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- Question : 1 - Most students will prefer to work in seconds, to avoid having to work with decimals or fractions.
- Question : 2 - Who? The individuals in the data set are students in a statistics class. What? There are eight variables: ID (a label, with no units); Exam1, Exam2, Homework, Final, and Project (in units in
- Question : 3 - Exam1 = 79, Exam2 = 88, Final = 88.
- Question : 4 - For this student, TotalPoints = 2
- Question : 5 - The cases are apartments. There are five variables: rent (quantitative), cable (categorical), pets (categorical), bedrooms (quantitative), distance to campus (quantitative).
- Question : 6 - (a) To find injuries per worker, divide the rates in Example 1.6 by 100,000 (or, redo the computations without multiplying by 100,000). For wage and salary workers, there are 0.000034 fatal injuries per worker. For self-employed workers, there are 0.000099 fatal injuries per worker. (b) These rates are 1/10 the size of those in Example 1.6, or 10,000 times larger than those in part (a): 0.34 fatal injuries per 10,000 wage/salary workers, and
- Question : 7 - Shown are two possible stemplots; the first uses split stems (described on page 11 of the text). The scores are slightly left-skewed; most range from 70 to the low 90s.
- Question : 8 - Preferences will vary. However, the stemplot in Figure 1.8 shows a bit more detail, which is useful for comparing the two distributions.
- Question : 9 - (a) The stemplot of the altered data is shown on the right. (b) Blank stems should always be retained (except at the beginning or end of the stemplot), because the gap in the distribution is an important piece of information about the data.
- Question : 10 - Student preferences will vary. The stemplot has the advantage of showing each individual score. Note that this histogram has the same shape as the second histogram in Exercise 1.7.
- Question : 11 - Student preferences may vary, but the larger classes in this histogram hide a lot of detail.
- Question : 12 - This histogram shows more details about the distribution (perhaps more detail than is useful). Note that this histogram has the same shape as the first histogram in the solution to Exercise 1.7.
- Question : 13 - Using either a stemplot or histogram, we see that the distribution is left-skewed, centered near 80, and spread from 55 to 98. (Of course, a histogram would not show the exact values of the maximum and minimum.)
- Question : 14 - (a) The cases are the individual employees. (b) The first four (employee identification number, last name, first name, and middle initial) are labels. Department and education level are categorical variables; number of years with the company, salary, and age are quantitative
- Question : 15 - A Web search for
- Question : 16 - Recall that categorical variables place individuals into groups or categories, while quantitative variables
- Question : 17 - Recall that categorical variables place individuals into groups or categories, while quantitative variables
- Question : 18 - Student answers will vary. A Web search for
- Question : 19 - For example, blue is by far the most popular choice; 70% of respondents chose 3 of the 10 options (blue, green, and purple).
- Question : 20 - For example, opinions about least-favorite color are somewhat more varied than favorite colors. Interestingly, purple is liked and disliked by about the same fractions of people.
- Question : 21 - a) There were 232 total respondents. The table that follows gives the percents; for example, 10 232 .= 4.31%. (b) The bar graph is on the following page. (c) For example, 87.5% of the group were between 19 and 50. (d) The age-group classes do not have equal width: The first is 18 years wide, the second is 6 years wide, the third is 11 years wide, etc.
- Question : 22 - (a) & (b) The bar graph and pie charts are shown below. (c) A clear majority (76%) agree or strongly agree that they browse more with the iPhone than with their previous phone. (d) Student preferences will vary. Some might prefer the pie chart because it is more familiar.
- Question : 23 - Ordering bars by decreasing height shows the models most affected by iPhone sales. However, because
- Question : 24 - (a) The weights add to 254.2 million tons, and the percents add to 99.9. (b) & (c) The bar graph and pie chart are shown below.
- Question : 25 - (a) & (b) Both bar graphs are shown below. (c) The ordered bars in the graph from (b) make it easier to identify those materials that are frequently recycled and those that are not. (d) Each percent represents part of a different whole. (For example, 2.6% of food scraps are recycled; 23.7% of glass is recycled, etc.)
- Question : 26 - (a) The bar graph is shown on the right. (b) The graph clearly illustrates the dominance of Google; its bar dwarfs those of the other search engines.
- Question : 27 - The two bar graphs are shown below.
