15.4. The Normal Distribution Objectives

15.4 The Normal Distribution Objectives 1. Understand the basic properties of the normal curve. 2. Relate the area under a normal curve to z-scores. 3. Make conversions between raw scores and z-scores. 4. Use the normal distribution to solve applied problems. If the shoe fits, wear it. Well-known proverb In addition to shoes, what about the fit of movie seats, men s ties, and headroom in cars? How do manufacturers decide how wide the seats at your favorite movie theater should be? Or, how long must a tie be so that it is neither too long nor too short for most men? How do we design a car so that most people will not bump their heads on its roof? Or, how high should a computer table be so that it does not put an unnecessary strain on your wrists and

15.4 y The Normal Distribution 751 arms? In the relatively new science of ergonomics, scientists gather data to answer questions such as these and to ensure most people fit comfortably into their environments. Much of the work of these scientists is based on a curve that you will study in this section called the normal curve, or normal distribution.* KEY POINT Many different types of data sets are normal distributions. FIGURE 15.12 A normal distribution. The Normal Distribution The normal distribution is the most common distribution in statistics and describes many real-life data sets. The histogram shown in Figure 15.12 will begin to give you an idea of the shape of a normal distribution. Distributions of such diverse data sets as SAT scores, heights of people, the number of miles before an automobile tire wears out, the size of hamburgers at a fast-food restaurant, and the number of hours of use before a certain brand of DVD player breaks down are all examples of normal distributions. There are several patterns in Figure 15.12 that are common to all normal distributions. First, if we were to take a wire and attach it to the tops of the bars in the histogram and then smooth out the wire to make a curve, the curve would have a bell-shaped appearance, as shown in Figure 15.13. This is why a normal distribution is often called a bell-shaped curve. Second, the, median, and mode of a normal distribution are the same. Third, the curve is symmetric with respect to the. This s that if you see some pattern in the graph on one side of the, then the acts as a mirror to give a reflection of that same pattern on the other side of the. Fourth, the area under a normal curve equals 1. In Figure 15.14, we have marked the and two points on the normal curve called inflection points. An inflection point is a point on the curve where the curve changes from being curved upward to being curved downward, or vice versa. For a normal curve, inflection points are located 1 standard deviation from the. Also, because a normal curve is symmetric with respect to the, one-half, or 50%, of the area under the curve is located on each side of the. In normal distributions, approximately 68% of the data values occur within 1 standard deviation of the, 95% of the data values lie within 2 standard deviations of the, and 99.7% of the data values lie within 3 standard deviations of the. We refer to these facts as the 68-95-99.7 rule. Figure 15.14 illustrates the 68-95-99.7 rule. In discussing normal distributions, we usually assume that we are dealing with an entire population rather than a sample, so in Figure 15.14 we represent the by m and the standard deviation by s (rather than xq and s). We summarize the properties of a normal distribution. = median = mode FIGURE 15.13 Smoothing a histogram of a normal distribution into a normal curve. Inflection points μ 3σ μ 2σ μ σ μ μ + σ μ + 2σ μ + 3σ 68% 95% 99.7% FIGURE 15.14 The 68-95-99.7 rule for a normal distribution. *Recall that a distribution is just another name for a set of data.

752 CHAPTER 15 y Descriptive Statistics PROPERTIES OF A NORMAL DISTRIBUTION 1. A normal curve is bell shaped. 2. The highest point on the curve is at the of the distribution. 3. The, median, and mode of the distribution are the same. 4. The curve is symmetric with respect to its. 5. The total area under the curve is 1. 6. Roughly 68% of the data values are within 1 standard deviation from the, 95% of the data values are within 2 standard deviations from the, and 99.7% of the data values are within 3 standard deviations from the.* We can use the 68-95-99.7 rule to estimate how many values we expect to fall within 1, 2, or 3 standard deviations of the of a normal distribution. EXAMPLE 1 The Normal Distribution and Intelligence Tests Suppose that the distribution of scores of 1,000 students who take a standardized intelligence test is a normal distribution. If the distribution s is 450 and its standard deviation is 25, a) how many scores do we expect to fall between 425 and 475? b) how many scores do we expect to fall above 500? SOLUTION: a) As you can see in Figure 15.15, the scores 425 and 475 are 1 standard deviation below and above the, respectively. 68% of scores 425 one standard deviation below 450 475 one standard deviation above FIGURE 15.15 Sixty-eight percent of the scores lie within 1 standard deviation of the. From the 68-95-99.7 rule, we know that 68%, or about 0.68 of the scores, lie within 1 standard deviation of the. Because we have 1,000 scores, we can expect that about 0.68 * 1,000 = 680 scores are in the range 425 to 475. b) Recall in Figure 15.16, that 95% of the scores in a normal distribution fall between 2 standard deviations below the and 2 standard deviations above the. *To keep this discussion simple, we are approximating these percentages. Shortly, we will use a table to slightly improve our estimate of the percentage of scores within 1 and 2 standard deviations from the.

