Maintaining Mathematical Proficiency (p. 3) 1. After School Activities. Pets Frequency 1 1 3 7 Number of activities 3. Students Favorite Subjects Math English Science History Frequency 1 1 1 3 Number of pets. Because there are responses, the sum of the frequencies is a + b =. If maybe is an option, the sum of the frequencies is a + b. Mathematical Practices (p. ) 1. Sample answer: U.S. Female Population 11.1 Explorations (p. ) 1. a. The total of the weights of the football players is about 1() + 19() + () + 1() + () + 3() + (3) + (3) + () + 7() + + 9() + 3(3) + 31() + 3() + 33 = 13,7 pounds. The mean of the weights of the football players is about 13,7 3.7, or about pounds. 33 = 9 1 = The weights of the football players vary from the mean of about pounds by up to 9 pounds. The total of the weights of the baseball players is about 1 + 17() + 1() + 19() + (9) + 1() + () + 3(3) + = 9 pounds. The mean of the weights of the football players is about 9 = 7., or about pounds. = 1 = The weights of the baseball players vary from the mean of about pounds by up to pounds. b. The weights of the football players vary from the mean much more than the weights of the baseball players. c. Each football position seems to be clustered around a similar weight. So, there does appear to be a correlation between body weights and the positions of players in professional football. Each baseball position is more spread out. So, there does not appear to be a correlation in professional baseball. Population (millions) Frequency (millions) 3 3 1 3 3 1 1 1 9 3 9 7 7 9 Ages (years) U.S. Population Male Female 1 1 9 3 9 7 7 9 Ages (years). Weights of Players on a Basketball Team 17 19 1 3 7 9 31 33 Weight (pounds) Power forwards Small forwards Centers Point guards Shooting guards It appears as though guards tend to be lighter and forwards tend to be heavier. So, there does appear to be a correlation between the body weights and the positions of players in professional football. 3. In order to describe the variation of a data set, you can describe how the data vary from the mean.. Sample answer: Histograms display frequencies in intervals, and double bar graphs display frequencies for each gender side-by-side. The populations are roughly equal for the first four age groups, but decrease significantly for the last two age groups and the female population is greater than the male population for the last two age groups. Copyright Big Ideas Learning, LLC Algebra 1 71
11.1 Monitoring Progress (pp. 9) 1. a. Mean: x = 1. +.7 +. +... +. + 9. +. 9. 9 9. So, the of the mean decreases slightly. Median:.,.,.,.,.,.7, 9., 9., 1.. The middle, or median, is.. So, the of the median decreases slightly. Mode:.,.,.,.,.,.7, 9., 9., 1... There are two modes now,. and., both of which are less than the mean and median. b. The mean is greater than most of the data and the modes are less than most of the data. So, the median best represents the data.. a. The $7, is much greater than the other wages. So, it is the outlier. Mean: with outlier: 3 + 1 + 3 + 7 + + 3 + + 3 x 33 = $.7 thousand, or $,7 without outlier: 3 + 1 + 3 + + 3 + + 3 x 7 $3.71 thousand, or about $3,71 So, the outlier increases the mean by $,7 $3,71 = $11. Median: with outlier: 3, 3, 3, 3, 1,,, 7 3 + 1 = $39. thousand, or $39, without outlier: 3, 3, 3, 3, 1,, $3 thousand, or $3, So, the outlier increases the median by $39, $3, = $1. Mode: with outlier: 3, 3, 3, 3, 1,,, 7 $3 thousand, or $3, without outlier: 3, 3, 3, 3, 1,, $3 thousand, or $3, So, the outlier does not affect the mode. b. Sample answer: The outlier could be the salary of the manager. 3. Show A: 19,,, 1,, 7, 7, 9, 9, 3, 31 So, the new range is still 31 19, or 1 years. Show B: 19,, 1,,,,,, 7, 7, 3 So, the new range is 3 19, or 13 years. For Show A the greatest and least s are the same. So, the range did not change. However, for Show B, the oldest contestant was voted off. So, the range is smaller, meaning the ages of the contestants are less spread out.. x = + + +... + + 1 + 1 31 1 = x x x x (x x ) 1 1 3 1 7 1 1 3 3 19 7 9 7 1 1 1 1 1 1 1 + 3 + 1 +... + 1 + + 1 7 1 7.7 = 7 1.3 The standard deviation is about 7.7 years. This means that the typical age of a contestant on Show B differs from the mean by about 7.7 years.. The standard deviation for Show B is greater. So, the ages are more spread out. 7 Algebra 1 Copyright Big Ideas Learning, LLC
9 1 + 1 + 1 + 1 1 + 1 1 + 9. Mean: 11.1 = 1. 1. + 1. = 3.3 So, the mean of the new data set is 3.3 miles. Median: 9, 1 1, 1, 1 1, 1 9, 1 1 + 1 1 1. = 1. 1. + 1. =.9 So, the median of the new data set is.9 miles. Mode: 9, 1 1, 1, 1 1, 1 9, 1 None of the data s are repeated. So, the data set has no mode. Range: 9, 1 1, 1, 1 1, 1 9, 1 1 9 =..9 = 3.3 So, the range of the old data set is 3.3 miles, and the range of the new data set is the same. Standard deviation: x x x x (x x ) 1.9 1... 1. 1.... 1..3. 1. 1..3.1 1. 1....9 1..9.9. +. +. +.1 +. +.9 7.17 1. 7.17 1.9 The standard deviation of the old data set is about 1.9 miles, and the standard deviation of the new data set is the same. 11.1 Exercises (pp. 9 9) Vocabulary and Core Concept Check 1. The measure of center represents the center, or typical, of the data. The measure of variation represents the distribution of the data, or how much the data s vary from the center. 3. Sample answer: The data set 3,,, 7, 9, 9 has two modes. The modes are and 9.. An advantage of using the range to describe a data set is that it is easy to calculate. The standard deviation, however, is considered a more reliable measure of variation than the range because all of the s of a data set are used in calculating the standard deviation. Monitoring Progress and Modeling with Mathematics. a. Mean: x 3 + + 1 + + 1 + 1 + + 3 + 1 9 3 9 = Median: 1, 1, 1,, 3, 3,,, 1 The middle is 3. Mode: 1, 1, 1,, 3, 3,,, 1 The data that occurs most often is 1. The mean is, the median is 3, and the mode is 1. b. The median best represents the data. The mode is less than most of the data and the mean is greater than most of the data.. a. Mean: 1 + 9 + 17 + 1 + x 3 = 1. Median: 9,, 1, 1, 17 The middle is 1. Mode: 9,, 1, 1, 17 All of the data s occur once. The mean is 1., the median is 1, and the data set has no mode. b. The mean best represents the data because there are no outliers. 7. a. Mean: 13 + 3 + 1 + 19 + + + + 31 x 17 = Median: 13, 1, 19,,,, 3, 31 The mean of the two middle s is + = 1. Mode: 13, 1, 19,,,, 3, 31 All of the data s occur once. The mean is, the median is 1, and the data set has no mode. b. The mean best represents the data because there are no outliers.. When the outlier of a data set is removed, the mean gets closer to the median. Copyright Big Ideas Learning, LLC Algebra 1 73
