Chapter 2 Full Solutions

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Section 2.1

Check Your Understanding, page 95: 1. c 2. Her daughter weighs more than 87% of girls her age and she is taller than 67% of girls her age. 3. About 65% of calls lasted less than 30 minutes. This means that about 35% of calls lasted 30 minutes or longer. 4. The first quartile (25th percentile) is at about Q1 = 13 minutes. The third quartile (75th percentile) is at about Q3 = 32 minutes. This suggests that the IQR =−=32 13 19 minutes.

Check Your Understanding, page 97: − 6762 1. z = −= 166.1 . Interpretation: Lynette’s height is 1.166 standard deviations below the mean height of 29.4 the class. 74− 76 2. Because Brent’s z-score is −0.85, we know that −=0.85 . Solving for σ we find that σ = 2.35 inches. σ

Check Your Understanding, page 103: 1. Converting the cost of the rides from dollars to cents will not change the shape. However, it will multiply the mean and the standard deviation by 100. 2. Adding 25 cents to the cost of each ride will not change the shape of the distribution, nor will it change the variability. It will, however, add 25 cents to the measures of center (mean, median). 3. Converting the costs to z-scores will not change the shape of the distribution. It will change the mean to 0 and the standard deviation to 1.

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Section 2.1 Exercises

2.1 (a) Because 17 of the 20 students (85%) own fewer pairs of shoes than Jackson (who owns 22 pairs of shoes), Jackson is at the 85th percentile in the number of pairs of shoes distribution.

(b) Interpretation: 45% of the boys had fewer pairs of shoes than Raul did. Raul was at the 45th percentile. This means that 45% of the 20 boys, or 9 boys, have fewer pairs of shoes. Therefore, Raul’s response is the 10th value in the ordered list. Putting the data in order we get: 4, 5, 5, 5, 6, 7, 7, 7, 7, 8, 10, 10, 10, 10, 11, 12, 14, 22, 35, and 38. Raul owns 8 pairs of shoes.

2.2 (a) Because 4 of the 50 states (8%) have smaller values for the percent of residents aged 65 and older than Colorado, Colorado is at the 8th percentile in the distribution. Interpretation: 8% of states have a smaller percent of residents aged 65 and older than the state of Colorado. (b) Interpretation: 80% of states have a smaller percent of residents aged 65 and older than the state of Rhode Island.

Rhode Island is at the 80th percentile. This means that 80% of the 50 states, or 40 states, have a smaller percent of residents aged 65 and older. Therefore, Rhode Island is the 41st value in the ordered list. Counting the observations in the stemplot allows us to determine that 13.9% of Rhode Island’s residents are aged 65 and older.

2.3 (a) The head circumferences (sorted in ascending order) are 20.8, 20.8, 21, 21.5, 21.6, 21.7, 21.9, 22, 22.2, 22.3, 22.4, 22.5, 22.5, 22.6, 22.6, 22.7, 22.7, 22.7, 23, 23, 23.1, 23.3, 23.4, 23.5, 23.5, 23.9, 23.9, 24, 24.2, 25.6. Because 10 of the 30 observations (33.3%) are below Antawn’s head circumference (22.4 inches), Antawn is at the 33.3rd percentile in the head circumference distribution. (b) The head circumferences (sorted in ascending order) are 20.8, 20.8, 21, 21.5, 21.6, 21.7, 21.9, 22, 22.2, 22.3, 22.4, 22.5, 22.5, 22.6, 22.6, 22.7, 22.7, 22.7, 23, 23, 23.1, 23.3, 23.4, 23.5, 23.5, 23.9, 23.9, 24, 24.2, 25.6. In this case, the 90th percentile means that 90% of the 30 players, or 27 players, will have a smaller head circumference than this player. Therefore, the player at the 90th percentile will have a head circumference that is the 28th value in the ordered list, or equivalently, the third largest head circumference. The player with a head circumference of 24 inches is at the 90th percentile of the distribution.

2.4 (a) The number of text messages sent (sorted in ascending order) are: 0, 0, 0, 1, 1, 3, 3, 5, 5, 7, 8, 8, 9, 14, 25, 25, 26, 29, 42, 44, 52, 72, 92, 98, 118. Because 18 of the 25 students (72%) sent fewer text messages than Sunny (who sent 42 texts), Sunny is at the 72nd percentile in the distribution of number of text messages sent in the past 24 hours. (b) The number of text messages sent (sorted in ascending order) are: 0, 0, 0, 1, 1, 3, 3, 5, 5, 7, 8, 8, 9, 14, 25, 25, 26, 29, 42, 44, 52, 72, 92, 98, 118. Joelle is at the 12th percentile. This means that 12% of the 25 students in the class, or 3 students sent fewer text messages than she did. Therefore, the number of texts Joelle sent will be the 4th value in the ordered list. Joelle sent 1 text message in the past 24 hours.

2.5 This means that the speed limit is set at such a speed that 85% of the vehicle speeds are slower than the posted speed.

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2.6 Larry’s wife should tell him that being at the 90th percentile for blood pressures is not a good thing. It means that 90% of men like him have a lower blood pressure than him! When it comes to blood pressure, a high number is not desirable. Larry may need treatment for his high blood pressure.

2.7 The girl in question weighs more than 48% of girls her age, but is taller than 78% of the girls her age. Because she is taller than 78% of girls, but only weighs more than 48% of girls, she is probably fairly thin.

2.8 Peter’s time was slower than 80% of his previous race times that season, but it was slower than only 50% of the racers at the league championship meet. Because this time was relatively slow for Peter but at the median for the runners in the league championship, Peter must be a good runner.

2.9 (a) No. A sprint time of 8 seconds is not unusually slow. A student with an 8 second sprint is at the 75th percentile, so 25% of the students took that long or longer. (b) The 20th percentile of the distribution is approximately 6.7 seconds. Interpretation: 20% of the students completed the 50-yard sprint in less than 6.7 seconds.

2.10 (a) No. North Dakota, with a median household income of $55,766, is not an unusually wealthy state. North Dakota is at the 65th percentile of median household income. That means that 35% of states have a median household income as large as or larger than North Dakota. (b) The 90th percentile of the distribution of median household income is approximately $65,000. Interpretation: 90% of the states have a median household income of less than $65,000.

2.11 (a) The IQR is calculated by subtracting Q1 from Q3. The first quartile is the 25th percentile. Find 25 on the y- axis, read over to the line and then down to the x-axis to get about Q1 = 4%. The 3rd quartile is the 75th percentile. Find 75 on the y-axis, read over to the line and then down to the x-axis to get about Q3 = 14%. Therefore, the IQR is approximately 14 – 4 = 10%. (b) Arizona, which had 15.1% foreign-born residents that year, is at the 90th percentile. (c) The graph is fairly flat between 20% and 27.5% foreign-born residents because there were very few states that had 20% to 27.5% foreign-born residents that year. (d)

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2.12 (a) The IQR is calculated by subtracting Q1 from Q3. The first quartile is the 25th percentile. Find 25 on the y- axis, read over to the line and then down to the x-axis to get about Q1 = $19. The 3rd quartile is the 75th percentile. Find 75 on the y-axis, read over to the line and then down to the x-axis to get about Q3 = $46. Therefore, the IQR is approximately $46−= $19 $27. (b) The person who spent $19.50 is just above what we have called the 25th percentile. It appears that $19.50 is at about the 26th percentile. (c) The graph is steepest between $10 and $30 because more shoppers spent amounts in this interval than any other interval. (d) The graph is below:

2.13 − 73.89.1 (a) The z-score for Montana is z = −= 12.1 . Interpretation: Montana’s percent of foreign-born 12.6 residents is 1.12 standard deviations below the mean percent of foreign-born residents for all states. (b) If we let x denote the percent of foreign-born residents in New York at that time, then we can solve for x in the x − 73.8 equation 10.2 = . Thus, x = 21.582% foreign-born residents. 12.6

2.14 − 44.742,51553,41 (a) The z-score for North Carolina is z = −= 24.1 . Interpretation: North Carolina’s median 64.210,8 household income is 1.24 standard deviations below the average median household income for all states. (b) If we let x denote the median household income for New Jersey, then we can solve for x in the equation x − 44.742,51 82.1 = . Thus, x = $66,685.80 is the median household income for New Jersey. 64.210,8

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2.15 (a) Interpretation: The number of pairs of shoes owned by Jackson is 1.10 standard deviations above the average number of pairs of shoes owned by the students in the sample. (b) If we let x denote the mean number of pairs of shoes owned by students in the sample, then we can solve for 22 − x x in the equation 10.1 = . Thus, x = 11.64 is the mean number of pairs of shoes owned by the students 9.42 in the sample.

2.16 (a) Interpretation: The number of texts sent by Alejandro is 1.89 standard deviations above the average number of texts sent by the students in the sample. (b) If we let x denote the mean number of texts sent in the sample, then we can solve for x in the equation 92 − x 89.1 = . Thus, x = 27.46 is the mean number of texts sent by the students in the sample. 34.15

2.17 (a) The fact that your standardized score is negative indicates that your bone density is below the average for your peer group. In fact, your bone density is about 1.5 standard deviations below average among 25-year-old women. (b) If we let σ denote the standard deviation of the bone density in Judy’s reference population, then we can solve 948− 956 for σ in the equation −=1.45 . Thus, σ = 5.52 g/cm2. σ

2.18 (a) Mary’s z-score (0.5) indicates that her bone density score is about half a standard deviation above the average score for all women her age. Because her z-score is higher than Judy’s (z = –1.45), Mary’s bones are healthier in comparison to other women in their age groups. (b) If we let σ denote the standard deviation of the bone density in Mary’s reference population, then we can solve 948− 944 for σ in the equation 0.5 = . Thus, σ = 8 g/cm2. There is more variability in the bone densities for older σ women. This isn’t surprising because as women get older there is more time for their good or bad health habits to have an effect, creating a wider range of bone densities.

2.19 680− 500 29− 21 Eleanor’s standardized score, z = =1.8 is higher than Gerald’s standardized score, z = =1.6 100 5

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2.20 The standardized batting averages (z-scores) for these three outstanding hitters are: Player z-score 0.420− 0.266 Cobb z = = 4.15 0.0371 0.406− 0.267 Williams z = = 4.26 0.0326 0.390− 0.261 Brett z = = 4.07 0.0317 All three hitters were at least 4 standard deviations above their peers, but Williams’ z-score is the highest.

2.21 (a) The shape of the distribution of corrected long jump distance will be the same as the original distribution of long jump distance: roughly symmetric with a single peak. (b) The mean of the distribution of corrected long jump distance is 577.3 – 20 = 557.3 centimeters. The median of the distribution of corrected long jump distance is 577 – 20 = 557 centimeters. Subtracting a constant from each value in a distribution decreases the measures of center by the same amount. (c) The standard deviation of the distribution of corrected long jump distance is the same as the standard deviation of the distribution of long jump distance, 4.713 centimeters. The IQR of the distribution of corrected long jump distance is the same as the IQR of the distribution of long jump distance, 7 centimeters. The standard deviation and IQR will not change because subtracting a constant from each value in a distribution does not change the variability.

2.22 (a) The shape of the distribution of distance will be the same as the original distribution of heights: slightly skewed right with several peaks. (b) The mean of the distribution of distance is 67 + 6 = 73 inches. The median of the distribution of distance is 66 + 6 = 72 inches. Adding a constant to each value in a distribution increases the measures of center by the same amount. (c) The standard deviation of the distribution of distance is the same as the standard deviation of the distribution of heights, 4.29 inches. The IQR of the distribution of distance is the same as the IQR of the distribution of heights, 7 inches. The standard deviation and IQR will not change because adding a constant to each value in a distribution does not change the variability.

2.23 (a) The shape of the new salary distribution will be the same as the shape of the original salary distribution. (b) The mean and median salaries will each increase by $1000. Adding a constant to each value in a distribution increases the measures of center by the same amount. (c) The standard deviation and IQR of the new salary distribution will each be the same as they were for the original salary distribution. The standard deviation will not change because adding a constant to each value in a distribution does not change the variability.

2.24 (a) The shape of the new price distribution will be the same as the shape of the original price distribution. (b) The mean and median of the new price distribution will each be $500 lower than they were for the original price distribution. Subtracting a constant from each value in a distribution decreases the measures of center by the same amount. (c) The standard deviation and IQR of the new price distribution will each be the same as they were for the original price distribution. The standard deviation and IQR will not change because subtracting a constant from each value in a distribution does not change the variability.

