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Chapter 1 – Independent and Dependent Events Answer Key CK-12 Basic Probability and Statistics Concepts 1 1.1 Venn Diagrams Answers 1. 2. 3. 0.45 + 0.55 = 1 4. A B = {5, 6, 8, 9, 10, 12, 13, 14} 5. A B = {7, 21}

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Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 1

1.1 Venn Diagrams

1.

2.

3. 0.45 + 0.55 = 1

4. A∪ B = {5, 6, 8, 9, 10, 12, 13, 14}

5. A∩ B = {7, 21}

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 2

6. The answer can be calculated as follows: 26 − (11 + 5 − 3) = 26 − 13 = 13

7. No, the ovals in the Venn diagram that represents this situation do not overlap, because you

cannot randomly choose a vowel and randomly choose a consonant at the same time. In

other words, a letter of the alphabet cannot be both a vowel and a consonant.

8. A ∪ B = {2, 4, 6, 7, 8, 10}

9. A∩ B = {5}

10. A = {5, 6, 10} and B = {5, 8, 9}

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 3

1.2 Independent Events and Sample Spaces

1. (a) and (b) are independent. They do not affect each other.

2. (a) and (c) are independent They do not affect each other.

3. The probability of getting tails and a 5 is calculated as follows: 1

1

6=

1

12=

0.083 = 8.3%

4. The probability can be calculated as follows: 0.09 × 0.4 = 0.036 = 3.6%

5. The probability that the baseball player will get a hit in 5 at-bats in a row can

be calculated as follows: 0.315 = 0.31 × 0.31 × 0.31 × 0.31 × 0.31 = 0.0029 =

0.29%.

6. To get 1 head and 1 tail, you can get either a head and then a tail or a tail and

then a head. Therefore the probability can be calculated as follows:

1

1

2+

1

1

2=

1

4+

1

4=

2

4=

1

2= 0.5 = 50%

7. The probability that both cards will be clubs can be calculated as follows:

13

52×

13

52=

1

1

4=

1

16= 0.0625 = 6.25%

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 4

8. The probability that both cards will be face cards can be calculated as follows:

12

52×

12

52=

3

13×

3

13=

9

169= 0.053 = 5.3%

9. To calculate the probability of not receiving any mail on a particular day, you

can subtract the probability of receiving mail on a particular day from 1.

Therefore, the probability of not receiving any mail for 3 days in a row can be

calculated as follows:

1 − 0.22 = 0.78

0.783 = 0.78 × 0.78 × 0.78 = 0.475 = 47.5%

10. To roll an 11, Johnathan can roll a 5 and a 6or a 6 and a 5. Therefore, the

probability can be calculated as follows: 1

1

6+

1

1

6=

1

36+

1

36=

2

36=

1

18= 0.056 =

5.6%

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 5

1.3 Dependent Events and Sample Spaces

1. (b) and (c) are dependent. The probability of the second event occurring

depends on the first event occurring.

2. (a) and (c) are dependent. The probability of the second event occurring

depends on the first event occurring.

3. The probability can be calculated as follows: 27

37×

10

36=

270

1332=

15

74= 0.203 = 20.3%

4. The probability can be calculated as follows: 10

37×

27

36=

270

1332=

15

74= 0.203 = 20.3%

5. The probability of drawing 2 face cards one after the other without

replacement can be calculated as follows: 12

52×

11

51=

132

2652=

11

221= 0.050 = 5.0%

6. The probability can be calculated as follows: 13

50×

3

49=

39

2450= 0.016 = 1.6%

7. The coins that are not dimes are quarters, nickels, and pennies. Therefore, the

probability can be calculated as follows:

quarters + nickels + pennies = 3 + 13 + 27 = 43 43

50×

42

49=

1806

2450= 0.737 = 73.7%

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 6

8. The doughnuts that are not jelly are glazed and plain. Therefore, the

probability can be calculated as follows:

5

4

3

2

1

2=

120

720=

1

6= 0.167 = 16.7%

9. The prime numbers from 1 through 20 are 2, 3, 5, 7, 11, 13, 17, and 19.

Therefore, the probability can be calculated as follows: 8

20×

7

19=

56

380=

14

95=

0.147 = 14.7%

10. For Steve’s cards to consist of a heart and a diamond, he can draw a heart

and then a diamond or a diamond and then a heart. Therefore, the probability

can be calculated as follows:

