Chapter Solutions Page 15 of 8.50 a. The median is 55. The mean is about 105. b. The median is a more representative average" than the median here. Notice in the stem-and-leaf plot on p.3 of the text that a clear majority of the students owned fewer than 105 CDs so the value 105 is not a representative "average." The large outlier has inflated the mean. c. Yes, the relationship between the mean and median is what you would expect. The data are skewed to the right and there is an extremely large outlier. In general, both of these characteristics cause the mean to be larger than the median..51 a. The five-number created using the methods described on p.3 of the text is shown below. There were n=0 ages, so the lower quartile is the median of the 30 lowest ages and the upper quartile is the median of the 30 highest ages. CEO ages (years) Median 50 Quartiles 45.5 57 Extremes 3 74 The five-number summary shows that the median age of the 0 CEOs of small companies is 50 years. The middle ½ of the CEOs have ages between 45.5 and 57 years. The youngest CEO is 3 years old. The oldest CEO is 74 years old..5 The boxplot can be made either with a horizontal axis (as shown here ) or a vertical axis.(as in Figure.8 of the text). Figure for Exercise.5.53 The mean of the CEO ages is 51.47 years. The median is 50 years. The mean and the median are similar. This is expected because the data are more or less symmetric in shape..54 a. Mean ± St. Dev is 7 ± 1.7, or 5.3 to 8.7. b. Mean ± St. Dev is 7 ± ()(1.7), or 3. to 10.4. c. Mean ± 3 St. Dev is 7 ± (3)(1.7), or 1.9 to 1.1..55 Figure for Exercise.55
Chapter Solutions Page 1 of 8.5 a. z = (00 170)/0 = 1.5. b. z = (140 170)/0 = 1.5. c. z = (170 170)/0 = 0. d. z = (30 170)/0 = 3..57 a. Mean =0; Standard deviation = 1.581. b. Mean =0; Standard deviation = 0. c. Mean = 0; Standard deviation = 33.09..58 a. 98-41 = 57. b. Standard deviation Range/ = 57/ = 9.5..59 The interval is.1 to.7 hours. The interval includes negative values, which are impossible times. Thus, the interval based an assumption of a bell-shaped curve would not reflect reality..0 The Empirical Rule says that 8% of values fall within 1 standard deviation of the mean, 95% fall within standard deviations of the mean, and 99.7% fall within 3 standard deviations of the mean. Of the 103 handspan measurements for women, 74 or 7% are within 1 standard deviation of the mean (18. to 1.8 cm). 100 of the 103 or 97% are within standard deviations. 101 of the 103 or 98% are within 3 standard deviations. This data seems to fit pretty well with the Empirical Rule..1 A histogram or dotplot of the ages at death for the first ladies shows that the data are approximately bellshaped. This may be a little surprising because the data are a mixture of many different distributions. Due to advancement in medicine and other areas, the mean age at death has been increasing over time. The mean age at death is higher now than in the 1800 s.. a. The First Ladies may constitute a population rather than a sample. They lived in unique circumstances, so it is hard to view these women as a representative sample from any larger population. And, they can't be considered to be a sample from a larger population of First Ladies because future First Ladies will have different circumstances affecting life expectancy. b. If the First Ladies are viewed as a population, the population standard deviation is σ = 14.7 years. In Excel, this can be found with the command "=STDEVP( )" and many calculators have a key for the population standard deviation. See p.43 of the text for a discussion of the population standard deviation. If the argument is made in part (a) that the First Ladies constitute a sample, the correct answer here is that the sample standard deviation is s = 14.97 years..3 Outliers affect the standard deviation. This happens because the calculation uses the deviation from the mean for every value. An outlier has a large deviation from the mean, so it inflates the standard deviation. Extreme values generally do not affect the quartiles, and consequently they generally don't affect the interquartile range. Remember that a quartile is determined by counting through the ordered data to a particular location, so the exact size of the largest or smallest observations doesn't matter..4 You expect women s heights to have a bell-shape curve because it is more common for a woman to have a height close to the mean than far from the mean. Generally, the further a height is from the mean (in either direction), the fewer the number of women with that height. The ages at marriage for women will probably not follow a bell-curve. Most of the ages will be in the 0 s, but the data will not be symmetric. The ages can only be as low as law permits 15, maybe. The other direction extends much farther from the mean some women do not get married until they are 40 or 50.