2.25 (a) The shape of the distribution of corrected long jump distance in meters will be the same as the distribution of corrected long jump distance in centimeters: roughly symmetric with a single peak. (b) The mean of the distribution of corrected long jump distance, in meters, is 577.3 – 20 = 557.3 cm ÷ 100 cm/m = 5.573 meters. (c) The standard deviation of the distribution of corrected long jump distance, in meters, is 4.713 centimeters ÷ 100 = 0.04713 meters.

2.26 (a) The shape of the distribution of distance, in feet, will be the same as the original distribution of height: slightly skewed right with several peaks. (b) The mean of the distribution of distance, in feet, is 67 + 6 = 73 inches ÷ 12 in/ft = 6.083 feet. (c) The standard deviation of the distribution of distance, in feet, is 4.29 inches ÷ 12 = 0.358 feet.

2.27 (a) The shape of the resulting salary distribution will be the same as the original distribution of salaries. (b) The median will increase by 5% because each value in the distribution is being multiplied by 1.05. Multiplying each value in a distribution by a constant multiplies measures of center by the same amount. (b) The IQR will increase by 5% because each value in the distribution is being multiplied by 1.05. Multiplying each value in a distribution by a constant multiplies measures of variability by the same amount.

2.28 (a) The shape of the resulting price distribution will be the same as the shape of the original price distribution. (b) The median of the new price distribution will be 0.90 times the median for the original price distribution because each value in the distribution is being multiplied by 0.90. Multiplying each value in a distribution by a constant multiplies measures of center by the same amount. (c) The IQR of the new price distribution will be 0.90 times the IQR for the original price distribution because each value in the distribution is being multiplied by 0.90. Multiplying each value in a distribution by a constant multiplies measures of variability by the same amount.

2.29 5 160 (a) To find the mean temperature in degrees Celsius, multiply the mean in Fahrenheit by and subtract . 9 9 5 160 Thus, the mean temperature reading in degrees Celsius = ()77 =− 25 degrees Celsius. 9 9 5 160 (b) To calculate the standard deviation, we just multiply by since subtracting just shifts the distribution 9 9 and does not affect the variability. Thus, standard deviation of the temperature reading in degrees Celsius = 5 ()3= 1.667 degrees Celsius. 9

2.30 (a) To find the correct measurement in centimeters we need to subtract the 0.2 inches that Clarence mistakenly added to each value. Then to transform that measurement to centimeters we need to multiply by 2.54. So, the mean of the corrected measurements, in centimeters, is ()3.2−= 0.2 (2.54) 7.62 cm. (b) To calculate the standard deviation of the corrected measurements in centimeters, we just multiply the old standard deviation by 2.54 because subtracting a constant from each value in a distribution does not affect the standard deviation. The standard deviation of the corrected measurements is 0.1() 2.54= 0.254 cm.

2.31 We are told that Mean(Fare) = 2.85 + 2.7 Mean(miles). To determine the mean of the lengths of his cab rides in miles we substitute $15.45 for the mean fare and solve for the mean number of miles. 15.45 = 2.85 + 2.7 Mean(miles) Mean(miles) = 4.667 miles To determine the standard deviation of the lengths of his cab rides in miles, we use the equation, SD(Fare) = 2.7 SD(miles). We substitute $10.20 for “SD(Fare)” and solve for “SD(miles)”. 10.20 = 2.7 SD(miles) SD(miles) = 3.778 miles The mean and standard deviation of the lengths of his cab rides, in miles, are 4.667 miles and 3.778 miles, respectively.

2.32 To make the standard deviation increase from 3 to 12, multiply each score by 4. This will make the standard deviation equal to 4(3) = 12 and the mean equal to 4(12) = 48. To make the mean increase from 48 to 75, add 27 to each adjusted score. Adding 27 does not change the variability, so the mean will be 48 + 27 = 75 and the standard deviation will still be 12.

2.33 c 2.34 a 2.35 d 2.36 b 2.37 c 2.38 d

2.39 The distribution is skewed to the right. The two largest values appear to be outliers. The data is centered roughly around a median of 15 minutes and the inter-quartile range is approximately 10 minutes.

2.40 (a) A bar graph is given below.

(b) Because 3/50 = 6% of the sample was left-handed, our best estimate of the percentage of the population that is left-handed is 6%.

Chapter 2: Modeling Distributions of Data 9