- Question : 28 - (a) The bar graph is below. (b) The number of Facebook users trails off rapidly after the top seven or so. (Of course, this is due in part to the variation in the populations of these countries. For example, that Norway has nearly half as many Facebook users as France is remarkable, because the 2008 populations of France and Norway were about 62.3 million and 4.8 million, respectively.)
- Question : 29 - (a) Most countries had moderate (single- or double-digit) increases in Facebook usages. Chile (2197%) is an extreme outlier, as are (maybe) Venezuela (683%) and Colombia (246%). (b) In the stemplot on the right, Chile and Venezuela have been omitted, and stems are split five ways. (c) One observation is that, even without the outliers, the distribution is right-skewed. (d) The stemplot can show some of the detail of the low part of the distribution, if the outliers are omitted.
- Question : 30 - (a) The given percentages refer to nine distinct groups (all M.B.A. degrees, all M.Ed. degrees, and so on) rather than one single group. (b) Bar graph shown on the right. Bars are ordered by height, as suggested by the text; students may forget to do this or might arrange in the opposite order (smallest to largest).
- Question : 31 - (a) The luxury car bar graph is below on the left; bars are in decreasing order of size (the order given in the table). (b) The intermediate car bar graph is below on the right. For this stand-alone graph, it seemed appropriate to re-order the bars by decreasing size. Students may leave the bars in the order
- Question : 32 - This distribution is skewed to the right, meaning that Shakespeare
- Question : 33 - Shown is the stemplot; as the text suggests, we have trimmed numbers (dropped the last digit) and split stems. 359 mg/dl appears to be an outlier. Overall, glucose levels are not under control: Only 4 of the 18 had levels in the desired range.
- Question : 34 - The back-to-back stemplot on the right suggests that the individual-instruction group was more consistent (their numbers have less spread) but not more successful (only two had numbers in the desired range).
- Question : 35 - The distribution is roughly symmetric, centered near 7 (or
- Question : 36 - (a) Totals emissions would almost certainly be higher for very large countries; for example, we would expect that even with great attempts to control emissions, China (with over 1 billion people) would have higher total emissions than the smallest countries in the data set. (b) A stemplot is shown;
- Question : 37 - To display the distribution, use either a stemplot or a histogram. DT scores are skewed to the right, centered near 5 or 6, spread from 0 to 18. There are no outliers. We might also note that only 11 of these 264 women (about 4%) scored 15 or higher.
- Question : 38 - (a) The first histogram shows two modes: 5
- Question : 39 - Graph (a) is studying time (Question 4); it is reasonable to expect this to be right-skewed (many students study little or not at all; a few study longer). Graph (d) is the histogram of student heights (Question 3): One would expect a fair amount of variation but no particular skewness to such a distribution.
- Question : 40 - Sketches will vary. The distribution of coin years would be left-skewed because newer coins are more common than older coins.
- Question : 41 - (a) Not only are most responses multiples of 10; many are multiples of 30 and 60. Most people will
- Question : 42 - The stemplot gives more information than a histogram (since all the original numbers can be read off the stemplot), but both give the same impression. The distribution is roughly symmetric with one value (4.88) that is somewhat low. The center of the distribution is between 5.4 and 5.5 (the median is 5.46, the mean is 5.448); if asked to give a single estimate for the
- Question : 43 - (a) There are four variables: GPA, IQ, and self-concept are quantitative, while gender is categorical. (OBS is not a variable, since it is not really a
- Question : 44 - Stemplot at right, with split stems. The distribution is fairly symmetric
- Question : 45 - Stemplot at right, with split stems. The distribution is skewed to the left, with center around 59.5. Most self-concept scores are between 35 and 73, with a few below that, and one high score of 80 (but not really high enough to be an outlier).
- Question : 46 - The time plot on the right shows that women
- Question : 47 - The total for the 24 countries was 897 days, so with Suriname, it is 897 + 694 = 1591 days, and the mean is x = 1591 25 = 63.64 days.
- Question : 48 - The mean score is x = 821 10 = 82.1.
- Question : 49 - To find the ordered list of times, start with the 24 times in Example 1.23, and add 694 to the end of the list. The ordered times (with median highlighted) are
- Question : 50 - The median of the service times is 103.5 seconds. (This is the average of the 40th and 41st numbers in the sorted list, but for a set of 80 numbers, we assume that most students will compute the median using software, which does not require that the data be sorted.)
