Copyright © 2009 Pearson Education, Inc. 5.2 Properties of the Normal Distribution LEARNING GOAL Know how to interpret the normal distribution in terms of the rule, standard scores, and percentiles.

Slide Copyright © 2009 Pearson Education, Inc. Consider a Consumer Reports survey in which participants were asked how long they owned their last TV set before they replaced it. The variable of interest in this survey is replacement time for television sets. Based on the survey, the distribution of replacement times has a mean of about 8.2 years, which we denote as  (the Greek letter mu). The standard deviation of the distribution is about 1.1 years, which we denote as  (the Greek letter sigma).

Slide Copyright © 2009 Pearson Education, Inc. Making the reasonable assumption that the distribution of TV replacement times is approximately normal, we can picture it as shown in Figure Figure 5.16 Normal distribution for replacement times for TV sets with a mean of m 8.2 years and a standard deviation of s 1.1 years.

Slide Copyright © 2009 Pearson Education, Inc. TIME OUT TO THINK Apply the four criteria for a normal distribution (see Section 5.1) to explain why the distribution of TV replacement times should be approximately normal.

Slide Copyright © 2009 Pearson Education, Inc. A simple rule, called the rule, gives precise guidelines for the percentage of data values that lie within 1, 2, and 3 standard deviations of the mean for any normal distribution. Figure 5.17 Normal distribution illustrating the rule.

Slide Copyright © 2009 Pearson Education, Inc. About 68% (more precisely, 68.3%), or just over two- thirds, of the data points fall within 1 standard deviation of the mean. About 95% (more precisely, 95.4%) of the data points fall within 2 standard deviations of the mean. About 99.7% of the data points fall within 3 standard deviations of the mean. The Rule for a Normal Distribution

Slide Copyright © 2009 Pearson Education, Inc. The tests that make up the verbal (critical reading) and mathematics SAT (and the GRE, LSAT, and GMAT) are designed so that their scores are normally distributed with a mean of  = 500 and a standard deviation of  = 100. Interpret this statement. EXAMPLE 1 SAT Scores Solution: From the rule, about 68% of students have scores within 1 standard deviation (100 points) of the mean of 500 points; that is, about 68% of students score between 400 and 600. About 95% of students score within 2 standard deviations (200 points) of the mean, or between 300 and 700. And about 99.7% of students score within 3 standard deviations (300 points) of the mean, or between 200 and 800.

Slide Copyright © 2009 Pearson Education, Inc. Solution: (cont.) Figure 5.18 shows this interpretation graphically; note that the horizontal axis shows both actual scores and distance from the mean in standard deviations. EXAMPLE 1 SAT Scores Figure 5.18 Normal distribution for SAT scores, showing the percentages associated with 1, 2, and 3 standard deviations.

Slide Copyright © 2009 Pearson Education, Inc. Vending machines can be adjusted to reject coins above and below certain weights. The weights of legal U.S. quarters have a normal distribution with a mean of 5.67 grams and a standard deviation of gram. If a vending machine is adjusted to reject quarters that weigh more than 5.81 grams and less than 5.53 grams, what percentage of legal quarters will be rejected by the machine? Solution: A weight of 5.81 is 0.14 gram, or 2 standard deviations, above the mean. A weight of 5.53 is 0.14 gram, or 2 standard deviations, below the mean. Therefore, by accepting only quarters within the weight range 5.53 to 5.81 grams, the machine accepts quarters that are within 2 standard deviations of the mean and rejects those that are more than 2 standard deviations from the mean. By the rule, 95% of legal quarters will be accepted and 5% of legal quarters will be rejected. EXAMPLE 2 Detecting Counterfeits

Slide Copyright © 2009 Pearson Education, Inc. Applying the Rule We can apply the rule to determine when data values lie 1, 2, or 3 standard deviations from the mean. For example, suppose that 1,000 students take an exam and the scores are normally distributed with a mean of  = 75 and a standard deviation of  = 7.

Slide Copyright © 2009 Pearson Education, Inc. Figure 5.19 A normal distribution of test scores with a mean of 75 and a standard deviation of 7. (a) 68% of the scores lie within 1 standard deviation of the mean. (b) 95% of the scores lie within 2 standard deviations of the mean.

Slide Copyright © 2009 Pearson Education, Inc. Identifying Unusual Results In statistics, we often need to distinguish values that are typical, or “usual,” from values that are “unusual.” By applying the rule, we find that about 95% of all values from a normal distribution lie within 2 standard deviations of the mean. This implies that, among all values, 5% lie more than 2 standard deviations away from the mean. We can use this property to identify values that are relatively “unusual”: Unusual values are values that are more than 2 standard deviations away from the mean.

