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CHAPTER 5 ANSWERS Section 5.1 Statistical Literacy and Critical Thinking 1 2 3 4 5 6 7 8 The word “normal” has a special meaning in statistics. It refers to a specific type of distributions which are characterized by their bell-shaped curve. The bell shape starts low, rises to a maximum, and then decreases, being symmetric around the point at which the maximum occurs. There is a formula that produces the shape known as the normal distribution. This distinguishes the normal curve from others which may be similar, but are not exactly normal. No. The ten different possible digits are all equally likely, so the graph of the distribution will be flat, not bell-shaped. Since the total area under a normal distribution curve is 1.0, if the area under the curve to the left a particular value is 0.2, then the area to the right of that value is 0.8. This statement is not sensible. A normal distribution is symmetric, so the mean and the median are equal. This statement is not sensible. A normal distribution has only one peak, so there can be only one mode. This statement is not sensible. Since the mean is 100 and the distribution is symmetric, 50% of the scores are greater than 100. The number of scores greater than 105 must be something less than 50%, not more. This statement makes sense. The actual volumes will vary a little bit, and it is common for such actual measurements to have a normal distribution. Her initial measurements of 100 cans could easily give her information about the nature of the distribution, so it makes sense that the distribution is normal. Concepts and Applications 9 10 11 12 13 14 Distribution (b) is not normal since it is not symmetric. Distribution (c) has the larger standard deviation since it is more spread out than distribution (a). Distribution (a) is not normal since it is not symmetric. Distribution (b) has the larger standard deviation since it is more spread out than distribution (c). Normal. It is common for manufactured products, such as CDs, to have a distribution that is normal. The weights typically vary above and below the mean weight by the same amounts, so the distribution has one peak and is symmetric. Not normal. Many people have relatively small incomes, while a smaller number have quite large incomes. You may have heard of expressions like the 90-10 rule, which states that 90 percent of the people have 10% of the income and 10% of the people have 90% of the income. The exact values are not important, but the idea behind them conveys the impression that incomes are not normally distributed or even symmetrically distributed. Not normal. The numbers 1 through 6 should occur with approximately equal frequency, so the distribution is uniform, not normal. A graph of the distribution of the distribution will be flat, not bell-shaped. Normal. There is a small number of scores that are very high and a small number that are very low, and the distribution is likely to peak at the value of the mean, which is around 1518. There is a possibility that the distribution might not be exactly symmetric (and not exactly normal) since the SAT is taken by students intending to go to college. Many of those students who do not take the SAT would score low on the test if they were to 86 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley SECTION 5.2, PROPERTIES OF THE NORMAL DISTRIBUTION 15 16 17 18 19 20 21 22 23 24 25 26 take it, so the lower end of the otherwise normal distribution may be “chopped off” or truncated. Normal. Such physical measurements generally tend to be normally distributed. A small number of males will have extremely high grip strength measurements, and a very small number will have low measurements. The distribution will tend to peak around the value of the mean. Not normal. Many flights experience little or no delay, and only a few flights experience lengthy delays, so the distribution of flight delay times will be quite right-skewed. Not normal. The waiting times will tend to be evenly distributed over a 10minute interval and will therefore have a uniform distribution. Not normal. Most cars will have no tickets. Those that do have tickets will most likely be subject to a small number of different amounts according to the types of parking violations (expired meter, too close to fire hydrant or crosswalk, loading zones, etc.). It is not possible to determine the exact shape of the distribution, but it will not be normal due to the number with no tickets. The movie lengths are not normally distributed because movies generally have a minimum length, but no maximum length. There are very few movies that are a lot shorter than a typical length, but there are some very long movies. Heart rates are nearly normal because data values are distributed symmetrically about the mean. Many physiological variables have normal distributions. The quarter weights should be nearly normal because the deviations from the mean are symmetrical above and