15.4 y The Normal Distribution 753 2.5 percent of scores here 95% of scores 2.5 percent of scores here Quiz Yourself Use the information given in Example 1 to determine the following. a) How many scores do we expect to fall between 400 and 500? b) How many scores do we expect to fall below 400? KEY POINT 11 Areas under a normal curve represent percentages of values in the distribution. This s that 5%, or 0.05, of the scores lie more than 2 standard deviations above or below the. Thus, we can expect to have 0.05, 2 = 0.025 of the scores to be above 500. Again multiplying by 1,000, we can expect that 0.025 * 1,000 = 25 scores to be above 500. Now try Exercises 5 to 16. ] 11 z-scores Two standard deviations below 400 450 500 Two standard deviations above FIGURE 15.16 Five percent of the scores lie more than 2 standard deviations from the. In Example 1, we estimated how many values were within 1 standard deviation of the. It is natural to ask whether we can predict how many values lie other distances from the. For example, knowing that the weights of women between ages 18 and 25 form a normal distribution, a clothing manufacturer may want to know what percentage of the female population is between 1.3 and 2.5 standard deviations above the weight for women. It is possible to find this by using Table 15.16, which is the table of areas under the standard normal curve. The standard normal distribution has a of 0 and a standard deviation of 1. Math in Your Life How Worried Are You About Your Next Test?* People often argue about how much of your physical, intellectual, and psychological makeup is due to your heredity and how much is due to your environment. Scientists have identified genes that affect characteristics such as height, body fat, blood pressure, and IQ. Surprisingly, even human anxiety has been associated with genetic makeup. It has been estimated that there are as many as 15 genes that can affect your anxiety level. If you have all these genes, then you have a greater tendency to be anxious, and if you have none of them, then you tend to be less anxious. If we were to plot a histogram of the number of anxiety genes present in a large number of people, we would find very few people have almost none of these genes, many people have a moderate number of these genes, and very few people have almost all of them. If you were to plot the distribution of anxiety genes present in the population, it would look very much like the normal curves that you are studying in this section. *This note is based on a series of PowerPoint slides by Dr. Lee Bardwell, professor of genetics at the University of California at Irvine. It is believed that 45% of the variance in human anxiety is due to genetic factors. The other 55% is due to other factors.

754 CHAPTER 15 y Descriptive Statistics z A z A z A z A z A z A.00.000.56.212 1.12.369 1.68.454 2.24.488 2.80.497.01.004.57.216 1.13.371 1.69.455 2.25.488 2.81.498.02.008.58.219 1.14.373 1.70.455 2.26.488 2.82.498.03.012.59.222 1.15.375 1.71.456 2.27.488 2.83.498.04.016.60.226 1.16.377 1.72.457 2.28.489 2.84.498.05.020.61.229 1.17.379 1.73.458 2.29.489 2.85.498.06.024.62.232 1.18.381 1.74.459 2.30.489 2.86.498.07.028.63.236 1.19.383 1.75.460 2.31.490 2.87.498.08.032.64.239 1.20.385 1.76.461 2.32.490 2.88.498.09.036.65.242 1.21.387 1.77.462 2.33.490 2.89.498.10.040.66.245 1.22.389 1.78.463 2.34.490 2.90.498.11.044.67.249 1.23.391 1.79.463 2.35.491 2.91.498.12.048.68.252 1.24.393 1.80.464 2.36.491 2.92.498.13.052.69.255 1.25.394 1.81.465 2.37.491 2.93.498.14.056.70.258 1.26.396 1.82.466 2.38.491 2.94.498.15.060.71.261 1.27.398 1.83.466 2.39.492 2.95.498.16.064.72.264 1.28.400 1.84.467 2.40.492 2.96.499.17.068.73.267 1.29.402 1.85.468 2.41.492 2.97.499.18.071.74.270 1.30.403 1.86.469 2.42.492 2.98.499.19.075.75.273 1.31.405 1.87.469 2.43.493 2.99.499.20.079.76.276 1.32.407 1.88.470 2.44.493 3.00.499.21.083.77.279 1.33.408 1.89.471 2.45.493 3.01.499.22.087.78.282 1.34.410 1.90.471 