. a. Mean: 1 + 1 + 3 + 1 + 1 + 1 + 1 + 1 + + 1 x 13 = 13. Median: 3,, 1, 1, 1, 1, 1, 1, 1, 1 The mean of the two middle s is 1 + 1 9 = 1.. Mode: 3,, 1, 1, 1, 1, 1, 1, 1, 1 The data s that occur most often are 1 and 1. The mean is 13., the median is 1., and the modes are 1 and 1. b. The median best represents the data. It is the mean of the two modes, and the mean is less than most of the data. 1 1 9. a. Mean: 3 + 3 + + 1 3 + 1 3 + + + 1 3 + 1 3 x 9 = 17 3 9 3 3 1 9 3 7 = 1 7 Median: 1 1 3, 1 3, 1 3, 1 3,,,, 1 3, 3 The middle is. Mode: 1 1 3, 1 3, 1 3, 1 3,,,, 1 3, 3 The data s that occur most often are 1 and. 3 The mean is 1 or about 1.9, the median is, and the 7 modes are 1 and. 3 b. The median best represents the data because the data are evenly distributed.. a. Mean: x = 1. +. +. +... + (.1) + 1.39 +. 1.. 1 Median: 13.7, 3.1,.1,.,.,.7, 1., 1.39,.3,.,.,. The mean of the two middle s is.7 + 1. 1.7 =.. Mode: 13.7, 3.1,.1,.,.,.7, 1., 1.39,.3,.,.,. All of the data s occur once. The mean is about., the median is., and the data set has no mode. b. The median best represents the data because the mean is less than most of the data and there is no mode. c. Mean: x = 1. +. +. +... + 1.39 +. +. 13.. 13 Median: 13.7, 3.1,.1,.,.,.7, 1., 1.39,.3,.,.,.,. The middle is 1.. Mode: 13.7, 3.1,.1,.,.,.7, 1., 1.39,.3,.,.,.,. All of the data s occur once. The mean is about., the median is 1., and the data set has no mode. Because $. is greater than the mean and median, they both increase. 11. + + 9 + 7 + + x = 3 + x = (3 + x) = 3 + x = 3 3 3 x = The of x is. 1. + ( ) + ( 7.) + x 1. = 11. + x = 11. ( + x) = 11. + x = + + x = 1 The of x is 1. 13. 9,, 1, x,, 1 + x = 1 (1 + x) = 1 1 + x = 1 1 x = 1 The of x is 1. 7 Algebra 1 Copyright Big Ideas Learning, LLC
1. 3,, x, + x = 1 ( + x) = 1 + x = x = 7 The of x is 7. 1. a. The is much less than the other masses. So, it is the outlier. Mean: With outlier: + 3 + + 3 + 71 + + 3 + 31 x 7 = 39. Without outlier: + 3 + + 3 + 71 + 3 + 31 x 7 1 = 7 So, the outlier decreases the mean by 39. =.. Median: With outlier:,, 31, 3, 3,, 71, 3 The mean of the two middle s is 3 + 3 7 = 31. Without outlier:, 31, 3, 3,, 71, 3 The middle is 3. So, the outlier decreases the median by 3 31 = 3. Mode: With outlier:,, 31, 3, 3,, 71, 3 Without outlier:, 31, 3, 3,, 71, 3 All of the data s occur once in each set. So, the outlier does not affect the mode. b. Sample answer: The outlier could be the mass of a baby polar bear. 1. a. The is much greater than the other sizes. So, it is the outlier. Mean: With outlier: x + 3 + + + 1 + + 3 + 7 + + 1 7 = 7. Without outlier: x + 3 + + + 1 + 3 + 7 + + 1 9 9.9 So, the outlier increases the mean by about 7..9 =.3. Median: With outlier: 1, 1,,,, 3, 3,, 7, The mean of the two middle s is + 3 =.. Without outlier: 1, 1,,,, 3, 3,, 7 The middle is. So, the outlier increases the median by. =.. Mode: With outlier: 1, 1,,,, 3, 3,, 7, Without outlier: 1, 1,,,, 3, 3,, 7 The mode is in each set. So, the outlier does not affect the mode. b. Sample answer: The email could have contained a picture. 17. Golfer A: 3,, 7,, 9, 9, 91, 9, 9, 9 So, the range is 9 3, or 1 strokes. Golfer B: 7,, 9, 9, 91, 9, 9, 93, 9, 9 So, the range is 9 7, or strokes. The range of the scores for Golfer A is 1 strokes, and the range of scores for Golfer B is strokes. So, the scores for Golfer A are more spread out. 1. Rookie season:,, 1,, 3, So, the range is, or home runs. This season:,,, 7,, 13 So, the range is 13, or 9 home runs. The range of the monthly home run totals for the player s rookie season is home runs, and the range this season is 9 home runs. So, the monthly home run totals for the player are more spread out this season than in the player s rookie season. 19. a. 3,,,, So, the range is 3, or. b. + 3 + + + x 3 = 7 x x x x (x x ) 7 7 9 3 7 1 1 7 7 7 13 19 9 + 1 + + + 19 3 = 9.7 The standard deviation is about 9.7. Copyright Big Ideas Learning, LLC Algebra 1 7
. a. 11, 117, 11, 1, 13, 11 So, the range is 11 11, or. b. 11 + 11 + 117 + 13 + 1 + 11 x x x x x (x x ) 11 1 1 11 1 117 1 9 1 13 1 9 1 1 1 11 1 + + 1 + 1 + + 1 9. The standard deviation is about 9.. 1. a.., 1., 1., 1.,,. So, the range is.., or. b.. +. +. + 1. + 1. + 1. x x x x x (x x ). 1. 1 1. 1.... 1. 1 1 1. 1. 1. 1... 1. 1. 1 +. + 1 + +. +.. The standard deviation is about.. = 1.33 = 7 = 1 = 9 = 1. =..17. a..,., 3.3,.,., 7.,.,.1 So, the range is.1., or 7.7. b.. +.1 +. +. +. +. + 7. + 3.3 x =. x x x x (x x )...7 7.9.1.. 1.1...9.1...7.9.. 3.1 9.1...1.1 7.. 1.. 3.3... 7.9 + 1.1 +.1 +.9 + 9.1 +.1 +. +.. =.77.. The standard deviation is about.. 3. a. Golfer A: 3 + + 91 + 9 + 9 + + 9 + 9 + 7 + 9 x 9 = 9 x x x x (x x ) 3 9 7 9 9 3 91 9 1 1 9 9 9 9 9 9 9 9 9 1 1 7 9 3 9 9 9 9 + 3 + 1 + + + + + 1 + 9 + 1 = 1. 1.. So, Golfer A s typical score differs from the mean by about. strokes. 7 Algebra 1 Copyright Big Ideas Learning, LLC
b. Golfer B: 9 + 93 + 9 + + 9 + 7 + 9 + 9 + 91 + 9 x 9 = 91 x x x x (x x ) 9 91 93 91 9 91 1 1 91 3 9 9 91 7 91 1 9 91 1 9 91 3 9 91 91 9 91 1 1 + + 1 + 9 + + 1 + 1 + 9 + + 1.. = =. So, Golfer B s typical score differs from the mean by about. strokes. c. The standard deviation for Golfer A is greater. So, Golfer A s scores are more spread out, and Golfer B is more consistent.. a. Rookie season: x 1 + + + + + 3 x x x x (x x ) 1 1 1 1 3 1 1 1 + + 1 + + + 1 =.33 = 1 =. So, the typical number of home runs differs from the mean by about. home runs. b. This season: x + + + + 7 + 13 x x x x (x x ) 7 3 9 7 1 1 7 3 9 7 1 1 7 7 13 7 3 9 + 1 + 9 + 1 + + 3 = 9.33 = = 7 3. So, the typical number of home runs differs from the mean by about 3. home runs. c. The standard deviation for this season is greater. So, the monthly home run totals are more spread out this season.. Mean: x = + = Median: The middle is 3 + = 7. Mode: The data that occurs most often is 1 + =. The new mean is, the new median is 7, and the new mode is.. Mean: x = 1. (1 +.) = 1. 1. = 1.1 Median: The middle is 1 (1 +.) = 1 1. = 1.. Mode: All of the data s occur once. The new mean is 1.1, the new median is 1., and the data set still has no mode. 7. Mean: x = + 1 = 7 Median: + 1 = 9 Mode: 9 + 1 = 3 The range and standard deviation stay the same. So, the new mean is 7, the new median is 9, the new mode is 3, the range is still, and the standard deviation is still 1... Mean: x = 3(.) = 1 Median: 3(.) = 1 Mode: none Range: (.) = Standard deviation: 7.(.) = 3.3 So, the new mean is 1, the new median is 1, the data set still has no mode, the new range is, and the new standard deviation is 3.3. Copyright Big Ideas Learning, LLC Algebra 1 77