below the mean. While the data values are distributed fairly symmetrically about the mean, the distribution does not have the left and right tails for the more extreme weights. In other samples, this variable could be normally distributed, but in this particular sample it is not. If the aspirin were taken from a bottle that had been purchased, it is possible that tablets that were either too light or too heavy were rejected in the packing process. This would explain the lack of tails. a) The total area under the curve is 1. b) 0.20 c) 0.80 d) 0.40 e) 0.45 a) The total area under the curve is 1. b) 0.50 c) 0.30 d) 0.70 e) 0.20 a) The mean is 115. b) 15% c) 50% d) 15% a) The mean is 155. b) 15% c) 15% d) 45% Section 5.2 Statistical Literacy and Critical Thinking 1 87 z = standard score = value - mean mean - mean = = 0 standard deviation standard deviation Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley 88 2 3 4 5 6 7 8 CHAPTER 5, A NORMAL WORLD Since the total percentage under the curve is 100%, the percentage of values to the right is 100% - 75% = 25%. No. The distribution of the outcomes from rolling a die is a uniform distribution, not normal. The rule applies only to normal distributions. A z core can be negative if the value is less than the mean. A percentile cannot be negative since percentiles represent the cumulative percentage of scores from the left. The smallest possible such percentage is 0%. This statement is not sensible. If we assume that the minimum test score is zero, then the standard deviation cannot be greater than the mean. If it were greater than the mean, at least 16% of the scores would be negative, those that are more than one standard deviation below the mean. The statement makes sense. For those who can’t visualize a 3420 g baby, this is about 7 lbs, 8 oz. The standard deviation of 495 g is about 1.09 lbs The statement makes sense. The given values are realistic and a bell-shaped distribution is plausible. This statement does not make sense. Marc’s score is less than one standard deviation above the mean. An unusually high score would be one that is at least two standard deviations above the mean. Concepts and Applications 9 a) b) c) d) e) f) g) h) i) j) 10 a) b) c) d) e) f) 50%. Half of all values are less than the mean. 0.84. 120 is 1 standard deviation to the right of the mean. There is 0.50 to the left of 100 and another 0.34 between 100 and 120. 97.5%. 140 is 2 standard deviations to the right of the mean. There is 0.50 to the left of 100 and another 0.475 between 100 and 140. 16%. 80 is 1 standard deviation to the left of the mean. The percent of the scores between 80 and the mean is 34%, so there is 16% to the left of 80. 0.025. 60 is two standard deviations to the left of the mean. The relative frequency of scores between 60 and the mean is 0.475, so there is 0.025 to the left of 60. 16%. 120 is 1 standard deviation to the right of the mean. The percent of the scores between 120 and the mean is 34%, so there is 16% to the right of 120. 2.5%. 140 is two standard deviations to the right of the mean. The percentage of scores between 140 and the mean is 47.5%, so there is 2.5% to the right of 140. 0.84. From part (d), 16% of the scores are less than 80, so 84% must be greater than 80. 68%. 80 and 120 are both 1 standard deviation from the mean. 81.5%. 80 is 1 standard deviation below the mean and 140 is 2 standard deviations above the mean. There is 34% between 80 and 100 and 47.5% between 100 and 140, a total of 81.5%. 50%. Half of all values are less than the mean. 16%. 55 is 1 standard deviation to the left of the mean. The percent of the scores between 55 and the mean is 34%, so there is 16% to the left of 55. 0.025. 40 is two standard deviations to the left of the mean. The relative frequency of scores between 40 and the mean is .475, so there is 0.025 to the left of 40. 0.84. 85 is 1 standard deviation to the right of the mean. There is 0.50 to the left of 70 and another 0.34 between 70 and 85. 0.975. 100 is 2 standard deviations to the right of the mean. There is 0.50 to the left of 70 and another 0.475 between 70 and 100, making a total of 0.975 to the left of 100. 16%. 85 is 1 standard deviation to the right of the mean. The Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley SECTION 5.2, PROPERTIES OF THE NORMAL DISTRIBUTION 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 89 percent of the scores between 85 and the mean is 34%, so there is 16% to the right of 85. g) 84%. From part (b), 16% of the scores are less than 55, so 84% must be greater than 55. h) 0.975. 40 is 2 standard deviations to the left of the mean. The relative frequency of scores between 40 and the mean is 0.475, so there is 0.025 to the left of 40 and 0.975 to the right of 40. i) 68%. 55 and 80 are both 1 standard deviation from the mean. j) 97.5%. 