2.46.493 3.02.499.23.091.79.285 1.35.412 1.91.472 2.47.493 3.03.499.24.095.80.288 1.36.413 1.92.473 2.48.493 3.04.499.25.099.81.291 1.37.415 1.93.473 2.49.494 3.05.499.26.103.82.294 1.38.416 1.94.474 2.50.494 3.06.499.27.106.83.297 1.39.418 1.95.474 2.51.494 3.07.499.28.110.84.300 1.40.419 1.96.475 2.52.494 3.08.499.29.114.85.302 1.41.421 1.97.476 2.53.494 3.09.499.30.118.86.305 1.42.422 1.98.476 2.54.495 3.10.499.31.122.87.308 1.43.424 1.99.477 2.55.495 3.11.499.32.126.88.311 1.44.425 2.00.477 2.56.495 3.12.499.33.129.89.313 1.45.427 2.01.478 2.57.495 3.13.499.34.133.90.316 1.46.428 2.02.478 2.58.495 3.14.499.35.137.91.319 1.47.429 2.03.479 2.59.495 3.15.499.36.141.92.321 1.48.431 2.04.479 2.60.495 3.16.499.37.144.93.324 1.49.432 2.05.480 2.61.496 3.17.499.38.148.94.326 1.50.433 2.06.480 2.62.496 3.18.499.39.152.95.329 1.51.435 2.07.481 2.63.496 3.19.499.40.155.96.332 1.52.436 2.08.481 2.64.496 3.20.499.41.159.97.334 1.53.437 2.09.482 2.65.496 3.21.499.42.163.98.337 1.54.438 2.10.482 2.66.496 3.22.499.43.166.99.339 1.55.439 2.11.483 2.67.496 3.23.499.44.170 1.00.341 1.56.441 2.12.483 2.68.496 3.24.499.45.174 1.01.344 1.57.442 2.13.483 2.69.496 3.25.499.46.177 1.02.346 1.58.443 2.14.484 2.70.497 3.26.499.47.181 1.03.349 1.59.444 2.15.484 2.71.497 3.27.500.48.184 1.04.351 1.60.445 2.16.485 2.72.497 3.28.500.49.188 1.05.353 1.61.446 2.17.485 2.73.497 3.29.500.50.192 1.06.355 1.62.447 2.18.485 2.74.497 3.30.500.51.195 1.07.358 1.63.449 2.19.486 2.75.497 3.31.500.52.199 1.08.360 1.64.450 2.20.486 2.76.497 3.32.500.53.202 1.09.362 1.65.451 2.21.487 2.77.497 3.33.500.54.205 1.10.364 1.66.452 2.22.487 2.78.497.55.209 1.11.367 1.67.453 2.23.487 2.79.497 TABLE 15.16 Standard normal distribution.

15.4 y The Normal Distribution 755 Table 15.16 gives the area under this curve between the and a number called a z- score. A z-score represents the number of standard deviations a data value is from the. In Example 1, we found that for a normal distribution with a of 450 and a standard deviation of 25, the value 500 was 2 standard deviations above the. Another way of saying this is that the value 500 corresponds to a z-score of 2. Notice that Table 15.16 gives areas only for positive z-scores that is, for data that lie above the. We use the symmetry of a normal curve to find areas corresponding to negative z-scores. Example 2 shows how to use Table 15.16. Because the total area under a normal curve is 1, we can interpret areas under the standard normal curve as percentages of the data values in the distribution. Also note that because the is 0 and the standard deviation for the standard normal curve is 1, the value of a z-score is also the same as the number of standard deviations that z-score is from the. After you have practiced working with the standard normal distribution, we will show you how to use the techniques that you have learned to solve problems involving real-life data. PROBLEM SOLVING The Three-Way Principle The Three-Way Principle in Section 1.1 suggests that you can often understand a problem graphically. This is certainly true in solving problems involving the normal curve. If you draw a good picture of the situation, it often becomes more clear to you as to what to do to set up and solve the problem. EXAMPLE 2 Finding Areas under the Standard Normal Curve Use Table 15.16 to find the percentage of the data (area under the curve) that lie in the following regions for a standard normal distribution: a) between z = 0 and z = 1.3 b) between z = 1.5 and z = 2.1 c) between z = 0 and z =-1.83 SOLUTION: a) The area under the curve between z = 0 and z = 1.3 is shown in Figure 15.17. We find this area by looking in Table 15.16 for the z-score 1.30. We see that A is 0.403 when z = 1.30. Thus, we expect 0.403, or 40.3%, of the data to fall between 0 and 1.3 standard deviations above the. The area of this region is 0.403. 0 1.3 z-score FIGURE 15.17 Area under the standard normal curve between z = 0 and z = 1.3.