9. The numbers should be put in numerical order before finding the median., 3,,,,, 7,, The middle is. So, the median is. 3. When a number is added to each in a data set, the range stays the same. 13, 1, 7,,, 13 So, the range is 13 ( 13), or. 31. a. Mean: Team A: 17 + 13 + 173 + 1 x 7 = 171.7 Team B: 13 + 1 + 1 + 19 x = 17 Team A has the greater mean score. So, Team A wins. Median Team A: 13, 17, 173, 1 The mean of the two middle s is 17 + 173 3 = 17.. Team B: 13, 1, 1, 19 The mean of the two middle s is 1 + 1 3 = 17. Team B has a greater median score. So, the result would be different if the rule were changed to say the team with the greater median score wins. b. Team A: 13, 17, 173, 1 So, the range is 1 13, or. x = 171.7 x x x x (x x ) 17 171.7.. 13 171.7 1.7 173. 173 171.7 1. 1. 1 171.7. 1.. + 173. + 1. + 1. 33.7 1.17 1.17 9. Team A s scores differ from the mean by about 9 pins. Team B: 13, 1, 1, 19 So, the range is 19 13, or. x = 17 x x x x (x x ) 13 17 3 11 1 17 1 19 1 17 19 17 11 + 19 + + 1 = 1. Team B s scores differ from the mean by about 1. pins. So, Team B is more consistent because their range and standard deviation are less than those of Team A. c. Mean: Team A: x = 171.7 + 1 = 1.7 Team B: x = 17(1 +.1) = 17(1.1) = 191. Team B s new mean score is greater than that of Team A. So, Team B wins. 3. Even though two data sets have the same range, the numbers in between may be distributed differently. So, your friend is incorrect. 33. The data is not numeric. So, the only measure of central tendency that can be used to describe the data is the mode. 3. Dot plot B has the greatest range, 1 1 =. So, the data s are spread out the most on this dot plot, and it will have the greatest standard deviation. Dot plot C has the smallest range, 1 1 =. So, the data s are spread out the least on this dot plot, and it will have the least standard deviation. 3. Mean: x = 7(3) + = 1 + = 9 Median: 3(3) + = 9 + = Mode: 1(3) + = + = Range: 1(3) = 13 Standard deviation: 9(3) = 7 So, the new mean is 9, the new median is, the new mode is, the new range is 13, and the new standard deviation is 7. 3. All the data s could be the same. Then, the standard deviation would be. The standard deviation cannot be negative, because it is a positive square root. 7 Algebra 1 Copyright Big Ideas Learning, LLC
37. a. Answers will vary. b. Sample answer: You would expect the mean, median, range, and standard deviation to increase, but because it is highly unlikely that another student is 7 feet tall, the mode will not be affected. 3. Sample answer: Find the geometric mean of the data from Exercises and, and compare them to their respective arithmetic means. Exercise x 3 + + 1 + + 1 + 1 + + 3 + 1 3 9 9 = 9 3()(1)()(1)(1)()(3)(1) = 9 7. Exercise 1 + 9 + 17 + 1 + x 1(9)(17)(1)() = 3 = 1. 7, 1. The geometric mean is always less than or equal to the arithmetic mean. 39. a. Mean: 1(.3 ) +... + 37(.1 ) 1(7) + 19() + () + 1() + 37() 1 + 11 + + + 7 37 = 19.37 Median: The median is the mean of the th and 1st ages, both of which are 19. So, the median age is 19 + 19 3 = 19. Mode: There are 7 students who are 1 years old. This is the most common age in the class. So, the mode is 1. The mean age is 19.37, the median age is 19, and the mode is 1 years. b. Two students are 37 years old, and they are significantly older than the other students in the class. So, the outliers are 37 and 37. Mean: = 1(.3 ) +... + 1(. ) 19 1(7) + 19() + () + 1() 19 1 + 11 + + 19 3 19 19.19 Median: The median is the mean of the 99th and th ages, both of which are still 19. So, the median age is 19 + 19 still 3 = 19. Mode: There are 7 students who are 1 years old. This is still the most common age in the class. So, the mode is 1. The outliers increase the mean but do not affect the median or mode. c. College Student Ages yr: 3% 1 yr: 1% 19 yr: 3% yr: % 3 yr: 1% Mean: 19.37 + 1 =.37 Median: 19 + 1 = Mode: 1 + 1 = 19 The new mean is.37, the new median is, and the new mode is 19. Maintaining Mathematical Proficiency. x + 1 x 9 x x x + 1 9 1 1 x x x The solution is x. 1. 3(3y ) < 1 9y 3(3y) 3( ) < 1 9y 9y + < 1 9y +9y +9y < 1 The statement < 1 is never true. So, the inequality has no solution.. (c ) (c + ) (c) + ( ) (c) + () c c + c c The statement is never true. So, the inequality has no solution. Copyright Big Ideas Learning, LLC Algebra 1 79
3. (3 w) > 3(w ) (3) (w) > 3(w) 3() 1 w > 1w 1 +w +w 1 > 1w 1 +1 +1 > 1w 1 > 1w 1 3 > w The solution is w < 3.. f (x) = x f (3) = 3 = ()()() = So, f(x) = when x = 3.. f(x) = 7 x f( ) = 7 1 7 1 9 So, f(x) 1 when x =. 9. f (x) = () x f () = () = () = 3 So, f(x) = 3 when x =. 7. f (x) = (3) x f () = (3) = (1) = 1 So, f(x) = 1 when x =. 11. Explorations (p. 93) 1. a. 1 3 3 3 3 9 1 13 13 1 1 1 19 3 3 median b. 1 3 3 3 3 9 least 1 13 13 1 1 1 19 3 3 The least is, the first is 3 + 1 + 1 the third is is. = =, 3 = 17, and the greatest c. The box represents the middle half of the data, and the whiskers represent the bottom and the top quarters of the data.. A box-and-whisker plot can be used to show the median, the first and third s, and the least and greatest s of a data set. 3. a. The least is 17, the first is 19, the median is 1, the third is, and the greatest is. So, a quarter of the students have a BMI between 17 and 19, half of the ninth-grade students have a BMI between 19 and, and the last quarter have a BMI between and. b. The least is 1, the first is 1, the median is 1, the third is, and the greatest is. So, a quarter of the roller coasters have a height between 1 and 1 feet, half of the roller coasters have a height between 1 and, and the remaining quarter have a height between and. 11. Monitoring Progress (pp. 9 9) 1. lower half upper half, 1, 1, 1,,,, 3 least first third first median Least : 1 + 1 First : 3 = 1 1 + Median: 3 = 1 + Third : = Greatest : 3 greatest third median greatest The median is. 1 1 3 3 Points 7 Algebra 1 Copyright Big Ideas Learning, LLC