40 and 100 are both 2 standard deviations from the mean. a) Since 55 seconds is the standard deviation, 68% of the times will be within 55 seconds of the mean of 184.0. b) Since 55 seconds is the standard deviation, 95% of the times will be within 110 seconds (2 standard deviations) of the mean of 184.0. c) Since 55 seconds is the standard deviation, 99.7% of the times will be within 165 seconds (3 standard deviations) of the mean of 184.0. a) From 24.4 in. to 29.6 in. is 27.0 + 2(1.3). Within two standard deviations of the mean, there are 95% of the women. b) 30.9 in. is three standard deviations (3.9 in.) above the mean. Since 99.7% of the women will have forward grip reaches within three standard deviations of the mean, only 0.3% will fall outside this range and half of those will be greater than 3 standard deviations above the mean. Thus, 0.15% of women will have forward grip reaches greater than 30.9 in. c) 68% of the women will have forward grip reaches within one standard deviation of the mean (27.0). Half of those (34%) will fall between the mean and 28.3 in. (one standard deviation above the mean). Since 100 is the mean, z = 0, and 50% of the scores are less than 100. z = (84 - 100)/16 = -1.00; from Table 5.1, 15.87% of scores are less than 84. z = (116 – 100)/16 = 1.00; from Table 5.1, 84.13% of scores are less than 116, so 100% - 84.13 = 15.87% of scores are greater than 116. z = (76 – 100)/16 = -1.50; from Table 5.1, 6.68% of scores are less than 76. z = (132 – 100)/16 = 2.00; from Table 5.1, 97.72% of scores are less than 132, so 100% - 97.72% = 2.28% of scores are greater than 132. z = (80 – 100)/16 = -1.25, which we’ll round to –1.30; from Table 5.1, 9.68% of z-scores are less than –1.3. z = (65 – 100)/16 = -2.2; from Table 5.1, 1.39% of scores are less than 65. z = (129 – 100)/16 = 1.80; from Table 5.1, 96.41% of scores are less than 129. Since 84 and 116 are each 1 standard deviation away from the mean, the ascores are –1 and +1. From Table 5.1, the area under the curve between these two values is 84.13 – 15.87 = 68.26%. For 76, the z-score is (76 – 100)/16 = -1.50. For 124, the z-score is (124 – 100)/16 = 1.50. From Table 5.1, the area between z = -1.5 and z = 1.5 is 93.32% - 6.68% = 86.64%. See Exercise 18. The z-score for 80 is –1.30. The z-score for 120 is +1.30. The area under the curve to the left of –1.30 is about 9.68%. The area to the left of +1.25 is 90.32%. Then the area between –1.30 and +1.30 is 90.32% - 9.68% = 80.64%. For 92, z = (92 – 100)/16 = -0.5; for 115, z = (115 – 100)/16 = 0.94. The closest table value of z is 0.95, so we’ll use that. The area between z = 0.50 and z = 0.95 is 82.89% - 30.85% = 52.04%. Since the mean is 162 cm, the percentage of heights greater than 162 cm is 50%. z = (168 – 162)/6 = 1.0; from Table 5.1, the percentage less than 168 is 84.13%. z = (156 – 162)/6 = -1.0; from Table 5.1, the percentage less than 156 is 15.87%. The percentage greater than 156 is 100% - 15.87% = 84.13%. Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley 90 28 29 30 31 32 33 34 35 36 37 38 CHAPTER 5, A NORMAL WORLD z = (171 – 162)/6 = 1.5; from Table 5.1, the percentage less than 171 is 93.32%. Therefore the percentage greater than 171 is 100% - 93.32% = 6.68%. z = (177 – 162)/6 = 2.5; from Table 5.1, the percentage less than 177 is 99.38%. z = (147 – 162)/6 = -2.5; from Table 5.1, the percentage less than 147 is 0.62%. z = (144 – 162)/6 = -3.0; from Table 5.1, the percentage less than 144 is 0.13%. z = (179 – 162)/6 = 2.80; from Table 5.1, the percentage less than 179 is 99.74%. Therefore the percentage greater than 179 is 100% - 99.74% = 0.26%. The two values of z are z = (156 – 162)/6 = -1.0 and z = (168 – 162)/6 = 1.0. From table 5.1, the percentage between 156 and 168 is 84.13% – 15.87% = 68.26%. The two values of z are z = (159 – 162)/6 = -0.50 and z = (165 – 162)/6 = 0.50. From table 5.1, the percentage between 159 and 165 is 69.15% – 30.85% = 38.30%. The two values of z are z = (148 – 162)/6 = -2.3 and z = (170 – 162)/6 = 1.3. From table 5.1, the percentage between 148 and 170 is 90.32% – 1.07% = 89.25%. The two values of z are z = (146 – 162)/6 = -2.7 and z = (156 – 162)/6 = 1.0. From table 5.1, the percentage between 156 and 168 is 15.87% – 0.35% = 15.52%. In all cases, 5% of coins are rejected. The ranges of weights that are acceptable to the vending machine are Cent 2.44-2.56 grams Nickel 4.88-5.12 grams Dime 2.208-2.328 grams Quarter 5.53-5.81 grams Half dollar 11.06-11.62 grams a) z value mean s tan dard deviation 250 268 15 1.20 From Table 5.1, the percentage lasting less than 250 days is 11.51%. b) z value mean s tan dard deviation 300 268 15 2.10 From Table 5.1, the percentage lasting less than 300 days is 98.21%. Therefore, the percentage lasting more than 300 days is 100% - 98.21% = 1.79%. c) z value mean s tan dard deviation 238 268 15 2.00 From Table 5.1, the percentage lasting less than 238 days is 2.28%. 