- Question : 51 - In order, the scores are: 55, 73, 75, 80, 80 , 85 , 90, 92, 93, 98 The middle two scores are 80 and 85, so the median is M = 80 + 85 2 = 82.5
- Question : 52 - See the ordered list given in the previous solution. The first quartile is Q1 = 75, the median of the first five numbers: 55, 73, 75 , 80, 80. Similarly, Q3 = 92, the median of the last five numbers: 85, 90, 92 , 93, 98.
- Question : 53 - The maximum and minimum can be found by inspecting the list. The sorted list (with quartile and median locations highlighted) is
- Question : 54 - The median and quartiles were found earlier; the minimum and maximum are easy to locate in the ordered list of scores (see the solutions to Exercises 1.51 and 1.52), so the five-number summary is Min = 55, Q1 = 75, M = 82.5, Q3 = 92, Max = 98.
- Question : 55 - Use the five-number summary from the solution to Exercise 1.54: Min = 55, Q1 = 75, M = 82.5, Q3 = 92, Max = 98
- Question : 56 - The interquartile range is IQR = Q3 ? Q1 = 92 ? 75 = 17, so the 1.5
- Question : 57 - The variance can be computed from the formula s2 = 1 n ? 1 (xi ? x)2; for example, the first term in the sum would be (80 ? 82.1)2 = 4.41. However, in practice,
- Question : 58 - In order to have s = 0, all 5 cases must be equal; for example, 1, 1, 1, 1, 1, or 12.5, 12.5, 12.5, 12.5, 12.5. (If any two numbers are different, then xi ? x would be nonzero for some i, so the sum of squared differences would be positive, so s2 > 0, so s > 0.)
- Question : 59 - Without Suriname, the quartiles are 23 and 46.5 days; with Suriname included, they are 23 and 53.5 days. Therefore, the IQR increases from 23.5 to 30.5 days
- Question : 60 - Divide total score by 4: 950 4 = 237.5 points.
- Question : 61 - (a) Use a stemplot or histogram. (b) Because the distribution is skewed, the five-number summary is the best choice; in millions of dollars, it is
- Question : 62 - (a) Either a stemplot or histogram can be used to display the distribution. Two stemplots are shown on the following page: one with all points, and one with the outlier mentioned in part (b) excluded. In the table are the mean and standard deviation, as well as the five-number summary, both with and without the outlier (all values are percents). The latter is preferable because of the outlier; in particular, note the outlier
- Question : 63 - All of these numbers are given in the table in the solution to the previous exercise. (a) x changes from 4.76% (with) to 4.81% (without); the median (4.7%) does not change. (b) s changes from 0.7523% to 0.5864%; Q1 changes from 4.3% to 4.35%, while Q3 = 5% does not change. (c) A low outlier decreases x; any kind of outlier increases s. Outliers have little or no effect on the median and quartiles.
- Question : 64 - (a) A stemplot or histogram can be used to display the distribution. Students may report either mean/standard deviation or the five-number summary (in units of calories):
- Question : 65 - Use a small data set with an odd number of points, so that the median is the middle number. After deleting the lowest observation, the median will be the average of that middle number and the next number after it; if that latter number is much larger, the median will change substantially. For example, start with 0, 1, 2 , 998, 1000; after removing 0, the median changes from 2 to 500.
- Question : 66 - Salary distributions (especially in professional sports) tend to be skewed to the right. This skew makes the mean higher than the median.
- Question : 67 - (a) The distribution is left-skewed. While the skew makes the five-number summary is preferable, some students might give the mean/standard deviation. In ounces, these statistics are:
- Question : 68 - (a) The five-number summary is Min = 2.2 cm, Q1 = 10.95 cm, M = 28.5 cm, Q3 = 41.9 cm, Max = 69.3 cm. (b) & (c) The boxplot and histogram are shown below. (Students might choose different interval widths for the histogram.) (d) Preferences will vary. Both plots reveal the right-skew of this distribution, but the boxplot does not show the two peaks visible in the histogram.
- Question : 69 - (a) The five-number summary is Min = 0 mg/l, Q1 = 0 mg/l, M = 5.085 mg/l, Q3 = 9.47 mg/l, Max = 73.2 mg/l. (b) & (c) The boxplot and histogram are shown below. (Students might choose different interval widths for the histogram.) (d) Preferences will vary. Both plots reveal the sharp right-skew of this distribution, but because Min = Q1, the boxplot looks somewhat strange. The histogram seems to convey the distribution better.