13

52×

13

51+

13

52×

13

51=

169

2652+

169

2652=

338

2652=

169

1326= 0.127 = 12.7%

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 7

1.4 Mutually Exclusive Events

1. 5

36

2. 10

36=

5

18

3. 21

36=

7

12

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 8

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 9

4. 26

36=

13

18

5. The probability of pulling out a blue or green jelly bean is calculated as follows: 15

37+

12

37=

27

37

6. These events are mutually exclusive. The probability can be calculated as follows: 13

52+

13

52=

26

52=

1

2

7. These events are mutually exclusive. The probability can be calculated as follows: 1

8+

1

8=

2

8=

1

4

8. The probability is 0, since getting 3 heads and getting 3 tails are mutually exclusive events.

9. Answers will vary. One possible answer is that they are not mutually exclusive events,

because a person who is ambidextrous is both left handed and right handed.

10. Answers will vary. One possible answer is randomly choosing a girl from the freshman class

of the high school, since you cannot randomly select a boy and randomly select a girl at the

same time.

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 10

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 11

1.5 Mutually Inclusive Events

1. 1

10

2. The probability can be calculated as follows: 5

10+

2

10−

1

10=

6

10=

3

5

3. 13

60+

17

60=

30

60=

1

2

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 12

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 13

4.

a. 32

71+

25

71−

4

71=

53

71

b. 18

71

5. These events are mutually inclusive. The probability can be calculated as follows: 13

52+

12

52−

3

52=

22

52=

11

26

6. The probability can be calculated as follows: 5

10+

5

10−

3

10=

7

10

7. The probability can be calculated as follows: 7

26+

5

26−

2

26=

10

26=

5

13

8. Answers will vary. One possible answer is that they are mutually inclusive events, since a person

who is a teacher can also be a father.

9. Answers will vary. One possible answer is getting an A on the test, since you can pass the test and

get an A on the test at the same time.

Chapter 1 – Independent and Dependent Events Answer Key

CK-12 Basic Probability and Statistics Concepts 14

10. The factors of 8 are 1, 2, 4, and 8, while the factors of 10 are 1, 2, 5, and 10. Therefore, the

probability can be calculated as follows: 4

10+

4

10−

2

10=

6

10=

3

5

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 15

2.1 Tree Diagrams

1. The answers are as follows:

a.

b. i. 3

3

7=

9

49

ii. 3

3

7+

4

3

7=

9

49+

12

49=

21

49=

3

7

2. The answers are as follows:

a.