39 a) z value mean s tan dard deviation 2000 1518 325 1.50 From Table 5.1, the percentage less than 2000 is 93.32%. Therefore, the percentage higher than 2000 is 100% - 93.32% = 6.68%. b) z value mean s tan dard deviation 1500 1518 325 0.05 From Table 5.1, the percentage less than 1500 is 48.01%. c) The two standard scores are Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley SECTION 5.2, PROPERTIES OF THE NORMAL DISTRIBUTION z value mean s tan dard deviation 1600 1518 325 0.25 z value mean s tan dard deviation 2100 1518 325 1.80 91 From Table 5.1, the percentage between 1600 and 2100 is 96.41% 59.87% = 36.54%. 40 a) z value mean s tan dard deviation 650 497 115 1.30 The percentage of scores less than 650 is 90.32, so a score of 650 is in the 90th percentile. 41 b) The 95th percentile corresponds to a standard score (from Table 5.1) of 1.70. Thus (value – 497)/115 = 1.70. Multiplying both sides of this equation by 115, we get value – 497 = 195.5. Now adding 497 to each side of the equation, we get value = 497 + 195.5 = 692.5 or 693. a) z value mean s tan dard deviation 31 30.4 0.23 2.60 From Table 5.1, the percentage less than 31 is 99.53%. Therefore, the percentage higher than 31 is 100% - 99.53% = 0.47%. [Note that of 50 barometers, this means that only 0.23 barometers read higher than 31, so most of the time, there will be no barometers reading over 31.] b) z value mean s tan dard deviation 30 30.4 0.23 1.70 From Table 5.1, the percentage less than 30 is 4.46%. c) For the low side, we have (value – 30.4)/0.23 = -1.50. Multiplying both sides of this equation by 0.23, we get value – 30.4 = -0.345. Now adding 30.4 to each side of the equation, we get value = 30.4 – 0.345 = 30.055. For the high side, we have (value – 30.4)/0.23 = 1.50. Multiplying both sides of this equation by 0.23, we get value – 30.4 = 0.345. Now adding 30.4 to each side of the equation, we get value = 30.4 + 0.345 = 30.745. Barometers will be rejected if they read below 30.055 or above 30.745/ d) The mean of the 50 barometers is the best measure of the actual atmospheric pressure at the time the barometers were read. The distribution of readings could be expected to be approximately normal, so the median could also be used if there were any outliers in the values. Outliers would not affect the median, but may have an effect of the mean. Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley 92 42 CHAPTER 5, A NORMAL WORLD z For the girls, value mean s tan dard deviation 75 71 6 0.70 From Table 5.1, the percentage less than 75 is 75.80%. Therefore 24.2% of the girls’ scores are above 75. Of the 60 girls, 24.2%, or 16, will go to the state bee. For the boys, z value mean s tan dard deviation 75 66 5 1.80 From Table 5.1, the percentage less than 75 is 96.41%. Therefore 3.59% of the boys’ scores are above 75. Of the 50 boys, 3.59%, or 2, will go to the state bee. Thus, 16 out of the 18, or 89%, of those going to the state bee are girls. 43 The two standard scores are z value mean s tan dard deviation 64 69 2.8 1.80 z value mean s tan dard deviation 78 69 2.8 3.20 From Table 5.1, the percentage between 64 and 78 is 99.87% - 3.59% = 96.28%. 44 a) z value mean s tan dard deviation 120 110.5 22.4 0.40 From Table 5.1, the percentage less than 2 hours (120 minutes) is 65.54%. Therefore, the percentage greater than 2 hours is 100% 65.54% = 34.46%. b) z value mean s tan dard deviation 90 110.5 22.4 0.90 From Table 5.1, the percentage less than 1 1/2 hours (90 minutes) is 18.41%. c) z value mean s tan dard deviation 150 110.5 22.4 1.80 From Table 5.1, the percentage less than 2 1/2 hours (150 minutes) is 96.41%, so the probability that a randomly selected movie will be less than 2.5 hours long is 0.9641. Section 5.3 Statistical Literacy and Critical Thinking 1 No. The sample is a convenience sample and is subject to a bia that would not be expected with a random sample. The cannot assume that her convenience sample has the same characteristics as a sample that was randomly selected. Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley SECTION 5.3, THE CENTRAL LIMIT THEOREM 2 93 3 The symbol represents the standard deviation of the population, and n represents the size of the sample. No. The Central Limit Theorem indicates that the sample means will be approximately normally distributed when sample sizes are large, but samples of size 2 cannot be considered large unless the population itself if normally distributed. 4 The standard deviation of the sample mean is V/ n V/ 4 V /2 . Concepts and Applications 5 a) The mean of the distribution of sample means is the population mean, 100. With n = 64, the standard deviation of the distribution of the b) V / n 16/ 64 2.0 . With a sample size of n = 100, the mean of the distribution of sample mean is 16/8 means is the population mean, 100. c) 6 a) distribution of the mean is V / n 16/ 100 16/10 1.6 . As the sample size n increases, the means tend to be closer together (less variation) and the value of the standard deviation of the mean decreases due to the presence of the square root of n in the denominator. The mean of the distribution of sample means is the population mean, 1518. With n = 100, the standard deviation of the