- Question : 70 - Answers depend on whether natural (base-e) or common (base-10) logarithms are used. Both sets of answers are shown here. If this exercise is assigned, it would probably be best for the sanity of both instructor and students to specify which logarithm to use. (a) The five-number summary is:
- Question : 71 - (a) The five-number summary (in units of ?mol/l) is Min = 0.24, Q1 = 0.355, M = 0.76, Q3 = 1.03, Max = 1.9. (b) & (c) The boxplot and histogram are shown below. (Students might choose different interval widths for the histogram.) (d) The distribution is right-skewed. A histogram (or stemplot) is preferable because it reveals an important feature not evident from a boxplot: This distribution has two peaks.
- Question : 72 - The mean and standard deviation for these ratings are x = 5.9 and s .= 3.7719; the five-number summary is Min = Q1 = 1, M = 6.5, Q3 = Max = 10. For a graphical presentation, a stemplot (or histogram) is better than a boxplot because the latter obscures details about the distribution. (With a little thought, one might realize that Min = Q1 = 1 and Q3 = Max = 10 means that there are lots of 1
- Question : 73 - The distribution of household net worth would almost surely be strongly skewed to the right: Most families would generally have accumulated little or modest wealth, but a few would have become rich. This strong skew pulls the mean to be higher than the median.
- Question : 74 - See also the solution to Exercise 1.36. (a) The fivenumber summary (in units of metric tons per person) is: Min = 0, Q1 = 0.75, M = 3.2, Q3 = 7.8, Max = 19.9 The evidence for the skew is in the large gaps between the higher numbers; that is, the differences Q3?M and Max?Q3
- Question : 75 - The total salary is $690,000, so the mean is x = $690,000 9 .= $76,667. Six of the nine employees earn less than the mean. The median is M = $35,000.
- Question : 76 - If three individuals earn $0, $0, and $20,000, the reported median is $20,000. If the two individuals with no income take jobs at $14,000 each, the median decreases to $14,000. The same thing can happen to the mean: In this example, the mean drops from $20,000 to $16,000.
- Question : 77 - The total salary is now $825,000, so the new mean is x = $825,000 9 .= $91,667. The median is unchanged.
- Question : 78 - Details at right.
- Question : 79 - The quote describes a distribution with a strong right skew: Lots of years with no losses to hurricane ($0), but very high numbers when they do occur. For example, if there is one hurricane in a 10-year period causing $1 million in damages, the
- Question : 80 - (a) x and s are appropriate for symmetric distributions with no outliers. (b) Both high numbers are flagged as outliers. For women, IQR = 60, so the upper 1.5
- Question : 81 - (a) & (b) See the table on the right. In both cases, the mean and median are quite similar.
- Question : 82 - See also the solution to Exercise 1.43. (a) The mean of this distribution appears to be higher than 100. (There is no substantial difference between the standard deviations.) (b) The mean and median are quite similar; the mean is slightly smaller due to the slight left skew of the data. (c) In addition to the mean and median, the standard deviation is shown for reference (the exercise did not ask for it).
- Question : 83 - With only two observations, the mean and median are always equal because the median is halfway between the middle two (in this case, the only two) numbers.
- Question : 84 - (a) The mean (green arrow) moves along with the moving point (in fact, it moves in the same direction as the moving point, at one-third the speed). At the same time, as long as the moving point remains to the right of the other two, the median (red arrow) points to the middle point (the rightmost nonmoving point). (b) The mean follows the moving point as before. When the moving point passes the rightmost fixed point, the median slides along with it until the moving point passes the leftmost fixed point, then the median stays there.
- Question : 85 - (a) There are several different answers, depending on the configuration of the first five points. Most students will likely assume that the first five points should be distinct (no repeats), in which case the sixth point must be placed at the median. This is because the median of 5 (sorted) points is the third, while the median of 6 points is the average of the third and fourth. If these are to be the same, the third and fourth points of the set of six must both equal the third point of the set of five.