. The range is 3 1 = 1 years. This means that the ages vary by no more than 1 years. IQR = Q3 Q1 = 9 3 = The inter range is years. This means that the middle half of the ages vary by no more than years. 3. % of the ages are between 1 and 3, % of the ages are between 3 and 9, and % of the ages are between 9 and 3.. For both shops, the right whisker is longer than the left whisker, and most of the data are on the left side of the plot. So, both distributions are skewed right. The range and inter range of the prices at Shop A are greater than the range and inter range at Shop B. So, the surfboard prices are more spread out at Shop A. 11. Exercises (pp. 97 9) Vocabulary and Core Concept Check 1. In order to find the first of the data set, order the data, find the median of the data set, and then find the median of the lower half.. The one that is different is Find the difference of the greatest and the least of the data set. This range is 1 1 = 3. The other three ask for the inter range, which is 11 = 9. Monitoring Progress and Modeling with Mathematics 3. The least is 3.. The greatest is 1.. 11. lower half 1, 17, 1,, 1,, 3, least first median Least : 1 17 + 1 First : 3 = 17. + 1 Median: 1 =. + 3 Third : =. Greatest : 1 1 Length (inches) upper half third greatest, 3, 3,,,, 1,,,, least lower half first Least : First : 3 Median: Third : Greatest : median upper half third greatest. The third is 11.. The first is. 7. The median is.. The range is 1 3 = 11. 9. lower half upper half,, 3,,,,, least first Least : First : + 3 Median: + Third : + Greatest : 1 3 7 = = median = =. = = Hours third greatest Elevation (feet) Copyright Big Ideas Learning, LLC Algebra 1 711 1. 9, 9,, 1, 11, 1, 1, 19, 3 least lower half first Least : 9 9 + First : Median: 11 1 +19 Third : Greatest : 3 9 13 17 9 median = = = 31 = 17 Price (dollars) upper half third greatest
13.,, 1, 1, 1, 1,,,,,,, 3, 3, 3,,,,, 7 least 1. lower half first Least : First : 1 + 1 Median: + Third : 3 + Greatest : 7 = = 1 3 7 lower half median = = 1 = 7 = 3. Hours upper half third greatest, 7,,, 9,,, 1, 1, 13, 1, 1, 17, 1, least first Least : First : Median: 1 Third : 1 Greatest : 1 1 median Length (inches) upper half third greatest 1. a. The range is 1..7 = 9.. This means that the prices of the entrées vary by no more than $9.. b. % of the entrées cost between $.7 and $., % of the entrées cost between $. and $1.7, and % of the entrées cost between $1.7 and $1.. c. The inter range is 1.7. =.. This means that the middle half of the prices vary by no more than $.. d. The right whisker is longer than the left whisker. So, the data above Q3 is more spread out than the data below Q1. 1. a. The range is 17 = 17 runs. This means that the number of runs vary by no more than 17. The inter range is 9 = 7 runs. This means that the middle half of the number of runs vary by no more than 7. b. % of the number of runs scored are between and, % of the number of runs scored are between and 9, and % of the number of runs scored are between 9 and 17. c. The box is wider between Q and Q3 because Q Q1 = = and Q3 Q = 9 =. So, the data are more spread out between Q and Q3. 17. a. For Sales Rep A, the whisker lengths are equal, and the median is in the middle of the plot. So, the distribution for Sales Rep A is symmetric. For Sales Rep B, the right whisker is longer than the left whisker, and most of the data are on the left side of the plot. So, the distribution for Sales Rep B is skewed right. b. The range and inter range of the monthly car sales for Sales Rep B are greater than the range and inter range for Sales Rep A. So, the monthly car sales for Sales Rep B are more spread out. c. The least is, and it belongs to Sales Rep B. So, Sales Rep B had the single worst sales month during the year. 1. When a distribution is skewed left, most of the data are not on the left side of the plot. The distribution is skewed left. So, most of the data are on the right side of the plot. 19. The range must be greater than the inter range. So, the range is 3, and the inter range is 1.. a. Because 11 is the greatest, it is always true that the data set contains the 11. b. The data set may not contain the. Sample answer: The data could be,,,,, 1. c. The right whisker is longer, and most of the data are on the left side of the plot. So, it is always true that the distribution is skewed right. d. The mean is not always the same as the median. So, the mean may not be. 1. a. For both brands, the right whiskers are longer than the left whiskers, and most of the data are on the left side of the plots. So, the distributions for both brands are skewed right. b. For Brand A, the range of the upper 7% of battery lives is about 7 3. = 3. hours. For Brand B, the range of the upper 7% of battery lives is about.7 3. =. hours. c. The inter range of Brand A is about.7 3. = 1. hours, and the inter range of Brand B is about. 3. = 1 hour. So, the inter range of Brand A is greater. d. The range of Brand A is greater. So, Brand A probably has a greater standard deviation. e. You should buy Brand A because 7% of the battery lives are greater than 3. hours.. The data set must have 3 as its least and 1 as its greatest. Also, the first must be, the median must be., and the third must be 1. Sample answer: A data set that fits this criteria and could be represented by the box-and-whisker plot shown: 3,,,, 11, 1, 1, 1. 71 Algebra 1 Copyright Big Ideas Learning, LLC