distribution of the V/ n 325/ 100 325/10 3.25 . With a sample size of n = 2500, the mean of the distribution of sample sample means is b) means is the population mean, 1518. c) 7 a) b) c) 8 a) b) 9 The standard deviation of the The standard deviation of the distribution of the sample means is V / n 325/ 2500 325/50 6.50 . As the sample size n increases, the means tend to be closer together (less variation) and the value of the standard deviation of the sample mean decreases due to the presence of the square root of n in the denominator. The mean of the distribution of sample means is the population mean, 6.5. With n = 81, the standard deviation of the distribution of the sample means is V / n 3.452/ 81 3.452/9 0.3836 . The mean of the distribution of sample means is the population mean, 6.5. With n = 100, the standard deviation of the distribution of the sample means is V / n 3.452/ 100 3.452/10 0.3452 . As the sample size n increases, the means tend to be closer together (less variation) and the value of the standard deviation of the sample mean decreases due to the presence of the square root of n in the denominator. The mean of the distribution of sample means is the population mean, 5.5. With n = 49, the standard deviation of the distribution of the sample means is V / n 2.872/ 49 2.872/7 0.4103 . The mean of the distribution of sample means is the population mean, 5.5. With n = 400, the standard deviation of the distribution of the sample means is. V / n 2.872/ 400 2.872/20 0.1436 c) As the sample size n increases, the means tend to be closer together (less variation) and the value of the standard deviation of the sample mean decreases due to the presence of the square root of n in the denominator. For individual aircraft, we use the population standard deviation, 7.9, to find the standard score. Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley 94 CHAPTER 5, A NORMAL WORLD z value mean s tan dard deviation 15 13.0 7.9 0.25 . From Table 5.1, the percentage with ages less than 15 years is 59.87%. Therefore, the percentage with ages greater than 15 years is 100% - 59.87% = 40.13%. For a sample of size 49, we use the standard deviation of the sample mean, V/ n z 10 7.9/ 49 7.9/7 value mean s tan dard deviation 1.1286 , to find the standard score. 15 13.0 1.80 . From Table 5.1, the percentage of 1.1286 sample means less than 15 years is 96.41%. Therefore, the percentage of sample means greater than 15 years is 100% - 96.41% = 3.59%. For individual aircraft, we use the population standard deviation, 7.9, to find the standard score. z value mean s tan dard deviation 10 13.0 7.9 0.40 . From Table 5.1, the percentage with ages less than 10 years is 34.46%. For a sample of size 84, we use the standard deviation of the sample mean, V/ n z 11 7.9/ 84 7.9/9.165 0.8620 , to find the standard score. value mean 10 13.0 3.48 . From Table 5.1, the percentage of s tan dard deviation 0.8620 sample means less than 10 years is 0.02%. For individual aircraft, we use the population standard deviation, 7.9, to find the standard scores. For 10, value mean s tan dard deviation z 10 13.0 7.9 0.40 . From Table 5.1, the percentage with ages less than 10 years is 34.46%. For 16, value mean s tan dard deviation z 16 13.0 7.9 0.40 . From Table 5.1, the percentage with ages less than 16 years is 65.54%. Thus the percentage with ages between 10 and 16 years is 65.54% – 34.46% = 31.08%. For a sample of size 81, we use the standard deviation of the sample mean, V/ n z z 12 7.9/ 81 7.9/9 value mean s tan dard deviation 0.8778 , to find the standard scores. 10 13.0 3.40 0.8778 value mean s tan dard deviation 16 13.0 0.8778 . 3.40 From Table 5.1, the percentage of sample means less than 10 years is 0.02% and the percentage less than 16 years is 99.98% Thus, the percentage of sample means between 10 and 16 is 99.98% - 0.02% = 99.96%. For individual aircraft, we use the population standard deviation, 7.9, to find the standard scores. For 12.5, z value mean s tan dard deviation 12.5 13.0 7.9 0.0633 . From Table 5.1, the percentage with ages less than 10 years is 48.01%. For 13.5, z value mean s tan dard deviation 13.5 13.0 7.9 0.0633 . percentage with ages less than 13.5 years is 51.99%. From Table 5.1, the Thus the percentage Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley SECTION 5.3, THE CENTRAL LIMIT THEOREM 95 with ages between 12.5 and 13.5 years is 51.99% – 48.01% = 3.98%. For a sample of size 400, we use the standard deviation of the sample mean, V/ n z z 13 7.9/ 400 7.9/20 0.395 , to find the standard scores. value mean 12.5 13.0 1.30 s tan dard deviation 0.395 . value mean 13.5 13.0 1.30 s tan dard deviation 0.395 From Table 5.1, the percentage of sample means less than 12.5 years is 9.68% and the percentage less than 13.5 years is 90.32% Thus, the percentage of sample means between 10 and 16 is 90.32% - 9.68% = 80.64%. a) For a sample of size 36, we use the standard deviation of the sample V / n 0.11/ 36 0.11/6 0.0183 , to find the standard score. 