Introductory Statistics 2e

# Solutions

1.
Stem Leaf
19 9 9
20 1 1 5 5 5 6 6 8 9
31 1 2 2 3 4 5 6 7 7 8 8 8 8
41 3 3
Table 2.85
3.
Stem Leaf
25 5 6 7 7 8
30 0 1 2 3 3 5 5 5 7 7 9
41 6 9
56 7 7
61
Table 2.86
5.
Figure 2.52
7.
Figure 2.53
9.
Figure 2.54
11.
Figure 2.55
13.

65

15.

The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value.

17.

Answers will vary. One possible histogram is shown:

Figure 2.56
19.

Find the midpoint for each class. These will be graphed on the x-axis. The frequency values will be graphed on the y-axis values.

Figure 2.57
21.
Figure 2.58
23.
1. The 40th percentile is 37 years.
2. The 78th percentile is 70 years.
25.

Jesse graduated 37th out of a class of 180 students. There are 180 – 37 = 143 students ranked below Jesse. There is one rank of 37.

x = 143 and y = 1. $x+0.5y n x+0.5y n$(100) = $143+0.5(1) 180 143+0.5(1) 180$(100) = 79.72. Jesse’s rank of 37 puts him at the 80th percentile.

27.
1. For runners in a race it is more desirable to have a high percentile for speed. A high percentile means a higher speed which is faster.
2. 40% of runners ran at speeds of 7.5 miles per hour or less (slower). 60% of runners ran at speeds of 7.5 miles per hour or more (faster).
29.

When waiting in line at the DMV, the 85th percentile would be a long wait time compared to the other people waiting. 85% of people had shorter wait times than Mina. In this context, Mina would prefer a wait time corresponding to a lower percentile. 85% of people at the DMV waited 32 minutes or less. 15% of people at the DMV waited 32 minutes or longer.

31.

The manufacturer and the consumer would be upset. This is a large repair cost for the damages, compared to the other cars in the sample. INTERPRETATION: 90% of the crash tested cars had damage repair costs of $1700 or less; only 10% had damage repair costs of$1700 or more.

33.

You can afford 34% of houses. 66% of the houses are too expensive for your budget. INTERPRETATION: 34% of houses cost $240,000 or less. 66% of houses cost$240,000 or more.

35.

4

37.

6 – 4 = 2

39.

6

41.

More than 25% of salespersons sell four cars in a typical week. You can see this concentration in the box plot because the first quartile is equal to the median. The top 25% and the bottom 25% are spread out evenly; the whiskers have the same length.

43.

Mean: 16 + 17 + 19 + 20 + 20 + 21 + 23 + 24 + 25 + 25 + 25 + 26 + 26 + 27 + 27 + 27 + 28 + 29 + 30 + 32 + 33 + 33 + 34 + 35 + 37 + 39 + 40 = 738;

$738 27 738 27$ = 27.33

45.

The most frequent lengths are 25 and 27, which occur three times. Mode = 25, 27

47.

4

49.

The data are symmetrical. The median is 3 and the mean is 2.85. They are close, and the mode lies close to the middle of the data, so the data are symmetrical.

51.

The data are skewed right. The median is 87.5 and the mean is 88.2. Even though they are close, the mode lies to the left of the middle of the data, and there are many more instances of 87 than any other number, so the data are skewed right.

53.

When the data are symmetrical, the mean and median are close or the same.

55.

The distribution is skewed right because it looks pulled out to the right.

57.

The mean is 4.1 and is slightly greater than the median, which is four.

59.

The mode and the median are the same. In this case, they are both five.

61.

The distribution is skewed left because it looks pulled out to the left.

63.

The mean and the median are both six.

65.

The mode is 12, the median is 12.5, and the mean is 15.1. The mean is the largest.

67.

The mean tends to reflect skewing the most because it is affected the most by outliers.

69.

s = 34.5

71.

For Fredo: z = = –0.67

For Karl: z = = –0.8

Fredo’s z-score of –0.67 is higher than Karl’s z-score of –0.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team.

73.
1. $sx =∑ fm2 n− x¯ 2= 193157.4530 −79.52 =10.88 sx =∑ fm2 n− x¯ 2= 193157.4530 −79.52 =10.88$
2. $sx =∑ fm2 n− x¯ 2 =380945.3 101− 60.942=7.62sx =∑ fm2 n− x¯ 2 =380945.3 101− 60.942=7.62$
3. $sx = ∑ fm 2n − x¯2 =440051.5 86 − 70.662 =11.14sx = ∑ fm 2n − x¯2 =440051.5 86 − 70.662 =11.14$
75.
1. Example solution for using the random number generator for the TI-84+ to generate a simple random sample of 8 states. Instructions are as follows.
• Number the entries in the table 1–51 (Includes Washington, DC; Numbered vertically)
• Press MATH
• Arrow over to PRB
• Press 5:randInt(
• Enter 51,1,8)

Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}.

Corresponding percents are {30.1, 22.2, 26.5, 27.1, 30.9, 34.0, 26.0, 25.1}.