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 16

b. 2

1

3+

1

1

3=

2

9+

1

9=

3

9=

1

3

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 17

3. 0.65

4. The probabilities can be calculated as follows:

𝑃(𝑔𝑟𝑒𝑒𝑛 𝑎𝑛𝑑 𝑦𝑒𝑙𝑙𝑜𝑤) = 0.75 × 0.45

𝑃(𝑔𝑟𝑒𝑒𝑛 𝑎𝑛𝑑 𝑦𝑒𝑙𝑙𝑜𝑤) = 0.3375

𝑃(𝑔𝑟𝑒𝑒𝑛 𝑎𝑛𝑑 𝑦𝑒𝑙𝑙𝑜𝑤) = 33.75%

𝑃(𝑔𝑟𝑒𝑒𝑛 𝑎𝑛𝑑 𝑟𝑒𝑑) = 0.75 × 0.55

𝑃(𝑔𝑟𝑒𝑒𝑛 𝑎𝑛𝑑 𝑟𝑒𝑑) = 0.4125

𝑃(𝑔𝑟𝑒𝑒𝑛 𝑎𝑛𝑑 𝑟𝑒𝑑) = 41.25%

𝑃(𝑏𝑙𝑢𝑒 𝑎𝑛𝑑 𝑦𝑒𝑙𝑙𝑜𝑤) = 0.25 × 0.35

𝑃(𝑏𝑙𝑢𝑒 𝑎𝑛𝑑 𝑦𝑒𝑙𝑙𝑜𝑤) = 0.0875

𝑃(𝑏𝑙𝑢𝑒 𝑎𝑛𝑑 𝑦𝑒𝑙𝑙𝑜𝑤) = 8.75%

𝑃(𝑏𝑙𝑢𝑒 𝑎𝑛𝑑 𝑟𝑒𝑑) = 0.25 × 0.65

𝑃(𝑏𝑙𝑢𝑒 𝑎𝑛𝑑 𝑟𝑒𝑑) = 0.1625

𝑃(𝑏𝑙𝑢𝑒 𝑎𝑛𝑑 𝑟𝑒𝑑) = 16.25%

Therefore, the students in BDF High School want green and red the most

5. The probability can be calculated as follows: 1

1

2+

1

1

2=

1

4+

1

4=

2

4=

1

2

6. The number of branches on the tree diagram can be calculated as follows: 24 = 2 × 2 × 2 × 2 = 16

7. The number of branches on the tree diagram can be calculated as follows: 62 = 6 × 6 = 36

8. The probability can be calculated as follows: 1 − 0.12 − 0.53 − 0.28 = 0.07

9. The probability can be calculated as follows: 0.1947

0.33= 0.59

10. The probability can be calculated as follows: 1 − 0.59 = 0.41

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 18

2.2 Permutations and Combinations Compared

1. The answer can be calculated as follows: 𝑃57 = 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 = 2,520

2. The answer can be calculated as follows:

3. The number of 4-digit numerals can be calculated as follows: 𝑃45 = 5 ∙ 4 ∙ 3 ∙ 2 =

120

4. The number of ways the students can be chosen can be calculated as follows:

5. The number of ways the TV station can fill the time slots can be calculated as follows:

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 19

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 20

6. The number of ways teams can finish in first, second, and third place can be calculated as follows:

7. The number of secret codes that are possible can be calculated as follows:

𝑃310 × 𝑃226

= (10 × 9 × 8) × (26 × 25) = 720 × 650 = 468,000

8. The number of ways Robert can do the matching can be calculated as follows:

𝑃1515 = 15! = 1.31 × 1012

9. The probability that the first boy born will be named Sam and the second boy born will be named

Mark can be calculated as follows:

10. In this case, the first boy born could have the name Peter and the second boy born could have the

name Paul, or vice versa. Therefore, the probability that one of the boys will be named Peter and the

other boy will be named Paul can be calculated as follows:

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 21

2.3 Permutation Problems

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 22

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 23

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 24

2.4 Permutations with Repetition

1. The answer is calculated as shown below:

The letters of the word REFERENCE can be arranged in 7,560 ways.

2. The answer is calculated as shown below:

The letters of the word MISSISSIPPI can be arranged in 34,650 ways.

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 25

3. The answer is calculated as shown below:

The letters of the word MATHEMATICS can be arranged in 4,989,600 ways.

4. The answer is calculated as shown below:

There are 37,837,800 answer sheets that are possible.

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 26

5. The number of 5-digit numerals that can be written can be calculated as follows:

There are 210 5-digit numerals that can be written using the 9 digits.

6. The number of 4-digit numerals that can be written can be calculated as follows:

There are 315 4-digit numerals that can be written using the 10 digits.

7. The number of 6-digit numerals that can be written can be calculated as follows:

There are 231 6-digit numerals that can be written using the 12 digits.

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 27

8. The number of orders in which the drinks can be consumed can be calculated as follows:

The drinks can be consumed in 27,720 different orders.

9. The number of orders in which the store can sell the shirts can be calculated as follows:

The store can sell the shirts in 360,360 different orders.

10. The number of orders in which the store can sell the remaining shirts can be calculated as follows:

The store can sell the remaining shirts in 4,620 different orders.

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 28

2.5 Combinations

1. The answers are as follows:

a. Yes, it would require calculating a combination.

b. No, it would require calculating a permutation.

c. No, it would require calculating a permutation.

2. The number of possible ways can you select 17 songs from a mix CD of a possible 38 songs can be

calculated as follows:

3. The number of different selections you can make can be calculated as follows:

4. The number of lines that can be formed using any 2 dots can be calculated as follows:

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 29

5. The number of different committees that can be formed can be calculated as follows:

6. The number of possible wraps that have 1 kind of meat and 3 veggies can be calculated as follows:

7. The answers can be calculated as follows:

a.

b.

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 30

8. The number of possible ways that Julia can pack t-shirts and shorts for her vacation can be

calculated as follows:

9. The number of groups of playoff teams that are possible in Major League Baseball can be calculated

as follows:

10. Since there are 12 months in a year, begin by calculating the number of ways 3 months can be

chosen from 12 months as follows:

Since there are 220 ways that 3 months can be chosen from 12 months, and since the first 3 months

of the year and the last 3 months of the year represent 2 of these ways, the probability that Rick

gives up junk food during the first 3 months or the last 3 months of the year can be calculated as

follows:

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 31

2.6 Combination Problems

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 32

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 33

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 34

2.7 Conditional Probability

1. 1

121

6

=6

12=

1

2

2. 1

121

6

=6

12=

1

2

3. 1

121

6

=6

12=

1

2

Chapter 2 – Conditional Probability Answer Key

CK-12 Basic Probability and Statistics Concepts 35

4. The probability can be calculated as follows:

1

26×

26

511

26

=26

51

5. The probability can be calculated as follows: 1

26×

26

511

26

=26

51

6. The probability can be calculated as follows:0.25

0.55= 0.45

7. The probability can be calculated as follows:0.15

0.45= 0.33

8. The probability can be calculated as follows:0.10

0.65= 0.15

9. Since the test is accurate 80% of the time, 80% of the people who have the disease will test positive

for the disease, and 80% of the people who don’t have the disease will test negative for the disease.