12.05 12.00 value mean 2.70 . s tan dard deviation 0.0183 mean, z b) 14 a) From Table 5.1, the percentage of sample means less than 12.05 oz is 99.65%. Thus, the percentage of sample means greater than 12.05 oz is 100% - 99.65% = 0.35%. If the true population mean is 12.00 oz, it is very unlikely that we would get a mean amount as high as 12.05 oz for a sample of 36 cans. More than likely, the cans are being filled with more than 12.00 oz. In this case, since the consumers are getting more than they thought they were getting, no one is likely to feel cheated. For individual times, we use the population standard deviation, 0.4, to find the standard scores. For 4.0, b) z value mean s tan dard deviation 4.0 3.0 0.4 2.5 . From Table 5.1, the percentage of times less than 4 seconds is 99.38%. Therefore, the percentage of times greater than 4.9 is 100% - 99.38% = 0.62%. For a sample of size 36, we use the standard deviation of the sample V / n 0.4/ 60 0.0516 , to find the standard score. value mean 4.0 3.0 19.4 . s tan dard deviation 0.0516 mean, z c) 15 a) From Table 5.1, the percentage of sample means less than 4.0 seconds is greater than 99.98%. Thus, the percentage of sample means greater than 4.0 seconds is less than 0.02%. The behavior of individual times is more important than the behavior of batches of 60 times since a pilot is only concerned about other pilots seeing one plane – his. For individual sizes, we use the population standard deviation, 1.0 inches, to find the standard scores. For 6.2, b) z value mean s tan dard deviation 6.2 6.0 1.0 0.2 . From Table 5.1, the percentage of male head breadths less than 6.2 inches is 57.93%%. For a sample of size 100, we use the standard deviation of the sample mean, V / n 1.0/ 100 0.1 , to find the standard score. Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley 96 CHAPTER 5, A NORMAL WORLD z c) 16 a) value mean s tan dard deviation z value mean s tan dard deviation z a) . 12 11.4 2.7 0.2222 . From Table 5.1, the percentage of those guests who shower less than 12 minutes is 57.93%. Therefore, the percentage of guests showering more than 12 minutes is 42.07%. There will not be enough hot water if the mean shower time is greater than 12 minutes. For a sample of size 84, we use the standard deviation of the sample mean, standard score. 17 2.0 From Table 5.1, the percentage of sample means less than 6.2 inches is 97.72%. Although the mean head breadth of 100 men is very likely to be less than 6.2 inches, from part (a), we see that only about 58% of men have head breadths less than 6.2 inches. Therefore, about 42% of men have head breadths larger than 6.2 inches and they could not use the helmets. For individual sizes, we use the population standard deviation, 2.7 minutes, to find the standard scores. For 12, b) 6.2 6.0 0.1 value mean s tan dard deviation 12 11.4 0.2946 V/ n 2.04 2.7/ 84 0.2946 , to find the . From Table 5.1, the percentage of sample means less than 12 minutes is 97.72%. Therefore, the percentage of means greater than 12 minutes is only 2.28%. The system appears to be effective because it will be overloaded only rarely. Considering that it is also highly unlikely that all 84 guests will shower at the same time, it is even less likely that the system will ever be overloaded. For individual women, we use the population standard deviation, 29 pounds, to find the standard scores. For 140 lbs, value mean s tan dard deviation z 140 143 29 0.10 . From Table 5.1, the percentage with weights less than 140 pounds is 46.02%. For 211 lbs,. b) z value mean s tan dard deviation 211 143 29 2.30 From Table 5.1, the percentage with weights less than 211 pounds is 98.93%. Thus the percentage with weights between 143 and 211 pounds is 98.93% – 46.02% = 52.91%. For a sample of size 36, we use the standard deviation of the sample V/ n 29/ 36 4.833 , to find the standard scores. 140 143 value mean 0.60 . For 140 lbs, z s tan dard deviation 4.833 value mean 211 143 14.0 . For 211 lbs, z s tan dard deviation 4.833 mean, From Table 5.1, the percentage of sample means less than 140 pounds is 27.43%. For 211 pounds, the standard score is off the chart, so at least 99.98% of the weights are less than 211 pounds. Therefore, Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley SECTION 5.3, THE CENTRAL LIMIT THEOREM c) 18 a) there is about a 99.98% - 27.43% = 72.55% chance that the mean weight is between 140 and 211 pounds. The result from part (a) is more important in redesigning the ejection seats since the seats are occupied by only one person, not a group of 36. For individual pieces of candy, we use the population standard deviation, 0.0518 g, to find the standard scores. z For 0.8535 g, b) 97 value mean s tan dard deviation 0.8535 0.8565 0.0518 0.05 . From Table 5.1, the percentage with weights less than 0.8535 g is 48.01%. It follows that 51.99% have weights greater than 0.8535 g. For a sample of size 465, we use the standard deviation of the sample V/ n 0.0518/ 465 