3. Yes, it is possible for the box-and-whisker plots to be different even though they have the same median, the same inter range, and the same range. Sample answer: A B 1 1 These two plots each have a median of, a range of 1, and an inter range of. Maintaining Mathematical Proficiency. The zeros of f are p = 9 and q = 3. The x-coordinate of the vertex is 9 + 3 f (x) = (x + 9)(x 3) f ( 3) = ( 3 + 9)( 3 3) = ()( ) = 7 = = 3. So, plot the x-intercepts ( 9, ) and (3, ) as well as the vertex ( 3, 7). Also, because a <, the parabola opens down. Draw a smooth curve through the points. 3 y. The zeros are p = and q =. The x-coordinate of the vertex is + ( ) y = 3(x )(x + ) = 3( )( + ) = 3( )() = 7 x = =. So, plot the x-intercepts (, ) and (, ) as well as the vertex (, 7). Also, because a >, the parabola opens up. Draw a smooth curve through the points. y. y = x 1x y = (x x 1) y = (x )(x + ) So, the zeros are p = and q =. The x-coordinate of the vertex is + ( ) y = x 1x = () 1() = () 3 = 1 3 = 1 = = =. So, plot the x-intercepts (, ) and (, ) as well as the vertex (, ). Also, because a >, the parabola opens up. Draw a smooth curve through the points. 1 7 y 1 3 7. h(x) = x + x + 1 h(x) = ( x x 1 ) h(x) = (x 7)(x + ) The zeros are p = 7 and q =. The x-coordinate of the vertex is 7 + ( ) h(x) = x + x + 1 = (.) + (.) + 1 =. + 1. + 1 =. + 1 =. x = =.. So, plot the x-intercepts (, ) and (7, ) as well as the vertex (.,.). Also, because a <, the parabola opens down. Draw a smooth curve through the points. 1 1 1 y 3 1 1 3 x 3 1 1 3 x Copyright Big Ideas Learning, LLC Algebra 1 713
11.3 Explorations (p. 99) 1. a. Sample answer: About 7 + 7 + + 91 + = 7 of the data s are within inches of the mean of inches. There are a total of 73 data s. So, about 7 73.7, or 7.%, of the data s that are within 1 standard deviation of the mean. b. Sample answer: About 193 + + 7 + 37 + 97 = of the data s are within inches of the mean of inches. There are a total of 73 data s. So, about 73.97, or 9.7%, of the data s that are within standard deviations of the mean. c. Sample answer: About + 9 + + + = 7 of the data s are within inches of the mean of inches. There are a total of 73 data s. So, about 7 73.993, or 99.3%, of the data s that are within 3 standard deviations of the mean.. a. The set of adult female heights has a smaller standard deviation. This means that the heights of adult females do not vary as far from the mean. b. Sample answer: About 19 + 3 + + 3 + 3 + + 3 = 17 of the adult males surveyed have a height that is between 7 inches and 73 inches. A total of males were surveyed. So, about 17 =.7, or about 7%, of the heights are between 7 inches and 73 inches. 3. Sample answer: You can characterize the basic shape of a distribution by making a histogram of the data and drawing a curve through the tops of the bars.. a. Most of the data are clustered around the mean in the center of the data, and the data on the right of the distribution are approximately a mirror image of the data on the left of the distribution. b. The curve connecting the tops of the bars looks like a bell. c. Sample answer: Two other real-life examples of symmetric distributions are standardized test scores and the length of daylight each day over a year. 11.3 Monitoring Progress (pp. 3) 1. Aluminum Cans Collected Frequency 3 3 1 1 11 1 3 31 1 1 Number of pounds The tail of the graph extends to the left, and most of the data are on the right. So, the distribution is skewed left.. a. Email Attachments Sent Frequency 1 1 3 1 9 1 1 1 1 Frequency 1 Email Attachments Sent 1 1 1 1 1 1 1 Number of attachments b. Because the data on the right of the distribution are approximately a mirror image of the data on the left of the distribution, the distribution is symmetric. So, use the mean to describe the center and the standard deviation to describe the variation. 3. Because most of the data are on the left of the distribution for football players and the tail of the graph extends to the right, the distribution is skewed right. So, the median and five-number summary best represent the distribution for professional football players. Because the data on the right of the distribution are approximately a mirror image of the data on the left, the distribution for company employees is symmetric. So, the mean and standard deviation best represent the distribution for company employees. The median number of years of experience for professional football players is probably in the 3 or interval, and the mean number of years of experience for the company employees is probably in the 9 11 or 1 1 interval. So, a typical company employee is more likely to have more years of experience than a typical professional football player. The data for the company employees are more variable than the data for the professional football players. This means that the number of years of experience tends to differ more from one company employee to the next.. The mean is greater than the median because the distribution is skewed right.. If more women are surveyed, you would expect about. = 3 of the women to own between and 1 pairs of shoes. 71 Algebra 1 Copyright Big Ideas Learning, LLC
11.3 Exercises (pp. ) Vocabulary and Core Concept Check 1. In a symmetric distribution, the data are evenly distributed to the left and right of the highest bar. In a distribution that is skewed left, most of the data are on the right. In a distribution that is skewed right, most of the data are on the left.. The mean and standard deviation best describe a symmetric distribution, and the median and five-number summary best describe skewed distributions. Monitoring Progress and Modeling with Mathematics 3. Monthly Student Volunteer Hours. Frequency 1 1 1 1 1 3 7 9 11 1 13 1 Number of volunteer hours Most of the data are in the center and the data on the right are approximately a mirror image of the data on the left. So, the distribution is symmetric. Frequency 3 3 1 Weekly Online Hours 3 7 11 1 1 1 19 3 7 Hours online Most of the data are on the right and the tail of the graph extends to the left. So, the distribution is skewed left. 7. Most of the data are in the center and the data on the right are approximately a mirror image of the data on the left, which means the distribution is symmetric. So, the mean and standard deviation best represent the data.. Most of the data are on the right and the tail of the graph extends to the left. So, the distribution is skewed left, which means that the median and five-number summary best represent the data. 9. a. ATM Withdrawals (dollars) Frequency 7 1 7 7 1 1 3 1 1 11 17 1 17 1 Frequency ATM Withdrawals 7 3 1 1 7 7 1 1 1 1 11 17 17 Amount of money (dollars) b. Most of the data are on the left and the tail of the graph extends to the right. So, the distribution is skewed right, which means that the median and five-number summary best represent the data. c. Most of the data s are on the left side of the graph, which represent withdrawals of less than $1. So, most people were charged a fee for their withdrawals.. Most of the data are in the first two stems and the tail of the data extends to the higher s. So, the distribution is skewed right.. Most of the data are in the center and the data in the top half are approximately a mirror image of the data in the bottom half. So, the distribution is symmetric. Copyright Big Ideas Learning, LLC Algebra 1 71
