pearson Education, Inc. Figure 6.4 Solution Example 6.1 a. The relative- frequency histogram shows that the distribution of heights has roughly the shape of a normal curve and, so “height” is approximately normally distributed for this population.

Slide Copyright © 2008 Pearson Education, Inc. Solution Example 6.1 b. The associated normal curve is the one whose parameters are the same as the mean and standard deviation of the variable, which are 64.4 and 2.4, respectively. Thus the required normal curve has parameters μ = 64.4 and σ = 2.4. It is superimposed on the histogram in Fig c. Consider, for instance, the students who are between 67 and 68 inches tall. According to Table 6.1, their exact percentage is 7.35%, or Note that also equals the area of the cross-hatched bar in Fig. 6.4 because the bar has height and width 1. And the area under the curve between 67 and 68, shaded in Fig. 6.4 approximates the area of the cross-hatched bar. Thus we can approximate the percentage of students between 67 and 68 inches tall by the area under the normal curve between 67 and 68. This result holds in general.

Slide Copyright © 2008 Pearson Education, Inc. Key Fact 6.1

Slide Copyright © 2008 Pearson Education, Inc. Definition 6.2

Slide Copyright © 2008 Pearson Education, Inc. Key Fact 6.2

Slide Copyright © 2008 Pearson Education, Inc. Figure 6.6 We can interpret Key Fact 6.2 in several ways. Theoretically, it says that standardizing converts all normal distributions to the standard normal distribution.

Slide Copyright © 2008 Pearson Education, Inc. Figure 6.7 Let x be a normally distributed variable with mean μ and standard deviation σ, and let a and b be real numbers with a < b. The percentage of all possible observations of x that lie between a and b is the same as the percentage of all possible observations of z that lie between (a −μ)/σ and (b−μ)/σ. This latter percentage equals the area under the standard normal curve between (a −μ)/σ and (b−μ)/σ.

Slide Copyright © 2008 Pearson Education, Inc. Key Fact 6.3

Slide Copyright © 2008 Pearson Education, Inc. Figure 6.12 Using Table II to find the area under the standard normal curve that lies (a) to the left of a specified z- score, (b) to the right of a specified z-score, and(c) between two specified z-scores

Slide Copyright © 2008 Pearson Education, Inc. Example 6.7 Use Table II (a portion of which is shown on the next slide) to find a.Z b.Z 0.05

Slide Copyright © 2008 Pearson Education, Inc. Table 6.2

Slide Copyright © 2008 Pearson Education, Inc. Figure 6.15 Figure 6.16 Solution Example 6.7 b. a.

Slide Copyright © 2008 Pearson Education, Inc. Procedure 6.1

Slide Copyright © 2008 Pearson Education, Inc. Example 6.9 Intelligence quotients (IQs) measured on the Stanford Revision of the Binet-Simon Intelligence Scale are normally distributed with a mean of 100 and a standard deviation of 16. Determine the percentage of people who have IQs between 115 and 140. Solution Step 1 Sketch the normal curve associated with the variable. Here μ =100 and σ = 16. Step 2 Shade the region of interest and mark its delimiting x-values.

Slide Copyright © 2008 Pearson Education, Inc. Figure 6.19 Solution Step 3 Compute z-scores: for x = 115, z = 0.94 and for x = 140, z = 2.50 Step 4 The area under the standard normal curve that lies between 0.94 and The area to the left of 0.94 is , and the area to the left of 2.50 is The required area, shaded in Fig.6.19, is therefore − = % of all people have IQs between 115 and 140.

Slide Copyright © 2008 Pearson Education, Inc. Key Fact 6.4 Figure 6.20

Slide Copyright © 2008 Pearson Education, Inc. Procedure 6.2

Slide Copyright © 2008 Pearson Education, Inc. Key Fact 6.5

Slide Copyright © 2008 Pearson Education, Inc. Table 6.3 Example 6.14 The Internal Revenue Service publishes data on federal individual income tax returns in Statistics of Income, Individual Income Tax Returns. A simple random sample of 12 returns from last year revealed the adjusted gross incomes, in thousands of dollars, shown in Table 6.3. Construct a normal probability plot for these data, and use the plot to assess the normality of adjusted gross incomes.

Slide Copyright © 2008 Pearson Education, Inc. Table 6.4 Solution Example 6.14 Here the variable is adjusted gross income, and the population consists of all last year’s federal individual income tax returns. To construct a normal probability plot, we first arrange the data in increasing order and obtain the normal scores from Table III. The ordered data are shown in the first column of Table 6.4; the normal scores, from the n = 12 column of Table III, are shown in the second column of Table 6.4.

Slide Copyright © 2008 Pearson Education, Inc. Figure 6.23 Next, we plot the points in Table 6.4, using the horizontal axis for the adjusted gross incomes and the vertical axis for the normal scores. This graph (Fig. 6.23) is the normal probability plot for the sample of adjusted gross incomes. Note that the normal probability plot is curved, not linear. Solution Example 6.14