0.0024 , to find the standard scores. value mean 0.8535 0.8565 1.25 . For 0.8535 g, z s tan dard deviation 0.0024 mean, c) 19 a) From Table 5.1, the percentage of sample means less than 0.8535 g is 11.51%. It follows that 88.49% of sample means are greater than 0.8535 g. If the company actually puts exactly 465 M&Ms in each bag, it appears that it is doing a good job of filling the bags since most (approximately 8 out of every 9) will exceed the weight printed on the bag. More likely, the bags will be filled by weight, not by count. Furthermore, the quality inspection process will probably result in underweight bags not reaching the market. For individual quarters, we use the population standard deviation, 0.062 g, to find the standard scores. For 5.550 g, z value mean s tan dard deviation 5.550 5.670 0.062 1.90 . From Table 5.1, the percentage with weights less than 5.550 g is 2.87%. For 5.790 g, b) z value mean s tan dard deviation 5.790 5.670 0.062 1.90 . From Table 5.1, the percentage with weights less than 5.790 g is 97.13%. Therefore 2.87% of the quarters will weigh more than 5.790 g. Altogether 5.74% of the quarters will be rejected. If 280 quarters are inserted in the vending machine, 5.74% of them, or 16, can be expected to be rejected. For a sample of size 280, we use the standard deviation of the sample V/ n 0.062/ 280 0.0037 , to find the standard scores. 5.550 5.670 value mean 32.43 . For 5.550 g, z s tan dard deviation 0.0037 mean, c) 20 a) From Table 5.1, the percentage of sample means less than 5.550 g is less than 0.02%. Similarly, for 5.790 g, z = 32.43, and the percentage of means less than 5.790 is greater than 99.98%. There is at least a 99.96% chance that the mean of 280 quarters is between the limits. The result from part (a) is more important. No one puts 280 quarters in a vending machine. Individual legitimate quarters rejected may mean the loss of a sale. However, the configuration is designed to prevent losing merchandise when someone puts in a slug or foreign coin about the size of a quarter. If his total weight is greater than 195 pounds, then he weighs more Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley 98 CHAPTER 5, A NORMAL WORLD than 175 pounds. z b) value mean s tan dard deviation V/ n From Table 5.1, 53.98% of men 29/ 213 For 175 pounds, z 1.9870 , to find the standard scores. value mean 175 172 1.50 . s tan dard deviation 1.9870 a) From Table 5.1, the percentage of sample means less than 175 is 93.32%, so there is a 6.68% chance that the mean weight of the men is over 175 pounds, making the mean total weight over 195 pounds. A pilot should be concerned because a 6.68% chance that such an overweight condition could occur is too high. However, there are other factors that contribute to the total weight being carried, including the weight of the checked baggage (which is normally weighed when it is checked in, and the amount of fuel carried, which may depend on the length of the flight and expected weather conditions. These factors are all taken into account in preparation for the flight to ensure that the plane is not overloaded. The standard deviation of the distribution of sample means will be b) 0.1 since it is found by V / n V / 100 V /10 . Sample means tend to cluster more closely around the population mean than do individual values. The smaller spread of the means is reflected in the smaller standard deviation. The standard deviation of the distribution of sample means for n = c) 22 0.10 . will weigh less than 175 pounds, so 46.02% will weigh more than 175 pounds, making his total weight greater than 195 pounds. For a sample of size 213, we use the standard deviation of the sample mean, 21 175 172 29 a) 1000 will be V / n V / 1000 V /31.62 0.0316V . Sample means tend to cluster even more closely around the population mean for samples of size 1000 than they do for samples of size 100, the standard deviation being about 1/3 the size it was for samples of size 100. The smaller spread of the means is reflected in the smaller standard deviation, which is now only about 3/100 of the size of the population standard deviation. As the sample size increases, the standard deviation of the sample means decreases due to the presence of n in the denominator of the formula for the standard deviation of the sample mean. For an individual, we use the population standard deviation, 13.1 mm, to find the standard scores. For 125 mm, b) z value mean s tan dard deviation 125 114.8 13.1 0.80 . From Table 5.1, the percentage with pressures less than 125 mm g is 78.81%. It follows that there is a 21.19% chance that the woman has a blood pressure over 125 mm. For a sample of size 300, we use the standard deviation of the sample V/ n 13.1/ 300 0.7563 , to find the standard scores. value mean 125 114.8 13.5 . For 125 mm, z s tan dard deviation 0.7563 mean, From Table 5.1, the percentage of sample means less than 125 is greater than 99.98%, so there is less than a 0.02% chance that the mean blood pressure of the women is over 125 mm. Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley CHAPTER 5 REVIEW EXERCISES c) z For 114 mm, value mean s tan dard deviation 114 114.8 0.7563 99 1.057 | 1.10 . From Table 5.1, the percentage of sample means less than 114 is 13.57%. Chapter 5 Review Exercises 1 a) b) c) 2 a) b) c) 3 The numbers will occur with equal likelihood. Therefore, they have a uniform distribution, not a normal distribution. Weights of a homogeneous population, such as Golden Retriever dogs, typically have a normal distribution. Reaction times of 18-year-old drivers may have a nearly normal distribution since the population is quite homogeneous. If they are not normally distributed, they are probably slightly right-skewed since the reaction times cannot go below zero, but there may be a few people who react slowly and have long reaction times. Since 9.6 is two standard deviations, about 95% of the act scores are within 9.6 of the mean of 21.1. Since 14.4 is three standard deviations, about 99.7% of the act scores are within 14.4 of the mean of 21.1. Yes. It is more than three standard deviations above the mean and therefore would happen less 0.3% of the time. [It would also be unusual because ACT scores are reported as whole numbers.] value mean s tan dard deviation 99.00 98.20 0.62 a) z b) c) d) percentile. From part (a), the standard score is 1.29. No, the data value lies less than 2 standard deviations from the mean. For a sample of size n = 50, the standard deviation of the 1.29 ; this corresponds to the 90th V / n 0.062/ 50 0.0877 . For 97.98, 97.98 98.20 2.50 . This corresponds to the 0.0877 distribution of means is z value mean s tan dard deviation 0.62 percentile, so the likelihood that the mean body temperature is 97.98 degrees is 0.0062. e) f) g) h) z For 101.00, value mean s tan dard deviation 101.00 98.20 0.62 4.52 . This is an unusual temperature since it lies more than two standard deviations above the mean. We conclude that the person has a fever. The 95th percentile is associated with a standard score of 1.65. Thus the temperature must be 1.65 standard deviations above the mean, or 1.65 x 0.62 = 1.02 degrees above 98.2 degrees, or 99.22. [If you choose to use the closest table value, use z = 1.6. Then the temperature is 98.20 + 0.99 = 99.19, not very different.] The 5th percentile is associated with a standard score of -1.65. Thus the temperature must be 1.65 standard deviations below the mean, or 1.02 degrees below 98.2 degrees, or 97.18. [Again, if you choose to use -1.60, the temperature will be 0.99 degrees below 98.20, or 97.21.] For 100.6, z value mean s tan dard deviation 100.60 98.20 0.62 3.87 . This corresponds to a percentile higher than 99.98. Thus fewer than 0.02% of healthy adults would be expected to have a temperature above 100.6. Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley 100 CHAPTER 5, A NORMAL WORLD i) It should be very safe to conclude that someone with a temperature of 100.6 or higher has a fever. On the other hand, using such a high cutoff point will certainly result in concluding that a number of people do not have fevers when, in fact, they do have a fever. For a sample of n = 106, the standard deviation of the distribution of V / n 0.062/ 106 0.0602 For a temperature of 98.2, value mean 98.20 98.60 6.64 . Thus the sample mean is s tan dard deviation 0.0602 sample means is z about 6.6 standard deviations below the mean; the chance of selecting such a sample is extremely small if the assumed mean is correct. The assumed mean (98.60) may be incorrect. Chapter 5 Quiz 1 2 3 4 Statements (b) and (e) are correct. Statement (a) is not correct because there are many distributions that are not unusual that are not normal (e.g., uniform). Statement (c) is not correct because the mean and median of a normal distribution are equal. Statement (d) is not correct because if the standard deviation were zero, all of the values would be the same. The distribution would not be bell-shaped. Twenty points is two standard deviations, so 95% of the scores fall within 20 points of the mean of 50. The mean of the sample means is always the same as the population mean, so in this case, it is 50. The standard deviation of the sample means is value mean s tan dard deviation value mean s tan dard deviation 70 50 10 40 50 10 V/ n 5/ 100 0.5 . 5 z 6 z 7 8 9 Since 50 is the mean, 50% of the scores are greater than 50. The percentage of scores greater than 70 is 100% - 97.72% = 2.28%. Since 70 is 2 standard deviations above the mean of 50 and 30 is 2 standard deviation below the mean of 50, the percentage of scores below 30 is the same as the percentage above 70, namely 2.28%. Since the sample size is largest in part (c), those means will have a distribution that is closest to normal. The outcomes in part (b) have a uniform distribution. The outcomes in part (b) have a distribution that looks like a triangle with its peak at 3.5. 10 2 1 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley