Solutions

65

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.

Answers will vary. One possible histogram is shown below.

Figure 2.57

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 2.58

1. The 40^th percentile is 37 years.
2. The 78^th percentile is 70 years.

Jesse graduated 37^th 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. $\frac{x+.5y}{n}$(100) = $\frac{143+.5(1)}{180}$(100) = 79.72. Jesse’s rank of 37 puts him at the 80^th percentile.

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 percent of runners ran at speeds of 7.5 miles per hour or less (slower), and 60 percent of runners ran at speeds of 7.5 miles per hour or more (faster).

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

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 percent of the crash-tested cars had damage repair costs of $1,700 or less; only 10 percent had damage repair costs of $1,700 or more.

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

4

6 – 4 = 2

6

More than 25 percent 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 percent and the bottom 25 percent are spread out evenly; the whiskers have the same length.

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;

$\frac{738}{27}$ = 27.33

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

4

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.

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.

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

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

The mean is 4.1 and is slightly greater than the median, which is 4.

The mode and the median are the same. In this case, both 5.

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

Both the mean and the median are 6.

The mode is 12, the median is 13.5, and the mean is 15.1. The mean is the largest.

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

sampling variability

induced variability

measurement variability

natural variability

s = 34.5

For Fredo: z = $\frac{.158\text{ – }.166}{.012}$ = –0.67.

For Karl: z = $\frac{.177\text{ – }.189}{.015}$ = –.8.

Fredo’s z score of –.67 is higher than Karl’s z score of –.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team.

1. $s_x=\sqrt{\frac{{\displaystyle \sum f{m}^2}}{n}-{\overline{x}}^2}=\sqrt{\frac{\mathrm{193,157.45}}{30}-{79.5}^2}=10.88$
2. $s_x=\sqrt{\frac{{\displaystyle \sum f{m}^2}}{n}-{\overline{x}}^2}=\sqrt{\frac{\mathrm{380,945.3}}{101}-{60.94}^2}=7.62$
3. $s_x=\sqrt{\frac{{\displaystyle \sum f{m}^2}}{n}-{\overline{x}}^2}=\sqrt{\frac{\mathrm{440,051.5}}{86}-{70.66}^2}=11.14$

1. Example solution for using the random number generator for the TI-84+ to generate a simple random sample of eight 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 2.60
2. 
   

   Figure 2.61

3. Figure 2.62

c

Answers will vary.

1. 1 – (.02+.09+.19+.26+.18+.17+.02+.01) = .06
2. .19+.26+.18 = .63
3. Check student’s solution.

4. 40^th percentile will fall between 30,000 and 40,000

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

5. Check student’s solution.

1. more children; the left whisker shows that 25 percent of the population are children 17 and younger; the right whisker shows that 25 percent of the population are adults 50 and older, so adults 65 and over represent less than 25 percent
2. 62.4 percent

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.

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 percent of buyers are more variable than the ages of the lower 50 percent.
2. The black sports car is most likely to have an outlier. It has the longest whisker.
3. Comparing the median ages, younger people tend to buy the black sports car, while older people tend to buy the white sports car. 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 are 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 percent fall between 41 and 64. Since 25 percent of values fall between 31 and 38, we know that fewer than 25 percent fall between 31 and 35.

the mean percentage, $\overline{x}=\frac{\mathrm{1,328.65}}{50}=26.75$

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 sixth number in order. Six years will have totals at or below the median.

474 FTES

919

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

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

For pianos, the cost of the piano is .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.

• $\overline{x}=23.32$
• Using the TI 83/84, we obtain a standard deviation of: $s_x=\mathrm{12.95.}$
• The obesity rate of the United States is 10.58 percent higher than the average obesity rate.
• Since the standard deviation is 12.95, we see that 23.32 + 12.95 = 36.27 is the disease percentage that is one standard deviation from the mean. The U.S. disease rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States, although 34 percent have the disease, does not have an unusually high percentage of people with the disease.

1. For graph, check student's solution.
2. 49.7 percent of the community is under the age of 35
3. Based on the information in the table, graph (a) most closely represents the data.

a

b

1. 1.48
2. 1.12

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
6. 205.5, 272.5
7. sample
8. population
9. 1. 236.34
   2. 37.50
   3. 161.34
   4. .84 standard deviations below the mean
10. young

1. true
2. true
3. true
4. false

1. Enrollment	Frequency
   1,000–5,000	10
   5,000–10,000	16
   10,000–15,000	3
   15,000–20,000	3
   20,000–25,000	1
   25,000–30,000	2

   Table 2.91
2. Check student’s solution.
3. mode
4. 8,628.74
5. 6,943.88
6. –0.09

a