- Question : 86 - The five-number summaries (all in millimeters) are:
- Question : 87 - (a) The means and standard deviations (all in millimeters) are:
- Question : 88 - (a) The mean is x = 15, and the standard deviation is s .= 5.4365. (b) The mean is still 15; the new standard deviation is 3.7417. (c) Using the mean as a substitute for missing data will not change the mean, but it decreases the standard deviation.
- Question : 89 - The minimum and maximum are easily determined to be 1 and 12 letters, and the quartiles and median can be found by adding up the bar heights. For example, the first two bars have total height 22.3% (less than 25%), and adding the third bar brings the total to 45%, so Q1 must equal 3 letters. Continuing this way, we find that the five-number summary, in units of letters, is:
- Question : 90 - Because the mean is to be 7, the five numbers must add up to 35. Also, the third number (in order from smallest to largest) must be 10 because that is the median. Beyond that, there is some freedom in how the numbers are chosen.
- Question : 91 - The simplest approach is to take (at least) six numbers
- Question : 92 - The algebra might be a bit of a stretch for some students:
- Question : 93 - (a) One possible answer is 1, 1, 1, 1. (b) 0, 0, 20, 20. (c) For (a), any set of four identical numbers will have s = 0. For (b), the answer is unique; here is a rough description of why. We want to maximize the
- Question : 94 - Answers will vary. Typical calculators will carry only about 12 to 15 digits; for example, a TI-83 fails (gives s = 0) for 14-digit numbers. Excel (at least the version I checked) also fails for 14-digit numbers, but it gives s = 262,144 rather than 0. The (very old) version of Minitab used to prepare these answers fails at 20,000,001 (eight digits), giving s = 2.
- Question : 95 - The table on the right reproduces the means and standard deviations from the solution to Exercise 1.87 and shows those values expressed in inches.
- Question : 96 - (a) x = 5.4479 and s = 0.2209. (b) The first measurement corresponds to 5.50
- Question : 97 - Convert from kilograms to pounds by multiplying by 2.2: x = (2.42 kg)(2.2 lb/kg) .= 5.32 lb and s = (1.18 kg)(2.2 lb/kg) .= 2.60 lb.
- Question : 98 - Variance is changed by a factor of 2.542 = 6.4516; generally, for a transformation xnew = a + bx, the new variance is b2 times the old variance.
- Question : 99 - There are 80 service times, so to find the 10% trimmed mean, remove the highest and lowest eight values (leaving 64). Remove the highest and lowest 16 values (leaving 48) for the 20% trimmed mean.
- Question : 100 - After changing the scale from centimeters to inches, the five-number summary values change by the same ratio (that is, they are multiplied by 0.39). The shape of the histogram might change slightly because of the change in class intervals. (a) The five-number
- Question : 101 - 1.101. Take the mean plus or minus two standard deviations: 572
- Question : 102 - 1.102. Take the mean plus or minus three standard deviations: 572
- Question : 103 - 1.103. The z-score is z = 620?572
- Question : 104 - The z-score is z = 510?572 51 =. ?1.22. This is negative because an ISTEP score of 510 is below average; specifically, it is 1.22 standard deviations below the mean
- Question : 105 - Using Table A, the proportion below 620 (z .= 0.94) is 0.8264 and the proportion at or above is 0.1736; these two proportions add to 1. The graph on the right illustrates this with a single curve; it conveys essentially the same idea as the
- Question : 106 - Using Table A, the proportion below 620 (z .= 0.94) is 0.8264, and the proportion below 660 (z .= 1.73) is 0.9582. Therefore:
- Question : 107 - Using Table A, this ISTEP score should correspond to a standard score of z .= 0.67 (software gives 0.6745), so the ISTEP score (unstandardized) is 572 + 0.67(51) .= 606.2 (software: 606.4).
- Question : 108 - Using Table A, x should correspond to a standard score of z =. ?0.84 (software gives ?0.8416), so the ISTEP score (unstandardized) is x = 572 ? 0.84(51) .= 529.2 (software: 529.1).
- Question : 109 - Of course, student sketches will not be as neat as the curves on the right, but they should have roughly the correct shape. (a) It is easiest to draw the curve first, and then mark the scale on the axis. (b) Draw a copy of the first curve, with the peak over 20. (c) The curve has the same shape, but is translated left or right.