. a. IQ Scores Frequency 11 1 3 17 1 13 19 199 1 1 1 3 Frequency IQ Scores 7 3 1 11 1 17 1 13 19 199 1 1 3 IQ score b. Most of the data are on the left and the tail of the graph extends to the right. So, the distribution is skewed right, which means that the median and five-number summary best represent the data. c. As you include more and more IQ scores in the data set, the shape of the distribution will become more symmetric. 11. When most of the data are on the right, the distribution is skewed left, not right. Most of the data are on the right and the tail of the graph extends to the left. So, the distribution is skewed left. 1. For a symmetric distribution, the mean is closest to the center. So, it is the most appropriate measure to use for describing the center of the data. Also for a symmetric distribution, the variation is approximately the same on each side of the center. So, the standard deviation is the most appropriate measure for describing the variation of the data. For a skewed distribution, the median is closest to the center of the data. So, it is the most appropriate measure to use for describing the center. Also for a skewed distribution, the variation is different on each side. So, the five-number summary gives the most accurate description of the variation. 1. Because most of the data are on the right of the distribution for Town A and the tail of the graph extends to the left, the distribution is skewed left. So, the median and five-number summary best represent the center and distribution for Town A. Because the data on the right of the distribution for Town B are approximately a mirror image of the data on the left of the distribution, the distribution is symmetric. So, the mean and standard deviation best represent the center and distribution for Town B. The median of the Town A data set is in the 7 79 interval, while the mean of the Town B data set is probably in the 9 interval. So, on a typical day, it is likely that the temperature in Town A is higher than the temperature in Town B. The data for Town B is more variable than the data for Town A. This means that the daily high temperature tends to differ more for Town B. 1. Use the standard deviation when the distribution of the data is symmetric, not skewed. Because the distribution is skewed, use the five-number summary to describe the variation of the data. 13. A stem-and-leaf plot requires that you list every data item, but a histogram only displays the frequency of data items in each range. So, a histogram would be more appropriate for a large data set. 71 Algebra 1 Copyright Big Ideas Learning, LLC
1. Entrée Prices 19. a. Freshmen Frequency Restaurant A Restaurant B 13 1 11 9 7 3 1 13 1 11 9 7 3 1 11 13 1 1 17 19 3 Price range (dollars) Because most of the data are on the left of the distribution for Restaurant A and the tail of the graph extends to the right, the distribution is skewed right. So, the median and five-number summary best represent the center and distribution for Restaurant A. Because most of the data are on the right of the distribution for Restaurant B and the tail of the graph extends to the left, the distribution is skewed left. So, the median and fivenumber summary best represent the center and distribution for Restaurant B. The median of the Restaurant A data set is in the 1 1 interval, while the median of the Restaurant B data set is between the 17 19 and intervals. So, entrée prices from Restaurant B are typically higher. Because most of the prices for Restaurant A seem to be concentrated in the two middle bars, the data for Restaurant B is more spread out than the data for Restaurant A. This means that the entrée prices for Restaurant B tend to vary more. 17. Sample answer: The salaries of employees at a large company tend to have a distribution that is skewed right. 1. Sample answer: The ages of residents and staff at a retirement home tend to have a distribution that is skewed left. Sophomores 1 1 17 3 Number of songs For freshmen, the whiskers are the same size, and the median is in the center of the plot. So, the distribution is symmetric. For sophomores, the whisker and box to the left of the median are larger than the whisker and box to the right of the median. So, most of the data is on the right, and the distribution is skewed left. b. The centers and spreads of the two data sets are quite different from each other. The median for sophomores is nearly twice the median for freshmen, and the mean for sophomores is significantly higher than that for freshmen. So, the number of songs downloaded by freshmen tends to be less than the number of songs downloaded by sophomores. Also, there is more variability in the number songs downloaded by sophomores. c. Assuming the symmetric distribution is bell-shaped, you know about % of the data lie within 1 standard deviation of the mean. Because the mean is 11, and the standard deviation is, the interval from 11 = 73 to 11 + = 17 represents about % of the data. So, you would expect about. 31 of the freshmen surveyed to have between 73 and 17 songs downloaded on their MP3 players. d. Assuming the symmetric distribution is bell-shaped, you know about 9% of the data lie within standard deviation of the mean. Because the mean is 11, and the standard deviation is, the interval from 11 () = 3 to 11 + () = 199 represents about 9% of the data. So, you would expect about.9 = 9 of the freshmen surveyed to have between 3 and 199 songs downloaded on their MP3 players.. a. Freshmen Sophomores 1 1 17 3 Number of songs The mean and median of this data set are less than the mean and median, respectively, of the data set for sophomores. So, the number of songs downloaded by freshmen still tends to be less than the number of songs downloaded by sophomores, but this data set has much more variability than the previous data set for freshmen. b. Because the whisker and box to the left of the median are longer than the median and box to the right of the median, the data are more concentrated on the right side of the plot and the distribution is skewed left. So, the median is greater than the mean for this group of freshmen. 1. The distribution will still be symmetric because the data will all be related proportionally. Because this is a data transformation using multiplication, the measures of center and distribution will each be doubled. Copyright Big Ideas Learning, LLC Algebra 1 717