Chapter Solutions Page 17 of 8.5 A categorical variable cannot have a bell-shaped distribution. A variable must be quantitative for it to be possible to have a distribution with any particular shape. For a categorical variable, the raw data are category labels without a meaningful numerical ordering. The ordering of bars in a bar chart is arbitrary and could be done in many different ways. So, with a categorical variable, there is no inherent shape to the distribution.. a. If the two possible outliers are ignored, the data appear to be more or less bell-shaped so the Empirical Rule may hold. b. The Empirical Rule implies that the range should span about 4 to standard deviations. About 95% of the data will be within standard deviations (plus or minus) of the mean and about 99.7% of a data set should be within 3 standard deviations (plus or minus) of the mean. Here, range = maximum minimum = 3.5 1.5 = 10.75 cm. This span is equal to 10.75/1.8 = 5.97 standard deviations so it is consistent with the Empirical Rule..7 a. If the two lowest values are deleted, the mean will increase and the standard deviation will decrease. b. The Empirical Rule for mean ± 3 standard deviations says that 99.7% of the values will be between 0. ± 3(1.45) or 15.85 and 4.55 cm. All of the data, or 100% of the values, are within this interval. c. Looking at the figures, it seems like the Empirical Rule should hold when the outliers are removed. The data looks pretty symmetric without those two values. If the outliers are not removed, the Empirical Rule may hold, but not as well, since the data seem more skewed to the left with those two points included. d. There may be justification for removing the outliers if a convincing argument can be made that they are errors. The value of 1.5 may really be an incorrect entry of 1.5. The value of 13 may really be an incorrect entry of 18 or 3. Assuming the original surveys were available, this could be checked. Or, you could see if either of these women is extremely short or had any other odd measurements. height mean.8 This will differ for each student. The calculation is z =. Use the mean and standard s deviation relevant to your gender. Note that the z-score will be negative if the height is less than the mean. Notice also that if the height equals the mean, the result is z = 0..9 a. If a z-score is 0, the value must equal the mean. b. Begin by setting the formula for a z-score equal to 1. observed value - mean = 1 standard deviation Two steps of algebra lead to observed value = mean + 1 standard deviation. Another strategy is to make observed value = mean + 1 standard deviation in the z-score formula. Algebraic simplification leads to z = 1. value mean 450 500.70 a. z = = = 0. 5, and the proportion below is.3085. standard deviation 100 value mean 3.5 34 b. z = = =. 5, and the proportion below is.9938. standard deviation 1 79 75 c. z = = 0. 5, and the proportion below is.915. 8 79 75 d. z = = 1, and the proportion below is.8413. 4.71 You should be more satisfied if the standard deviation was 5. This would mean you scored standard deviations above the mean and, if scores are bell-shaped, only about.5% of students are expected to score higher..7 The only possible set of numbers is {50, 50, 50, 50, 50, 50, 50} because a standard deviation of 0 means there is no variability.
Chapter 8 Solutions Page of 15 8.38 The answers for this exercise can be found using any of the methods discussed in Section 8.4, including the use of Minitab or Excel. a. P(X = 4) =.051 b. P(X 4) = 1 P(X 3) = 1.49 =.3504 c. P(X 3) =.49 d. P(X = 0) =.5905 e. P(X 1) = 1 P(X = 0) = 1.5905 =.4095 8.39 a. Note that 1/4 of 1,000 is 50 so the desired probability is P(X 50). n = 1000 and p = the proportion of adults in the United States living with a partner, but not married at the time of the sampling. The value of p is not known. b. The desired probability is P(X 110), n = 500, and p =.0. c. Note that 70% of 0 is 14 so the desired probability is P(X 14). n = 0, and p =.50. 8.40 The formulas are µ =np and σ = np( 1 p) a. µ = 10(1/) = 5 and σ = 10 (.5)(1.5) = 1. 5811 b. µ = 100(1/4) = 5 and σ = 100 (.5)(1.5) = 4. 33 c. µ = 500(1/5) = 500 and σ = 500 (.)(1.) = 0 d. µ = 1(1/10) =.1 and σ = 1 (.1)(1.1) =. 3 e. µ = 30(.4) = 1 and σ = 30 (.4)(1.4) =. 83 8.41 For n = and p =.5, P(X = 0) =.5, P(X = 1) =.5, and P(X = ) =.5. These can be found in several ways. One way is to list possible outcomes, which are {SS, SF, FS, and FF}, recognize that all outcomes are equally likely, and then tabulate the distribution of X = number of successes. µ = E(X) = xp (x) = (0.5) + (1.5)+ (0.5) = 1. σ = ( x µ ) p( x) = (0 1) (.5) + (1 1) (.5) + ( 1) (.5 ) = 0.5 =.7071. 8.4 a. P(0 X 30) =.5 (because the interval from 0 to 30 is one-half of the interval of possible outcomes (0 to 0) and the distribution is uniform. b. P(30 X 0) =.5 by the same reasoning as in part (a). 8.43 1.5 0 a. = 1.5. 1 b. 4 10 = 1. c. 0 10 =. 5 d. 5 ( 10) = 1. 15 8.44 Table A.1 can be used to find the answers. a..5000. b..33. c..38. d..9750. e..0099. f..9951. g..9505.
Chapter 8 Solutions Page 7 of 15 00 180 8.45 a. Answer =.8413. For 00 lbs, z = = 1. P(Z 1) =.8413. 0 15 180 b. Answer =.. For 15 lbs, z = = 0. 75. P(Z 0.75) =.. 0 c. Answer =.7734. This is the opposite event to part (b), so calculation is 1. =.7734. 8.4 a. X is a uniform random variable (and it is continuous). b. X ranges from 0 to 100 and the area under any density curve is 1, so f(x) = 1/100=.01 for all x between 0 and 100. This creates a rectangle (with area=1) similar to Figure 8.. Note: f(x) = 0 for any x not between 0 and 100. c. P(X 15 seconds) is the area of the rectangle from 0 to 15 seconds. The interval width is 15 and the height is 1/100, so the answer is (15)(1/100) =.15. d. P(X 40 seconds) is the area of the rectangle between 40 and 100. The interval width is 0 and the height is 1/100 so the answer is (0)(1/100)=.0. e. Figure for Exercise 8.4e f. The expected value or mean is 50. The distribution is symmetric, so the mean equals the median. For a uniform random variable, the median is at the middle of the range of possible values. 8.47 This will differ for each student. 8.48 a. The rectangle has height =1/10=0.1 because the range of X is 0 10=10. Figure for Exercise 8.48a
Chapter 8 Solutions Page 8 of 15 b. Figure for Exercise 8.48b Note: The range of this normal curve was determined using the fact that about 99.7% of the area will be in the range mean ± 3 standard deviation. c. Figure for Exercise 8.48c Note: The range of this normal curve was determined using the fact that about 99.7% of the area will be in the range mean ± 3 standard deviation. 8.49 a. P(Z 1.4) =.0808 b. P(Z 1.4) =.919 c. P( 1.4 Z 1.4) = P(Z 1.4) P(Z 1.4) =.919.0808 =.8384 d. P(Z 1.4) = 1 P(Z 1.4) = 1.919 =.0808. Equivalently, P(Z 1.4) = P(Z 1.4) =.0808. 8.50 a. 0001. Use the In the Extreme portion of Table A.1. b. P( 3.7 Z 3.7) = P(Z 3.7) P(Z 3.7) =.9999.0001 =.9998. c. About 0. This is far beyond the usual range of a standard normal curve. 8.51 a. z = 1.9. If using Table A.1, look for.05 within the interior part of the table. b. z =1.9. If using Table A.1, look for.975 within the interior part of the table. Or, note that the area to the right of z must be.05, so by the symmetry of the standard normal curve the answer is the positive version of the answer for part (a). c. z =1.9 because if.95 is in the central area,.975 must be the area to the left of z. This means the answer is the same as for part (b). 8.5 a. Define event A as Z a ; thus, A c is Z > a. So, P(Z > a) = 1 P(Z a) by Rule 1. b. Define event A as Z a. Define the event B as a Z b. Events A and B are mutually exclusive because a value cannot be both less than a and between a and b at the same time, assuming a is less than b. By Rule b, P(A or B) = P(A) + P(B) = P(Z a) + P(a Z b). The event Z a or a Z b is equivalent to Z b so P(Z b) = P(Z a) + P(a Z b). One step of algebra leads to P(a Z b) = P(Z b) P(Z a).
Chapter 8 Solutions Page 9 of 15 8.53 a. Note that 500 is the mean, and the distribution is symmetric, so P(X 500) =.5 (because the probability is.5 on each side of the mean). 50 500 b. For 50, z = = 1. 5 so P(X 50) = P(Z 1.5) =.933 100 700 500 c. For 700, z = = so P(X 700) = P(Z ) = 1 P(Z ) = 1.977 =.08. 100 Equivalently, P(Z ) = P(Z ) =.08 d. P(500 X 700) = P(0 Z ) = P(Z ) P(Z 0) =.977.5 =.477 5 8.54 a. For 5, z = 0 because 5 is the mean while for, z = = 1. 11..7 So, P( X 5) = P( 1.11 Z 0) = P(Z 0) P(Z 1.11) =.5.1335 =.35 0 5 70 5 b. For 0, z = = 1. 85 while for 70, z = = 1. 85..7.7 So, P(0 X 70) = P( 1.85 Z 1.85) = P(Z 1.85) P(Z 1.85) =.978.03 =.935 c. P(X 70) = P(Z 1.85) =.978 d. P(X 0) = P(Z 1.85) = P(Z 1.85) =.978 e. X is either less than or equal to 0 or greater than or equal to 70 so the answer can be computed as P(X 0) + P(X 70) = P(Z 1.85) + P(Z 1.85) =.03 +.03 =.044 8.55 The value of z for which P(Z z ) =.10 is about 1.8. (If 10% are taller, then 90% are shorter so if using Table A.1, look for.90 in the interior part of the table.) The answer is 1.8 standard deviations above the mean, which is (1.8.7) + 5 = 8.5 inches. The percentile ranking for a height of 8.5 inches is.90 or 90%. 8.5 The value of z for which P(Z z ) =.5 is about 0.75 which means the answer is 0.75 standard deviations below the mean. This height is ( 0.75.7) + 5 = 3. inches. The percentile ranking for a height of 3. inches is.10 or 10%. 8.57 a. This will differ for each student. Suppose, for example, that the student is a male with a right 3.5 handspan of 3 cm. In that case, z = = 0. 33 1.5 b. The answer will differ for each student. For the example given in the solution for part (a), the proportion of males with a handspan smaller than 3 cm. is P(Z 0.33) =.93. 10 18 8.58 a. For 10, z = = 1. 33 so P(X < 10) = P(Z < 1.33) =.0918 30 18 b. For 30, z = = so P(X > 30) = P(Z > ) = 1 P(Z ) = 1.977=.08. Equivalently, P(Z > ) = P(Z < ) =.08. 1 18 c. For 1, z = = 0. 5while for 15, z = +0.5. So, P(15 X 1) = P( 0.5 Z 0.5) = P(Z 0.5) P(Z 0.5) =.915.3085 =.3830 35 18 d. P(X > 35) = P(Z > ) = P(Z >.83) = 1 P(Z.83) = 1.9977 =.003. Equivalently, P(Z >.83) = P(Z <.83) =.003.