756 CHAPTER 15 y Descriptive Statistics TI Screen showing area under standard normal curve between z = 1.5 and z = 2.1. b) Figure 15.18 shows the area we want. Areas in Table 15.16 correspond to regions from z = 0 to the given z-score. Therefore, to find the area under the curve between z = 1.5 and z = 2.1, we must first find the area from z = 0 to z = 2.1 and then subtract the area from z = 0 to z = 1.5. From Table 15.16, we see that when z = 2.1, A = 0.482, and when z = 1.5, A = 0.433. To finish this problem, we can set up our calculations as follows: larger area from z 0 to z 2.1 0.482 smaller area from z 0 to z 1.5 0.433 0.049 subtract areas, not z-scores area we want This s that in the standard normal distribution, the area under the curve between z = 1.5 and z = 2.1 is 0.049, or 4.9%. A = 0.433 A = 0.482 0.433 A = 0.466 A = 0.466 0 1.5 2.1 FIGURE 15.18 Finding the area under the standard normal curve between z = 1.5 and z = 2.1. 1.83 0 1.83 FIGURE 15.19 Finding the area under the standard normal curve between z =-1.83 and z = 0. Quiz Yourself 12 Use Table 15.16 to find the following areas under the standard normal curve: a) between z = 0 and z = 1.45 b) between z = 1.23 and z = 1.85 c) between z = 0 and z =-1.35 KEY POINT We convert values in a nonstandard normal distribution to z-scores. c) Due to the symmetry of the normal distribution, the area between z = 0 and z =-1.83 is the same as the area between z = 0 and z = 1.83 (Figure 15.19). From Table 15.16 we see that when z = 1.83, A = 0.466. Therefore, 46.6% of the data values lie between 0 and -1.83. Now try Exercises 17 to 34. ] Some Good Advice A common mistake in solving a problem such as Example 2(b) is to subtract 1.5 from 2.1 to get 0.6 and then wrongly use this for the z-score in Table 15.16. If you think about the graph of the standard normal curve, clearly the area between z = 0 and z = 0.6 is not the same as the area between z = 1.5 and z = 2.1. Converting Raw Scores to z-scores 12 A real-life normal distribution, such as the set of all weights of women between ages 18 and 25, may have a of 120 pounds and a standard deviation of 25 pounds. Such a distribution will have the properties that we stated earlier for a normal distribution, but because the distribution does not have a of 0 and a standard deviation of 1, we cannot use Table 15.16 directly, as we did in Example 2. We can use Table 15.16, however, if we first convert nonstandard values, called raw scores, to z-scores. The following formula shows the desired relationship between values in a nonstandard normal distribution and z-scores. FORMULA FOR CONVERTING RAW SCORES TO Z -SCORES Assume a normal distribution has a of m and a standard deviation of s. We use the equation x z - m = to convert a value x in the nonstandard distribution to a z-score. s

15.4 y The Normal Distribution 757 EXAMPLE 3 Converting Raw Scores to z -Scores Suppose the of a normal distribution is 20 and its standard deviation is 3. a) Find the z-score corresponding to the raw score 25. b) Find the z-score corresponding to the raw score 16. SOLUTION: a) Figure 15.20 gives us a picture of this situation. We will use the conversion formula z = x - m s, with the raw score x = 25, the m = 20, and the standard deviation 25-20 s = 3. Making the substitutions, we have, z = 3 = 5 3 = 1.67. We can interpret this result as telling us that in this distribution, 25 is 1.67 standard deviations above the. x 16 raw score 20 3 standard deviation x 25 raw score Quiz Yourself 13 FIGURE 15.20 Normal distribution with of 20 and standard deviation of 3. Suppose a normal distribution has a of 50 and a standard deviation of 7. Convert each raw score to a z-score. a) 59 b) 50 c) 38 b) We will use Figure 15.20 and the conversion formula again, but this time with x = 16, 16-20 m = 20, and s = 3. Therefore, the corresponding z-score is z = 3 = - 4 3 =-1.33. This tells us that 16 is 1.33 standard deviations below the. Now try Exercises 43 to 48. ] 13 0.08 0.06 0.04 0.02 Slide 20 10 0 10 20 30 FIGURE 15.21 (a) First slide the distribution 20 units to the left to get a of 0. Squeeze 0.4 0.3 0.2 0.1 Squeeze 15 10 5 0 5 10 15 FIGURE 15.21 (b) Then squeeze the distribution to get a standard deviation of 1. The Three-Way Principle in Section 1.1 suggests that you can remember the formula for converting raw scores to z-scores by interpreting the formula geometrically. In Example 3, we had a normal distribution with a of 20 and standard deviation 3. In order to make the of 20 in the nonstandard distribution correspond to the of 0 in the standard normal distribution, you can think of sliding the distribution to the left 20 units, as we show in Figure 15.21(a). That is why we compute an expression of the form x - 20 in the numerator of the z-score formula. Because the standard deviation was 3, we divided x - 20 by 3 to narrow the distribution so that it would have a standard deviation of 1, as you can see in Figure 15.21(b), and therefore allow us to use Table 15.16.

758 CHAPTER 15 y Descriptive Statistics Applications Normal distributions have many practical applications. EXAMPLE 4 Interpreting the Significance of an Exam Score Suppose that to qualify for a management training program offered by your employer you must score in the top 10% of those employees who take a standardized test. Assume that the distribution of scores is normal and you received a score of 72 on the test, which had a of 65 and a standard deviation of 4. What percentage of those who took this test had a score below yours? SOLUTION: We will find the number of standard deviations your score is above the and then determine what percentage of scores fall below this number. First, we calculate the z-score, which corresponds to your 72: z = 72-65 4 = 7 4 = 1.75 Now, looking in Table 15.16 for z = 1.75, we find that A = 0.460. Therefore, 46% of the scores fall between the and your score (Figure 15.22). A 0.500 A 0.460 65 4 standard deviation x 75 (raw score) z 1.75 FIGURE 15.22 Normal distribution with of 65 and standard deviation of 4. However, we must not forget the scores that fall below the. Because a normal curve is symmetric, another 50% of the scores fall below the. So, there are 50% + 46% = 96% of the scores below yours. Congratulations! You qualify for the program. ] We can use the ideas we have been discussing about the normal curve to compare data. EXAMPLE 5 Using z -Scores to Compare Data Consider the following information: Ty Cobb hit.420 in 1911. Ted Williams hit.406 in 1941. George Brett hit.390 in 1980. In the 1910s, the batting average was.266 and the standard deviation was.0371. In the 1940s, the batting average was.267 and the standard deviation was.0326. In the 1970s, the batting average was.261 and the standard deviation was.0317. Assume that the batting averages were normally distributed in each of these three decades. Use z-scores to determine which of the three batters was ranked the highest in relationship to his contemporaries.

15.4 y The Normal Distribution 759 SOLUTION: We will convert each of the three batting averages to a corresponding z-score for the distribution of batting averages for the decade in which the athlete played. In the 1910s, Ty Cobb s average of.420 corresponded to a z-score of In the 1940s, Ted Williams s average of.406 corresponded to a z-score of In the 1970s, George Brett s average of.390 corresponded to a z-score of We see that in using the standard deviation to compare each hitter with his contemporaries, Ted Williams was ranked as the best hitter. Now try Exercises 69 and 70. ] We conclude our discussion of normal distributions by explaining how manufacturers can use information regarding the life of their products as a basis for warranties. EXAMPLE 6.420 -.266.0371.406 -.267.0326.390 -.261.0317 =.154.0371 = 4.1509. =.139.0326 = 4.2638. =.129.0317 = 4.0694. Using the Normal Distribution to Write Warranties To increase sales, a manufacturer plans to offer a warranty on a new model of the ipod Touch. In testing the ipod Touch, quality control engineers found that it has a time to failure of 3,000 hours with a standard deviation of 500 hours. Assume that the typical purchaser will use the device for 4 hours per day. If the manufacturer does not want more than 5% to be returned as defective within the warranty period, how long should the warranty period be to guarantee this? SOLUTION: In solving this problem, we first work with the standard normal distribution and then convert the answer to fit our nonstandard normal distribution. Figure 15.23 gives a picture of the situation for the standard normal curve. 5% 45% A = 45% =.45 5% z = 1.64 0 z = 1.64 FIGURE 15.23 Finding the area under the standard normal curve for a negative z-score. In Figure 15.23, we see that we need to find a z-score such that at least 95% of the entire area is beyond this point. Observe that this score is to the left of the and is negative. Therefore, we use symmetry and find the z-score such that 95% of the entire area is below this score. However, Table 15.16 gives the areas only for the upper half of the distribution. However, this is not a problem because we know that 50% of the entire area lies

760 CHAPTER 15 y Descriptive Statistics Quiz Yourself 14 Redo Example 6, but now assume that the time to failure is 4,200 hours, with a standard deviation of 600, and the manufacturer wants no more than 2% to be returned as defective within the warranty period. below the. Therefore, our problem reduces to finding a z-score greater than 0 such that 45% of the area lies between the and that z-score. In Table 15.16, we see that if A = 0.450, the corresponding z-score is 1.64. This s that 95% of the area underneath the standard normal curve falls below z = 1.64. By symmetry, we also can say that 95% of the values lie above -1.64. We must now interpret this z-score in terms of original normal distribution of raw scores. Recall the equation that relates values in a distribution to z-scores, namely, Previously we used this equation to convert raw scores to z-scores. We now use the equation in answering the reverse question. Specifically, what is x when m = 3,000, s = 500, and z =-1.64? Substituting these numbers in equation (1), we get We solve for x in this equation by multiplying both sides by 500: -1.64 # x - 3,000 (500) = # (500) 500 Simplifying the equation, we get z = x - m s. -1.64 = x - 3,000. 500-820 = x - 3,000. Last, we add 3,000 to both sides to obtain x = 2,180 hours. We expect 95% of the breakdowns to occur beyond 2,180 hours. Because we know that owners use the ipod Touch about 4 hours per day, we divide 2,180 545 2,180 by 4 to get 4 = 545 days. This is equivalent to 31 = 17.58 months if we use 31 days per month. Therefore, if the manufacturer wants to have no more than 5% of the breakdowns occur within the warranty period, the warranty should be for roughly 18 months. ] 14 (1) (2) Exercises 15.4 Looking Back* These exercises follow the general outline of the topics presented in this section and will give you a good overview of the material that you have just studied. 1. List six properties of the normal distribution. 2. What was the common error that we warned you against making in solving Example 2(b)? 3. What property of the normal distribution makes it unnecessary to have negative z-scores in Table 15.16? 4. State the equation that allows you to convert raw scores to z-scores? Sharpening Your Skills Assume that all distributions in this exercise set are normal. Assume that the distribution in Exercises 5 10 has a of 10 and a standard deviation of 2. Use the 68-95-99.7 rule to find the percentage of values in the distribution that we describe. 5. Between 10 and 12 6. Between 12 and 14 7. Above 14 8. Below 8 9. Above 12 10. Below 10 Assume that the distribution in Exercises 11 16 has a of 12 and a standard deviation of 3. Use the 68-95-99.7 rule to find the percentage of values in the distribution that we describe. 11. Below 9 12. Between 12 and 15 13. Above 6 14. Below 12 15. Between 15 and 18 16. Above 18 Use Table 15.16 to find the percentage of area under the standard normal curve that is specified. 17. Between z = 0 and z = 1.23 18. Between z = 0 and z = 2.06 19. Between z = 0.89 and z = 1.43 *Before doing these exercises, you may find it useful to review the note How to Succeed at Mathematics on page xix.