- Question : 110 - (a) As in the previous exercise, draw the curve first, and then mark the scale on the axis. (b) In order to have a standard deviation of 1, the curve should be 1/3 as wide, and three times taller. (c) The curve is centered at the same place (the mean), but its height and width change. Specifically, increasing the standard deviation makes the curve wider and shorter; decreasing the standard deviation makes the curve narrower and taller.
- Question : 111 - Sketches will vary.
- Question : 112 - (a) The table on the right gives the ranges for women; for example, about 68% of women speak between 7856 and 20,738 words per day.
- Question : 113 - (a) Ranges are given in the table on the right. In both cases, some of the lower limits are negative, which does not make sense; this happens because the women
- Question : 114 - (a) For example, 68?70 10 = ?0.2. The complete list is given on the right. (b) The cut-off for an A is the 85th percentile for the N(0, 1) distribution. From Table A, this is approximately 1.04; software gives 1.0364. (c) The top two students (with scores of 92 and 98) received A
- Question : 115 - a) We need the 5th, 15th, 55th, and 85th percentiles for a N(0, 1) distribution. These are given in the table on the right. (b) To convert to actual scores, take the standard-score cut-off z and compute 10z + 70. (c) Opinions will vary.
- Question : 116 - (a) The curve forms a 1
- Question : 117 - (a) The height should be 14 since the area under the curve must be 1. The density curve is on the right. (b) P(X ? 1) = 14 = 0.25. (c) P(0.5 < X < 2.5) = 0.5.
- Question : 118 - The mean and median both equal 0.5; the quartiles are Q1 = 0.25 and Q3 = 0.75.
- Question : 119 - (a) Mean is C, median is B (the right skew pulls the mean to the right). (b) Mean A, median A. (c) Mean A, median B (the left skew pulls the mean to the left).
- Question : 120 - Hint: It is best to draw the curve first, then place the numbers below it. Students may at first make mistakes like drawing a half-circle instead of the correct
- Question : 121 - (a) The applet shows an area of 0.6826 between ?1.000 and 1.000, while the 68
- Question : 122 - See the sketch of the curve in the solution to Exercise 1.120. (a) The middle 95% fall within two standard deviations of the mean: 266
- Question : 123 - (a) 99.7% of horse pregnancies fall within three standard deviations of the mean: 336
- Question : 124 - Because the quartiles of any distribution have 50% of observations between them, we seek to place the flags so that the reported area is 0.5. The closest the applet gets is an area of 0.5034, between ?0.680 and 0.680. Thus, the quartiles of any Normal distribution are about 0.68 standard deviations above and below the mean. Note: Table A places the quartiles at about
- Question : 125 - The mean and standard deviation are x = 5.4256 and s = 0.5379. About 67.62% (71/105 .= 0.6476) of the pH measurements are in the range x
- Question : 126 - Using values from Table A: (a) Z > 1.65: 0.0495. (b) Z < 1.65: 0.9505. (c) Z > ?0.76: 0.7764. (d) ?0.76 < Z < 1.65: 0.9505 ? 0.2236 = 0.7269.
- Question : 127 - Using values from Table A: (a) Z ? ?1.8: 0.0359. (b) Z ? ?1.8: 0.9641. (c) Z > 1.6: 0.0548. (d) ?1.8 < Z < 1.6: 0.9452 ? 0.0359 = 0.9093.
- Question : 128 - (a) 22% of the observations fall below ?0.7722. (This is the 22nd percentile of the standard Normal distribution.) (b) 40% of the observations fall above 0.2533 (the 60th percentile of the standard Normal distribution).
- Question : 129 - (a) z = 0.3853 has cumulative proportion 0.65 (that is, 0.3853 is the 65th percentile of the standard Normal distribution). (b) If z = 0.1257, then Z > z has proportion 0.45 (0.1257 is the 55th percentile).
- Question : 130 - 70 is two standard deviations below the mean (that is, it has standard score z = ?2), so about 2.5% (half of the outer 5%) of adults would have WAIS scores below 70.
- Question : 131 - 130 is two standard deviations above the mean (that is, it has standard score z = 2), so about 2.5% of adults would score at least 130.