Figure 2.59

2. Figure 2.60
3. Figure 2.61
77.
Amount($) Frequency Relative Frequency 51–100 5 0.08 101–150 10 0.17 151–200 15 0.25 201–250 15 0.25 251–300 10 0.17 301–350 5 0.08 Table 2.87 Singles Amount($) Frequency Relative Frequency
100–150 5 0.07
201–250 5 0.07
251–300 5 0.07
301–350 5 0.07
351–400 10 0.14
401–450 10 0.14
451–500 10 0.14
501–550 10 0.14
551–600 5 0.07
601–650 5 0.07
Table 2.88 Couples
1. See Table 2.87 and Table 2.88.
2. In the following histogram data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where both boundary values are included).
Figure 2.62
3. In the following histogram, the data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where values on both boundaries are included).
Figure 2.63
4. Compare the two graphs:
• Both graphs have a single peak.
• Both graphs use class intervals with width equal to \$50.
• The couples graph has a class interval with no values.
• It takes almost twice as many class intervals to display the data for couples.
3. Answers may vary. Possible answers include: The graphs are more similar than different because the overall patterns for the graphs are the same.
6. Compare the graph for the Singles with the new graph for the Couples:
• Both graphs have a single peak.
• Both graphs display 6 class intervals.
• Both graphs show the same general pattern.
1. Answers may vary. Possible answers include: Although the width of the class intervals for couples is double that of the class intervals for singles, the graphs are more similar than they are different.
7. Answers may vary. Possible answers include: You are able to compare the graphs interval by interval. It is easier to compare the overall patterns with the new scale on the Couples graph. Because a couple represents two individuals, the new scale leads to a more accurate comparison.
8. Answers may vary. Possible answers include: Based on the histograms, it seems that spending does not vary much from singles to individuals who are part of a couple. The overall patterns are the same. The range of spending for couples is approximately double the range for individuals.
79.

c

81.

83.
1. 1 – (0.02+0.09+0.19+0.26+0.18+0.17+0.02+0.01) = 0.06
2. 0.19+0.26+0.18 = 0.63
4. 40th percentile will fall between 30,000 and 40,000

80th percentile will fall between 50,000 and 75,000

85.
1. more children; the left whisker shows that 25% of the population are children 17 and younger. The right whisker shows that 25% of the population are adults 50 and older, so adults 65 and over represent less than 25%.
2. 62.4%
87.
1. Answers will vary. Possible answer: State University conducted a survey to see how involved its students are in community service. The box plot shows the number of community service hours logged by participants over the past year.
2. Because the first and second quartiles are close, the data in this quarter is very similar. There is not much variation in the values. The data in the third quarter is much more variable, or spread out. This is clear because the second quartile is so far away from the third quartile.
89.
1. Each box plot is spread out more in the greater values. Each plot is skewed to the right, so the ages of the top 50% of buyers are more variable than the ages of the lower 50%.
2. The BMW 3 series is most likely to have an outlier. It has the longest whisker.
3. Comparing the median ages, younger people tend to buy the BMW 3 series, while older people tend to buy the BMW 7 series. However, this is not a rule, because there is so much variability in each data set.
4. The second quarter has the smallest spread. There seems to be only a three-year difference between the first quartile and the median.
5. The third quarter has the largest spread. There seems to be approximately a 14-year difference between the median and the third quartile.
6. IQR ~ 17 years
7. There is not enough information to tell. Each interval lies within a quarter, so we cannot tell exactly where the data in that quarter is concentrated.
8. The interval from 31 to 35 years has the fewest data values. Twenty-five percent of the values fall in the interval 38 to 41, and 25% fall between 41 and 64. Since 25% of values fall between 31 and 38, we know that fewer than 25% fall between 31 and 35.
92.

The mean percentage, $x ¯ = 1328.65 50 =26.75 x ¯ = 1328.65 50 =26.75$

94.

The median value is the middle value in the ordered list of data values. The median value of a set of 11 will be the 6th number in order. Six years will have totals at or below the median.

96.

474 FTES

98.

919

100.
• mean = 1,809.3
• median = 1,812.5
• standard deviation = 151.2
• first quartile = 1,690
• third quartile = 1,935
• IQR = 245
102.

Hint: Think about the number of years covered by each time period and what happened to higher education during those periods.

104.

For pianos, the cost of the piano is 0.4 standard deviations BELOW the mean. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar costs the most in comparison to the cost of other instruments of the same type.

106.
• $x ¯ =23.32 x ¯ =23.32$
• Using the TI 83/84, we obtain a standard deviation of: $s x =12.95. s x =12.95.$
• The poverty rate of the United States is 10.58% higher than the average rate.
• Since the standard deviation is 12.95, we see that 23.32 + 12.95 = 36.27 is the poverty percentage that is one standard deviation from the mean. The United States rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States does not have an unusually high percentage of people experiencing poverty.
108.
1. For graph, answers may vary.
2. 49.7% of the community is under the age of 35.
3. Based on the information in the table, graph (a) most closely represents the data.
110.

a

112.

b

113.
1. 1.48
2. 1.12
115.
1. 174; 177; 178; 184; 185; 185; 185; 185; 188; 190; 200; 205; 205; 206; 210; 210; 210; 212; 212; 215; 215; 220; 223; 228; 230; 232; 241; 241; 242; 245; 247; 250; 250; 259; 260; 260; 265; 265; 270; 272; 273; 275; 276; 278; 280; 280; 285; 285; 286; 290; 290; 295; 302
2. 241
3. 205.5
4. 272.5
5. 205.5, 272.5
6. sample
1. 236.34
2. 37.50
3. 161.34
4. 0.84 std. dev. below the mean
7. Young
117.
1. True
2. True
3. True
4. False
119.
1. EnrollmentFrequency
1000-500010
5000-1000016
10000-150003
15000-200003
20000-250001
25000-300002
Table 2.89
3. mode
4. 8628.74
5. 6943.88
6. –0.09
121.

a