. a. C; The data on the left are approximately a mirror image of the data on the right. So, the distribution is symmetric. This matches box-and-whisker plot C, which also shows a symmetric distribution, because the whisker and box to the right of the median are approximately the same size as the whisker and box, respectively, to the left of the median. b. A; Most of the data are on the left of the histogram and the tail extends to the right. So, the distribution is skewed right. This matches box-and-whisker plot A, which also shows a distribution that is skewed right, because most of the data are concentrated on the left side of the plot. c. B; Most of the data are on the right of the histogram, and the tail extends to the left. So, the distribution is skewed left. This matches box-and-whisker plot B, which also shows a distribution that is skewed left, because most of the data are concentrated on the right side of the plot. 3. a. Waiting Times Frequency 9 19 9 7 3 39 9 3 Frequency Restaurant Waiting Times 7 3 1 9 19 9 3 39 9 Time (minutes) The data on the right side of the graph are approximately a mirror image of the data on the left side. So, the distribution is approximately symmetric. b. Waiting Times Frequency 1 9 1 1 1 1 19 3 3 9 3 3 3 39 3 9 Frequency Restaurant Waiting Times 3 1 9 1 1 19 9 3 3 3 39 9 Time (minutes) This graph shows that there are more data on the right side of the graph, and that a tail extends to the left. So, when the number of intervals is increased, the distribution becomes skewed left. c. More intervals show the spread of the data more accurately. So, the histogram in part (b) best represents the data.. Sample answer: The age at which people receive diplomas or degrees is a bimodal distribution, because peaks in frequency occur both at the typical high school graduation age and again at the typical college graduation age. Maintaining Mathematical Proficiency. x + x So, the domain of the function is x.. x x x So, the domain of the function is x. 7. x 7 +7 +7 x 7 So, the domain of the function is x 7. 71 Algebra 1 Copyright Big Ideas Learning, LLC
11.1 11.3 What Did You Learn? (p. 7) 1. In Exercise 1, the outlier is much less than the rest of the data s, and it decreases the mean. In Exercise 1, the outlier is much greater than the rest of the data s, and it increases the mean. When a number is much less, it decreases the total of the data items, and when a data is much greater, it increases the total. Because the mean depends on the sum of the data s, it makes sense that an outlier that is much less will decrease the mean and an outlier that is much greater will increase the mean.. Sample answer: The residents would be much older than the staff, and the residents would significantly outnumber the staff. So, most data s would be on the right. 11.1 11.3 Quiz (p. ) 1 3 1. a. Mean: + 3 + + 3 1 + 1 + 1 x = 1 = 3 Median: 1, 1, 3, 3 1, 3 1, The mean of the two middle numbers is 1 3 + 3 = 1 13 1 13 = 3 1. Mode: 1, 1, 3, 3 1, 3 1, The data that occurs most often is 3 1. The mean is 3, the median is 3 1, and the mode is 3 1. The median best represents the data, because it is in the center of the data, and it is the average of the mean and the mode.. a. Mean: x = + 1 + +... + 13 + + 1 9 11,3 = 131 9 9 Median:,, 1, 11, 1191, 1, 17, 13, The middle number is 1191. Mode:,, 1, 11, 1191, 1, 17, 13, The data that occurs most often is. The mean is 131, the median is 1191, and the mode is 9. Because the data have an outlier of, which is much greater than the other data s, the mean is greater than most of the data. Also, the mode is less than most of the data. So, the median best represents the data. 3. Female students:, 3,,, So, the range is = absences. x + + + 3 + 19 = 3. x x x x (x x ) 3... 3. 1. 3. 3... 3 3... 3.... + 3. +. +. +.. = 1.7 1.7 1.3 The standard deviation is about 1.3. Male students: 3,,,, 9 So, the range is 9 3 = absences. x + 3 + + + 9 = 9 =. x x x x (x x )... 3.. 7....... 9. 3... + 7. +. +. +. 3.7 1.9 The standard deviation is about 1.9. = 1. = 3.7 The data have a greater range and standard deviation for the male students. So, the numbers of absences are more spread out for the male students. Copyright Big Ideas Learning, LLC Algebra 1 719
. Juniors:, 1, 1, 1, 1, 19,, 1 So, the range is 1 = 11 points. 19 + 1 + + + 1 + 1 + 1 + 1 x 13 = 1. x x x x (x x ) 19 1... 1 1. 1.. 1. 3. 1. 1... 1 1... 1 1... 1 1. 1.. 1 1. 1... +. + 1. +. +. +. +. +. 9 11.7 11.7 3. The standard deviation is about 3.. Seniors: 1, 19, 19,,, 9, 3, 3 So, the range is 3 1 = 17 points. + 19 + 9 + 3 + 1 + + 3 + 19 x 19 = x x x x (x x ) 19 9 3 1 9 1 3 3 19 + + + + 1 + + 3 + 33.7 The standard deviation is about.7. = = 33 The data have a greater range and standard deviation for the seniors. So, the numbers of points scored are more spread out for the seniors... lower half 1,,, 3,,,, 7 least first median Least : 1 + First : 3 + Median: 7 = 3 + Third : Greatest : 7 1 3 lower half = =. = 11 = 7. Age upper half third greatest,,, 3, 3,,,,,,, least 7. first median Least : + 3 First : + Median: = + Third : Greatest : 3 Frequency 1 1 1 1 Quiz Scores 3 Score = = = = Minutes 9 11 1 1 upper half third Most of the data are on the right and the tail of the graph extends to the left. So, the distribution is skewed left. greatest 7 Algebra 1 Copyright Big Ideas Learning, LLC