15.4 y Exercises 761 20. Between z = 0.76 and z = 1.52 21. Between z = 0 and z =-0.75 22. Between z = 0 and z =-1.35 23. Between z = 1.25 and z = 1.95 24. Between z = 0.37 and z = 1.23 25. Between z =-0.38 and z =-0.76 26. Between z =-1.55 and z =-2.13 27. Above z = 1.45 28. Below z =-0.64 29. Below z =-1.40 30. Above z = 0.78 31. Below z = 1.33 32. Above z =-0.46 33. Above z =-0.84 34. Below z = 1.22 35. Find a z-score such that 10% of the area under the standard normal curve is above that score. 36. Find a z-score such that 20% of the area under the standard normal curve is above that score. 37. Find a z-score such that 12% of the area under the standard normal curve is below that score. 38. Find a z-score such that 24% of the area under the standard normal curve is below that score. 39. Find a z-score such that 60% of the area is below that score. 40. Find a z-score such that 90% of the area is below that score. 41. Find a z-score such that 75% of the area is above that score. 42. Find a z-score such that 60% of the area is above that score. In Exercises 43 48, we give you a, a standard deviation, and a raw score. Find the corresponding z-score. 43. Mean 80, standard deviation 5, x = 87 44. Mean 100, standard deviation 15, x = 117 45. Mean 21, standard deviation 4, x = 14 46. Mean 52, standard deviation 7.5, x = 61 47. Mean 38, standard deviation 10.3, x = 48 48. Mean 8, standard deviation 2.4, x = 6.2 In Exercises 49 54, we give you a, a standard deviation, and a z-score. Find the corresponding raw score. 49. Mean 60, standard deviation 5, z = 0.84 50. Mean 20, standard deviation 6, z = 1.32 51. Mean 35, standard deviation 3, z =-0.45 52. Mean 62, standard deviation 7.5, z =-1.40 53. Mean 28, standard deviation 2.25, z = 1.64 54. Mean 8, standard deviation 3.5, z =-1.25 Applying What You ve Learned Due to random variations in the operation of an automatic coffee machine, not every cup is filled with the same amount of coffee. Assume that the amount of coffee dispensed is 8 ounces and the standard deviation is 0.5 ounce. Use Figure 15.14 to solve Exercises 55 and 56. 55. Analyzing vending machines. a. What percentage of the cups should have at least 8 ounces of coffee? b. What percentage of the cups should have less than 7.5 ounces of coffee? 56. Analyzing vending machines. a. What percentage of the cups should have at least 8.5 ounces of coffee? b. What percentage of the cups should have less than 8 ounces of coffee? A machine fills bags of candy. Due to slight irregularities in the operation of the machine, not every bag gets exactly the same number of pieces. Assume that the number of pieces per bag has a of 200 with a standard deviation of 2. Use Figure 15.14 to solve Exercises 57 and 58. 57. Analyzing vending machines. a. How many of the bags do we expect to have between 200 and 202 pieces in them? b. How many of the bags do we expect to have at least 202 pieces in them? 58. Analyzing vending machines. a. How many of the bags do we expect to have between 198 and 200 pieces in them? b. How many of the bags do we expect to have at least 196 pieces in them? 59. Product reliability. Suppose that for a particular brand of LCD television set, the distribution of failures has a of 120 months with a standard deviation of 12 months. With respect to this distribution, 140 months corresponds to what z-score? 60. Product reliability. Redo Exercise 59 using 130 months instead of 140. 61. Distribution of heights. Assume that in the NBA the distribution of heights has a of 6 feet, 8 inches, and a standard deviation of 3 inches. If there are 324 players in the league, how many players do you expect to be over 7 feet tall? 62. Distribution of heights. Redo Exercise 61, but now we want to know how many players you would expect to be over 6 feet, 6 inches. 63. Distribution of heart rates. Assume that the distribution of 21-year-old women s heart rates at rest is 68 beats per minute with a standard deviation of 4 beats per minute. If 200 women are examined, how many would you expect to have a heart rate of less than 70? 64. Distribution of heart rates. Redo Exercise 63, but now we ask how many you would expect to have a heart rate less than 75. In Exercises 65 and 66, we assume that the length of a human pregnancy is 268 days. 65. Birth statistics. If 95% of all human pregnancies last between 250 and 286 days, what is the standard deviation of the distribution of lengths of human pregnancies? What percentage of human pregnancies do we expect to last at least 275 days?