- Question : 132 - Tonya
- Question : 133 - Jacob
- Question : 134 - Jose
- Question : 135 - Maria
- Question : 136 - Maria
- Question : 137 - Jacob
- Question : 138 - 1920 and above: The top 10% corresponds to a standard score of z = 1.2816, which in turn corresponds to a score of 1509 + 1.2816
- Question : 139 - 1239 and below: The bottom 20% corresponds to a standard score of z = ?0.8416, which in turn corresponds to a score of 1509 ? 0.8416
- Question : 140 - The quartiles of a Normal distribution are
- Question : 141 - The quintiles of the SAT score distribution are 1509 ? 0.8416
- Question : 142 - For a Normal distribution with mean 55 mg/dl and standard deviation 15.5 mg/dl: (a) 40 mg/dl standardizes to z = 40?55 15.5 =. ?0.9677. Using Table A, 16.60% of women fall below this level (software: 16.66%). (b) 60 mg/dl standardizes to z = 60?55 15.5 .= 0.3226. Using Table A, 37.45(c) Subtract the answers from (a) and (b) from 100%: Table A gives 45.95% (software: 45.99%), so about 46% of women fall in the intermediate range.
- Question : 143 - For a Normal distribution with mean 46 mg/dl and standard deviation 13.6 mg/dl: (a) 40 mg/dl standardizes to z = 40?46 13.6 =. ?0.4412. Using Table A, 33% of men fall below this level (software: 32.95%). (b) 60 mg/dl standardizes to z = 60?46 13.6 .= 1.0294. Using Table A, 15.15(c) Subtract the answers from (a) and (b) from 100%: Table A gives 51.85% (software: 51.88%), so about 52% of men fall in the intermediate range.
- Question : 144 - (a) About 0.6% of healthy young adults have osteoporosis (the cumulative probability below a standard score of ?2.5 is 0.0062). (b) About 31% of this population of older women has osteoporosis: The BMD level which is 2.5 standard deviations below the young adult mean would standardize to ?0.5 for these older women, and the cumulative probability for this standard score is 0.3085.
- Question : 145 - (a) About 5.2%: x < 240 corresponds to z < ?1.625. Table A gives 5.16% for ?1.63 and 5.26% for ?1.62. Software (or averaging the two table values) gives 5.21%. (b) About 54.7%: 240 < x < 270 corresponds to ?1.625 < z < 0.25. The area to the left of 0.25 is 0.5987; subtracting the answer from part (a) leaves about 54.7%. (c) About 279 days or longer: Searching Table A for 0.80 leads to z > 0.84, which corresponds to x > 266 + 0.84(16) = 279.44. (Using the software value z > 0.8416 gives x > 279.47.)
- Question : 146 - (a) The quartiles for a standard Normal distribution are
- Question : 147 - (a) As the quartiles for a standard Normal distribution are
- Question : 148 - In the previous two exercises, we found that for a N(?, ?) distribution, Q1 = ? ? 0.6745?, Q3 = ? + 0.6745?, and IQR = 1.3490?. Therefore, 1.5
- Question : 149 - (a) The first and last deciles for a standard Normal distribution are
- Question : 150 - The shape of the quantile plot suggests that the data are right-skewed (as was observed in Exercises 1.36 and 1.74). This can be seen in the flat section in the lower left
- Question : 151 - (a) The plot is reasonably linear except for the point in the upper right, so this distribution is roughly Normal, but with a high outlier. (b) The plot is fairly linear, so the distribution is roughly Normal. (c) The plot curves up to the right
- Question : 152 - See also the solution to Exercise 1.42. The plot suggests no major deviations from Normality, although the three lowest measurements do not quite fall in line with the other points.
- Question : 153 - (a) All three quantile plots are below; the yellow variety is the nearest to a straight line. (b) The other two distributions are slightly right-skewed (the lower-left portion of the graph is somewhat flat); additionally, the bihai variety appears to have a couple of high outliers.
- Question : 154 - Shown are a histogram and quantile plot for one sample of 200 simulated N(0, 1) points. Histograms will vary slightly but should suggest a bell curve. The Normal quantile plot shows something fairly close to a line but illustrates that, even for actual Normal data, the tails may deviate slightly from a line.
- Question : 155 - Shown are a histogram and quantile plot for one sample of 200 simulated uniform data points. Histograms will vary slightly but should suggest the density curve of Figure 1.34 (but with more variation than students might expect). The Normal quantile plot shows that, compared to a Normal distribution, the uniform distribution does not extend as low or as high (not surprising, since all observations are between 0 and 1).
