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.
Find the midpoint for each class. These will be graphed on the xaxis. The frequency values will be graphed on the yaxis values.
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. $\frac{x+0.5y}{n}$(100) = $\frac{143+0.5(1)}{180}$(100) = 79.72. Jesse’s rank of 37 puts him at the 80^{th} percentile.
 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.
 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).
When waiting in line at the DMV, the 85^{th} 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.
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.
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.
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 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.
For Fredo: z = $\frac{0.158\text{\u2013}0.166}{0.012}$ = –0.67
For Karl: z = $\frac{0.177\text{\u2013}0.189}{0.015}$ = –0.8
Fredo’s zscore of –0.67 is higher than Karl’s zscore of –0.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team.
 ${s}_{x}=\sqrt{\frac{{\displaystyle \sum f{m}^{2}}}{n}{\stackrel{\u2013}{x}}^{2}}=\sqrt{\frac{193157.45}{30}{79.5}^{2}}=10.88$
 ${s}_{x}=\sqrt{\frac{{\displaystyle \sum f{m}^{2}}}{n}{\stackrel{\u2013}{x}}^{2}}=\sqrt{\frac{380945.3}{101}{60.94}^{2}}=7.62$
 ${s}_{x}=\sqrt{\frac{{\displaystyle \sum f{m}^{2}}}{n}{\stackrel{\u2013}{x}}^{2}}=\sqrt{\frac{440051.5}{86}{70.66}^{2}}=11.14$
 Example solution for using the random number generator for the TI84+ 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}.
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 
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 
 See Table 2.87 and Table 2.88.
 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).
 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).
 Compare the two graphs:
 Answers may vary. Possible answers include:
 Both graphs have a single peak.
 Both graphs use class intervals with width equal to $50.
 Answers may vary. Possible answers include:
 The couples graph has a class interval with no values.
 It takes almost twice as many class intervals to display the data for couples.
 Answers may vary. Possible answers include: The graphs are more similar than different because the overall patterns for the graphs are the same.
 Answers may vary. Possible answers include:
 Check student's solution.
 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.
 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.
 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.
 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.
 1 – (0.02+0.09+0.19+0.26+0.18+0.17+0.02+0.01) = 0.06
 0.19+0.26+0.18 = 0.63
 Check student’s solution.
40^{th} percentile will fall between 30,000 and 40,000
80^{th} percentile will fall between 50,000 and 75,000
 Check student’s solution.
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.
 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 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.
 $\stackrel{\u2013}{x}=23.32$
 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% higher than the average obesity rate.
 Since the standard deviation is 12.95, we see that 23.32 + 12.95 = 36.27 is the obesity percentage that is one standard deviation from the mean. The United States obesity rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States, while 34% obese, does not hav e an unusually high percentage of obese people.
 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
 241
 205.5
 272.5
 205.5, 272.5
 sample

 236.34
 37.50
 161.34
 0.84 std. dev. below the mean
 Young

Enrollment Frequency 10005000 10 500010000 16 1000015000 3 1500020000 3 2000025000 1 2500030000 2  Check student’s solution.
 mode
 8628.74
 6943.88
 –0.09