This means that 20% of the people who don’t have the disease will test positive for the disease. Also,

since 2.5% of the population has the disease, 97.5% of the population doesn’t have the disease.

Therefore, the probability of testing positive for the disease can be calculated as follows:

(0.025)(0.8) + (0.975)(0.2) = 0.02 + 0.195 = 0.215 = 21.5% Since the probability of testing positive

for the disease and having the disease is (0.025)(0.8) = 0.02 = 2%, the probability the you have the

disease, given that you tested positive, can be calculated as follows: 0.02

0.215= 0.093 = 9.3%

10. Since the test is accurate 80% of the time, 80% of the people who have the disease will test positive

for the disease, and 80% of the people who don’t have the disease will test negative for the disease.

This means that 20% of the people who have the disease will test negative for the disease. Also,

since 2.5% of the population has the disease, 97.5% of the population doesn’t have the disease.

Therefore, the probability of testing negative for the disease can be calculated as follows:

(0.025)(0.2) + (0.975)(0.8) = 0.005 + 0.78 = 0.785 = 78.5% Since the probability of testing negative

for the disease and not having the disease is (0.975)(0.8) = 0.78 = 0.78%, the probability the you

don’t have the disease, given that you tested negative, can be calculated as follows: 0.78

0.785= 0.994 =

99.4%

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 36

3.1 Discrete Random Variables

3. Answers will vary, depending on who is doing the problem and where he or she lives.

4. Answers will vary, depending on who is doing the problem and where he or she lives.

5. The amount of lemonade in a pitcher cannot be represented by a discrete random variable, because

the amount can take on any value within a certain range. For example, it could be 1 liter, 1.1 liters,

1.11 liters, 1.111 liters, and so on.

6. The number of musicians in an orchestra can be represented by a discrete random variable, because

the number will always be an integer.

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 37

7. The weights of the alligators in a swamp cannot be represented by a discrete random variable,

because the weights can take on any values within a certain range. For example, a weight could be

800 pounds, 800.8 pounds, 800.88 pounds, 800.888 pounds, and so on.

8. The number of tickets sold to a movie can be represented by a discrete random variable, because the

number will always be an integer.

9. Answers will vary. One possible answer is the number of trees in a forest, because the number will

always be an integer.

10. Answers will vary. One possible answer is the temperature in an oven, because the temperature

could take on any value within a certain range.

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 38

3.2 Probability Distribution

1. No, it does not. ∑ 𝑃(𝑋) = 0.2 + 0.4 + 0.6 + 0.8 = 2.0

2. No, it does not. ∑ 𝑃(𝑋) = 0.202 + 0.174 + 0.096 + 0.078 + 0.055 = 0.605

3. Yes, it does. ∑ 𝑃(𝑋) = 0.302 + 0.251 + 0.174 + 0.109 + 0.097 + 0.067 = 1.0

4. The probability that the spinner doesn’t land on green on either of the spins can be calculated as

follows: 1

9+

1

9+

1

9+

1

9=

4

9

5. The probability that the spinner lands on green on 1 of the spins can be calculated as follows: 1

9+

1

9+

1

9+

1

9=

4

9

6. The probability that the spinner lands on green on both of the spins is as follows: 1

9

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 39

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 40

3.3 Binomial Distributions and Probability

1. The probability of scoring below 80% can be calculated as follows: 1−0.20 = 0.80

2. The probability of not getting a job after university can be calculated as follows: 1−0.85 = 0.15

3. The probabilities are as follows:

a. 1

6= 0.167

b.

4. The probabilities are as follows:

a. 1

6= 0.167

b.

5. The probabilities are as follows:

a. 1

6= 0.167

b.

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 41

6.

a. 11

36= 0.306

b.

7.

a. 11

36= 0.306

b.

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 42

8. The probabilities are as follows:

a. 1

2= 0.5

b.

9. The probabilities are as follows:

a. 1

2= 0.5

b.