. a. Mean: 9 + 119 + 9 + 11 + 13 + 9 + + 1 x 97 = 1 Median: 9, 9, 9,, 119, 1, 13, 11 The mean of the two middle numbers is + 119 19 = 9.. Mode: 9, 9, 9,, 119, 1, 13, 11 The data that occurs most often is 9. The range is 11 9 = 11. x x x x (x x ) 9 1 7 119 1 3 9 9 1 7 79 11 1 9 791 13 1 9 1 7 1 1 1 3 9 7 + 9 + 79 + 791 + + 7 + + 9,3 = 19 19 = 3 So, the mean is $1, the median is $9., the mode is $9, the range is $11, and the standard deviation is $3. b. The price of $11 is significantly greater than the other prices. So, it is the outlier. Mean: 9 + 119 + 9 + 13 + 9 + + 1 x 7 7 7 9.9 Median: 9, 9, 9,, 119, 1, 13 The middle number is. Mode: 9, 9, 9,, 119, 1, 13 The data that occurs most often is still 9. So, the outlier increases the mean by about $1 $9.9 = $1.71 and it increases the median by $9. $ = $9., but the outlier does not affect the mode. c. lower half 9, 9, 9,, 119, 1, 13, 11 least first median upper half third Least : 9 9 + 9 First : 19 = 9 + 119 Median: 19 = 9. 1 + 13 Third : = 17. Greatest : 11 1 1 1 1 Price greatest The inter range is 17. 9 = 9.. This means that the middle half of the prices vary by no more than $9.. Because most of the data items are on the left side of the plot and the right whisker is longer than the left whisker, the distribution is skewed right. d. Mean: x = 1(1.) = 1(.9) = 11.9 Median: 9.(.9) =. Mode: 9(1.) = 9(.9) = 93.1 Range: 11(1.) = 11(.9) = 1. Standard deviation: 3(1.) = 3(.9) = 3. 9. a. Time (minutes) Frequency 3 9 11 1 1 3 1 17 1 Frequency 1 Presentation Times 3 9 11 1 1 1 17 Time (minutes) b. Because the data on the right are approximately a mirror image of the data on the left, the distribution is symmetric. So, the mean and standard deviation best represent the data. c. Because most of the data are in the 9 11 range or close to it, most of the presentations are about the right length. Copyright Big Ideas Learning, LLC Algebra 1 71
11. Explorations (p. 9) 1. a. Color Beginning of T-Shirt Size season S M L XL XXL Total blue/white 7 9 blue/gold 7 9 red/white 7 9 black/white 7 9 black/gold 7 9 Total 3 3 3 1 T-Shirt Size End of season S M L XL XXL Total blue/white 1 1 blue/gold 3 1 red/white 1 3 1 black/white 3 1 1 11 black/gold 3 1 Total 1 1 Color b. Sample answer: Next season, for each shirt category, you could order the quantity equal to the difference between the number at the beginning of this season and the number at the end of this season. From the table, you can see that you should order fewer t-shirts that are sizes small and medium, and large especially. You may even want to order a couple more of the t-shirts that were sold out by the end of the season, such as the blue/white, size XL.. a. Sample answer: Hours Gender Males Females Total hours per week 1 1 9 1 7 hours per week 1 11 + hours per week 7 11 1 Total 3 37 7 b. Sample answer: More females work + hours per week than males. Approximately equal numbers of males and females work 1 7 hours per week and hours per week. 11. Monitoring Progress (pp. 13) 1. Tablet Computer Yes No Total Yes 3 1 1 No 1 7 Total 191 3. 3.. Cell Phone So, students own a tablet, and 191 students do not own a tablet. Also, 1 students own a cell phone, and students do not own a cell phone. In total, 3 students were surveyed. Gender Gender Summer Job Yes No Total Male 7 1 7 Female 1 7 Total 3 13 Summer Job Yes No Total Male.3.1.7 Female.3.9.3 Total.77.3 1 So, about.3, or 3%, of the students are not getting a summer job. Class Major in Medical Field Yes No Junior.3. Senior.3.7 So, given that a student is a senior, the conditional relative frequency that he or she is planning to major in a medical field is about.3, or 3%. 3. Sample answer: Rows and columns represent different categories, each entry represents the number in both categories. 7 Algebra 1 Copyright Big Ideas Learning, LLC
. Cell Phone Yes 3 Tablet Computer Yes No. 1 191. No 1.3 7 191.3 The conditional relative frequencies for students who have cell phones are about the same whether they have a tablet computer or not. Similarly, the conditional relative frequencies for students who do not have cell phones are about the same whether they have a tablet computer or not. So, there is not an association between owning a tablet computer and owning a cell phone. 11. Exercises (pp. 1 1) Vocabulary and Core Concept Check 1. Each entry in a two-way table is called a joint frequency.. It is appropriate to use a two-way table to organize data when data are collected from one source and belong to two different categories. 3. A marginal relative frequency is the sum of the joint relative frequencies in a row or column. A conditional relative frequency is the ratio of a joint relative frequency to the marginal relative frequency.. In order to find conditional relative frequencies, you can use the marginal relative frequency of each row or of each column. Or, you can find the ratio of a joint relative frequency to the corresponding marginal relative frequency. Monitoring Progress and Modeling with Mathematics Table for Exercises Class Buy Lunch at School Yes No Total Freshmen 9 17 Sophomore 11 1 Total 13 3. A total of 17 freshmen were surveyed.. A total of 1 sophomores were surveyed. 7. A total of students buy lunch at school.. A total of 13 students do not buy lunch at school. 9.. 11. Class 1. Gender Gender Set Academic Goals Yes No Total Male 1 3 Female 1 19 Total 11 3 So, 3 male students and 19 female students were asked if they set academic goals. A total of 11 students set academic goals, and 3 students have not. A total of students were asked if they set academic goals. Dog Cat Yes No Total Yes 31 No 1 9 Total 9 3 9 So, 31 people have a dog, while people do not. Also, 9 people have a cat, while 3 people do not. A total of 9 people were asked whether they have a cat or dog. Participate in Spirit Week Yes No Undecided Total Freshmen 11 Sophomore 9 3 19 Total 1 1 So, freshmen and 19 sophomores responded to the survey. A total of students say they will participate in school spirit week, 1 students say they will not, and students are undecided. A total of 1 students were surveyed. Type of Degree Associate s Bachelor s Master s Total Male 1 Female 11 Total 1 9 So, males and females responded to the survey. A total of 1 seniors plan to receive an associate s degree, seniors plan to receive a bachelor s degree, and 9 seniors plan to receive a master s degree. A total of seniors were surveyed. Copyright Big Ideas Learning, LLC Algebra 1 73