- Question : 156 - Shown is a back-to-back stemplot; the distributions could also be compared with histograms or boxplots. Either mean/standard deviation or the five-number summary could be used; both are given below. Both the graphical and numerical descriptions reveal that hatchbacks generally have higher fuel efficiency (and also are more variable).
- Question : 157 - (a) The distribution appears to be roughly Normal. (b) One could justify using either the mean and standard deviation or the five-number summary:
- Question : 158 - (a) The stemplot on the right suggests that there are two groups of states: the under-23% and over-23% groups. Additionally, while they do not qualify as outliers, Oklahoma (16.3%) and Vermont (30%) stand out as notably low and high. (b) One could justify using either the mean and standard deviation or the five-number summary:
- Question : 159 - Students might compare color preferences using a stacked bar graph like that shown on the right, or side-by-side bars like those below. (They could also make six pie charts, but comparing slices across pies is difficult.) Possible observations: white is considerably less popular in Europe, and gray is less common in China.
- Question : 160 - Students might compare color preferences using a stacked bar graph like that shown on the right, or side-by-side bars like those below. (They could also make six pie charts, but comparing slices across pies is difficult.) Possible observations: white is considerably less popular in Europe, and gray is less common in China.
- Question : 161 - Students might compare color preferences using a stacked bar graph like that shown on the right, or side-by-side bars like those below. (They could also make six pie charts, but comparing slices across pies is difficult.) Possible observations: white is considerably less popular in Europe, and gray is less common in China.
- Question : 162 - Using either a histogram or stemplot, we see that this distribution is sharply rightskewed. For this reason, the five-number summary is preferred
- Question : 163 - The distribution is somewhat right-skewed (although considerably less than the distribution with all countries) with only one country (Bosnia and Herzegovina) in the 20
- Question : 164 - (a) & (b) The graphs are below. Bars are shown in alphabetical order by city name (as the data were given in the table). (c) For Baltimore, for example, this rate is 5091 651 .= 7.82. The complete table is shown on the right. (d) & (e) Graphs below. Note that the text does not specify whether the bars should be ordered by increasing or decreasing rate. (f) Preferences may vary, but the ordered bars make comparisons easier.
- Question : 165 - The given description is true on the average, but the curves (and a few calculations) give a more complete picture. For example, a score of about 675 is about the 97.5th percentile for both genders, so the top boys and girls have very similar scores.
- Question : 166 - (a) & (b) Answers will vary. Definitions might be as simple as
- Question : 167 - Shown is a stemplot; a histogram should look similar to this. This distribution is relatively symmetric apart from one high outlier. Because of the outlier, the five-number summary (in hours) is
- Question : 168 - Gender and automobile preference are categorical; age and household income are quantitative.
- Question : 169 - Either a bar graph or a pie chart could be used. The given numbers sum to 66.7, so the
- Question : 170 - Women
- Question : 171 - (a) For car makes (a categorical variable), use either a bar graph or pie chart. For car age (a quantitative variable), use a histogram, stemplot, or boxplot. (b) Study time is quantitative, so use a histogram, stemplot, or boxplot. To show change over time, use a time plot (average hours studied against time). (c) Use a bar graph or pie chart to show radio station preferences. (d) Use a Normal quantile plot to see whether the measurements follow a Normal distribution.
- Question : 172 - The counts given add to 6067, so the others received 626 spam messages. Either a bar graph or a pie chart would be appropriate. What students learn from this graph will vary; one observation might be that AA and BB (and perhaps some others) might need some advice on how to reduce the amount of spam they receive.
- Question : 173 - No, and no: It is easy to imagine examples of many different data sets with mean 0 and standard deviation 1
- Question : 174 - The time plot is shown below; because of the great detail in this plot, it is larger than other plots. Ruth
- Question : 175 - Bonds
- Question : 176 - Recall the text
- Question : 177 - Results will vary. One set of 20 samples gave the results at the right (Normal quantile plots are not shown). Theoretically, x will have a Normal distribution with mean 25 and standard deviation 8/ ? 30 .= 1.46, so that about 99.7% of the time, one should find x between 20.6 and 29.4. Meanwhile, the theoretical distribution of s is nearly Normal (slightly skewed) with mean .= 7.9313 and standard deviation .= 1.0458; about 99.7% of the time, s will be between 4.8 and 11.1.

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