Slide Copyright © 2009 Pearson Education, Inc. You measure your resting heart rate at noon every day for a year and record the data. You discover that the data have a normal distribution with a mean of 66 and a standard deviation of 4. On how many days was your heart rate below 58 beats per minute? Solution: A heart rate of 58 is 8 (or 2 standard deviations) below the mean. According to the rule, about 95% of the data points are within 2 standard deviations of the mean. Therefore, 2.5% of the data points are more than 2 standard deviations below the mean, and 2.5% of the data points are more than 2 standard deviations above the mean. On 2.5% of 365 days, or about 9 days, your measured heart rate was below 58 beats per minute. EXAMPLE 4 Normal Heart Rate

Slide Copyright © 2009 Pearson Education, Inc. TIME OUT TO THINK As Example 4 suggests, many measurements of the resting heart rate of a single individual are normally distributed. Would you expect the average resting heart rates of many individuals to be normally distributed? Which distribution would you expect to have the larger standard deviation? Why?

Slide Copyright © 2009 Pearson Education, Inc. On a visit to the doctor’s office, your fourth-grade daughter is told that her height is 1 standard deviation above the mean for her age and sex. What is her percentile for height? Assume that heights of fourth-grade girls are normally distributed. Solution: Recall that a data value lies in the nth percentile of a distribution if n% of the data values are less than or equal to it (see Section 4.3). According to the rule, 68% of the heights are within 1 standard deviation of the mean. Therefore, 34% of the heights (half of 68%) are between 0 and 1 standard deviation above the mean. We also know that, because the distribution is symmetric, 50% of all heights are below the mean. Therefore, 50% + 34% = 84% of all heights are less than 1 standard deviation above the mean (Figure 5.21). Your daughter is in the 84th percentile for heights among fourth-grade girls. EXAMPLE 5 Finding a Percentile

Slide Copyright © 2009 Pearson Education, Inc. Figure 5.21 Normal distribution curve showing 84% of scores less than 1 standard deviation above the mean.

Slide Copyright © 2009 Pearson Education, Inc. Standard Scores Computing Standard Scores The number of standard deviations a data value lies above or below the mean is called its standard score (or z-score), defined by z = standard score = The standard score is positive for data values above the mean and negative for data values below the mean. data value – mean standard deviation

Slide Copyright © 2009 Pearson Education, Inc. The Stanford-Binet IQ test is scaled so that scores have a mean of 100 and a standard deviation of 16. Find the standard scores for IQs of 85, 100, and 125. Solution: We calculate the standard scores for these IQs by using the standard score formula with a mean of 100 and standard deviation of 16. standard score for 125: z = = standard score for 100: z = = 0.00 EXAMPLE 6 Finding Standard Scores 85 – –

Slide Copyright © 2009 Pearson Education, Inc. Solution: (cont.) standard score for 125: z = = 1.56 We can interpret these standard scores as follows: 85 is 0.94 standard deviation below the mean, 100 is equal to the mean, and 125 is 1.56 standard deviations above the mean. EXAMPLE 6 Finding Standard Scores 125 –

Slide Copyright © 2009 Pearson Education, Inc. Figure 5.22 shows the values on the distribution of IQ scores from Example 6. Figure 5.22 Standard scores for IQ scores of 85, 100, and 125.

Slide Copyright © 2009 Pearson Education, Inc. Standard Scores and Percentiles Once we know the standard score of a data value, the properties of the normal distribution allow us to find its percentile in the distribution. This is usually done with a standard score table, such as Table 5.1 (next slide). (Appendix A has a more detailed standard score table.)

Slide Copyright © 2009 Pearson Education, Inc.

Slide Copyright © 2009 Pearson Education, Inc. Cholesterol levels in men 18 to 24 years of age are normally distributed with a mean of 178 and a standard deviation of 41. a. What is the percentile for a 20-year-old man with a cholesterol level of 190? EXAMPLE 7 Cholesterol Levels Solution: a.The standard score for a cholesterol level of 190 is z = standard score = = ≈ 0.29 Table 5.1 shows that a standard score of 0.29 corresponds to about the 61st percentile. 190 – data value – mean standard deviation

Solution: b. Table 5.1 shows that 90.32% of all data values have a standard score less than 1.3. Thus, the 90th percentile is about 1.3 standard deviations above the mean. Given the mean cholesterol level of 178 and the standard deviation of 41, a cholesterol level 1.3 standard deviations above the mean is A cholesterol level of about 231 corresponds to the 90th percentile. Slide Copyright © 2009 Pearson Education, Inc. Cholesterol levels in men 18 to 24 years of age are normally distributed with a mean of 178 and a standard deviation of 41. b. What cholesterol level corresponds to the 90th percentile, the level at which treatment may be necessary? EXAMPLE 7 Cholesterol Levels

Slide Copyright © 2009 Pearson Education, Inc. Toward Probability Suppose you pick a baby at random and ask whether the baby was born more than 15 days prior to his or her due date. Because births are normally distributed around the due date with a standard deviation of 15 days, we know that 16% of all births occur more than 15 days prior to the due date (see Example 3). For an individual baby chosen at random, we can therefore say that there’s a 0.16 chance (about 1 in 6) that the baby was born more than 15 days early. In other words, the properties of the normal distribution allow us to make a probability statement about an individual. In this case, our statement is that the probability of a birth occurring more than 15 days early is This example shows that the properties of the normal distribution can be restated in terms of ideas of probability.

Slide Copyright © 2009 Pearson Education, Inc. The End