762 CHAPTER 15 y Descriptive Statistics 66. Birth statistics. Babies born before the 37th week (that is, before day 252 of a pregnancy) are considered premature. What percentage of births do we expect to be premature? (See Exercise 65.) 67. Analyzing Internet use. Comcast has analyzed the amount of time that its customers are online per session. It found that the distribution of connect times has a of 37 minutes and a standard deviation of 11 minutes. For this distribution, find the raw score that corresponds to a z-score of 1.5. 68. Analyzing customer service. The distribution of times that customers spend waiting in line in a supermarket has a of 3.6 minutes and a standard deviation of 1.2 minutes. For this distribution, find the raw score that corresponds to a z-score of -1.3. In Exercises 69 and 70, use the information and methods of Example 5 to make the indicated comparisons. 69. Comparing athletes. In 1949, Jackie Robinson hit.342 for the Brooklyn Dodgers; in 1973, Rod Carew hit.350 for the Minnesota Twins. Using z-scores to compare each batter with his contemporaries, determine which batting average was more impressive. 70. Comparing athletes. In 1940, Joe DiMaggio hit.352 for the New York Yankees; in 1975, Bill Madlock hit.354 for the Pittsburgh Pirates. Using z-scores to compare each batter with his contemporaries, determine which batting average was more impressive. 71. Writing a warranty. A manufacturer plans to provide a warranty on Wii Fit. In testing, the manufacturer has found that the set of failure times for Wii Fit is normally distributed with a time of 2,000 hours and a standard deviation of 800 hours. Assume that the typical purchaser will use Wii Fit for 2 hours per day. If the manufacturer does not want more than 4% returned as defective within the warranty period, how long should the warranty period be to guarantee this? (Assume that there are 31 days in each month.) 72. Writing a warranty. Redo Exercise 71, but now assume that the of the distribution of failure times is 2,500 hours, the standard deviation is 600 hours, and the manufacturer wants no more than 2.5% returned during the warranty period. 73. Analyzing investments. A certain mutual fund over the last 15 years has a yearly return of 7.8% with a standard deviation of 1.3%. a. If you had invested in this fund over the past 15 years, in how many of these years would you expect to earn at least 9% on your investment? b. In how many years would you expect to earn less than 6%? 74. Analyzing investments. A certain bond fund over the last 12 years has a yearly return of 5.9% with a standard deviation of 1.8%. a. If you had invested in this fund over the past 12 years, in how many of these years would you expect to earn at least 8% on your investment? b. In how many years would you expect to earn less than 4%? 75. Analyzing the SATs. Assume that the math SAT scores are normally distributed with a of 500 and a standard deviation of 100. If you score 480 on this exam, what percentage of those taking the test scored below you? 76. Analyzing the SATs. Assume that the verbal SAT scores are normally distributed with a of 500 and a standard deviation of 100. If you score 520 on this exam, what percentage of those taking the test scored below you? Communicating Mathematics 77. If a raw score corresponds to a z-score of 1.75, what does that tell you about that score in relationship to the of the distribution? 78. If a raw score corresponds to a z-score of -0.85, what does that tell you about that score in relationship to the of the distribution? 79. Explain why, if you want to compute the area under a normal curve between z = 1.3 and z = 2.0, it is not correct to subtract 2.0-1.3 = 0.7 and then use the z-score of 0.7 to find the area. 80. Without looking at Table 15.16, can you determine whether the area under the standard normal curve between z = 0.5 and z = 1.0 is greater than or less than the area between z = 1.5 and z = 2.0? Explain. 81. Explain how you would estimate the and the standard deviation of the two normal distributions in the given figure. Distribution 1 Distribution 2 5 15 20 25 30 82. Explain how you would estimate the and the standard deviation of the two normal distributions in the given figure. Using Technology to Investigate Mathematics Distribution 1 Distribution 2 3 6 9 12 15 18 21 24 27 30 36 39 41 83. You can find many interactive programs that illustrate various properties of the normal distribution. Search the Internet for normal distribution applets, run some, and report on your findings. 84. Search the Internet for some interesting applications of the normal distribution. While searching, I found many applications in medicine and even some in music. Report on your findings.

15.5 y Looking Deeper: Linear Correlation 763 For Extra Credit 85. If you had a large collection of data, how might you go about determining whether the distribution was normal? 86. Discuss whether you feel that the method we used to compare Ty Cobb, Ted Williams, and George Brett would be a valid way to compare the great African American athletes who played in the Negro leagues with the great white athletes of the same era. The nth percentile in a distribution is a number x such that n% of the values in that distribution fall below x. For example, 20% of the values in the distribution fall below the 20th percentile. 87. What z-score in the standard normal distribution is the 80th percentile? 88. What z-score in the standard normal distribution is the 30th percentile? 89. If a distribution has a of 40 and a standard deviation of 4, what is the 75th percentile? 90. If a distribution has a of 80 and a standard deviation of 6, what is the 35th percentile? 91. Finding percentiles. If you score a 75 on a commercial pilot s exam that has a score of 68 and a standard deviation of 4, to what percentile does your score correspond? 92. Finding percentiles. If you know that the salary for your profession is $53,000 with a standard deviation of $2,500, to what percentile does your salary of $57,000 correspond?