10. When tossing 2 coins, there are 4 possibilities: HH, HT, TH, and TT. Therefore, the probabilities are

as follows:

a. 3

4= 0.75

b.

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 43

3.4 Multinomial Distributions

1. The probability can be calculated as follows:

2. The probability can be calculated as follows:

3. The probability can be calculated as follows:

4. The probability can be calculated as follows:

5. The probability can be calculated as follows:

6. The probability can be calculated as follows:

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 44

7. The probability can be calculated as follows:

8. The probability can be calculated as follows:

9. The probability can be calculated as follows:

10. The probability can be calculated as follows:

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 45

3.5 Theoretical and Experimental Spinners

1. Answers will vary, but they should be somewhat similar to the following definitions: Theoretical

Probability is the calculated probability of an event. Experimental Probability is the actual

probability of an event found through experimentation.

2. Answers will vary, but they should be somewhat similar to the following:

a. The difference between theoretical and experimental probability is that theoretical probability is

calculated and experimental probability is actual. In theoretical probability, you are analyzing

equally likely outcomes.

b. In general, the more data that is collected, the closer the experimental probability gets to the

theoretical probability.

c. Yes, spinning 1 spinner 100 times is the same as spinning 100 spinners 1 time, because each

spin of a spinner (whether it’s the same spinner or different spinners) is independent of any

other spin.

3. The experimental probability of landing on blue can be calculated as follows:193

750= 0.257 = 25.7%

4. The experimental probability of landing on purple can be calculated as follows:197

750= 0.263 =

26.3%

5. The experimental probability of landing on green can be calculated as follows:190

750= 0.253 =

25.3%

6. The experimental probability of landing on red can be calculated as follows:170

750= 0.227 = 22.7%

7. The experimental probability of landing on blue can be calculated as follows:197

900= 0.219 = 21.9%

8. The experimental probability of landing on purple can be calculated as follows:240

900= 0.267 =

26.7%

9. The experimental probability of landing on green can be calculated as follows:222

900= 0.247 =

24.7%

10. . The experimental probability of landing on red can be calculated as follows:241

900= 0.268 = 26.8%

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 46

3.6 Theoretical and Experimental Coin Tosses

1. The probability will vary, depending on what the calculator returns.

2. The probability will vary, depending on what the calculator returns.

3. The probability can be calculated as follows: 𝐶310

210 =120

1024= 0.117

Chapter 3 – Introduction to Discrete Random Variables Answer Key

CK-12 Basic Probability and Statistics Concepts 47

4. The probability will vary, depending on what the calculator returns.

5. The probability can be calculated as follows: 𝐶812

212 =495

4096= 0.121

6. The probability can be calculated as follows: 𝐶714

214 =3432

16384= 0.209

7. The experimental probability of getting 0 heads can be calculated as follows: 58

500= 0.116 = 11.6%

8. The experimental probability of getting 1 head can be calculated as follows:187

500= 0.374 = 37.4%

9. The experimental probability of getting 2 heads can be calculated as follows: 193

500= 0.386 = 38.6%

10. The experimental probability of getting 3 heads can be calculated as follows: 62

500= 0.124 = 12.4%

Chapter 4 – Probability Distributions Answer Key

CK-12 Basic Probability and Statistics Concepts 48

4.1 Normal Distributions

1. The data gives the following histogram:

The data does not appear to be normally distributed but does appear to have a good spread of the

data. For a quiz out of 25, the lowest score was 2 and the highest was 25. The mean score was 16.04.

2. The data the manager collected can be plotted to give the following histogram:

This data does appear to be normally distributed about the mean.

The movie is rated PG, with an additional violence warning. A mean age of 15.56 years for movie

goers tells the manager that there are a number of very young people attending. When he looks at the

survey results, he can see that there are indeed children as young as 5 going to the movie.

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3. The data collected from the survey can be plotted to give the following histogram:

This data is obviously not normally distributed, but is in two distinct groups. It appears that there is

one group growing well, with a mean height between 21 and 25 feet (upper part of the histogram),

and one group not growing so well, with a mean height between 6 and 10 feet (lower part of the

histogram). In this part of the park, the park warden may need to look to see if there is a problem,

such as a beetle or other pest that is invading the trees.

4. No, the data points cannot be connected when graphed, because this is discrete data.

5. Yes, the data points can be connected when graphed, because this is continuous data.

6. No, the data points cannot be connected when graphed, because this is discrete data.

7. Yes, the data points can be connected when graphed, because this is continuous data.

8. Yes, the data points can be connected when graphed, because this is continuous data.

9. No, the data points cannot be connected when graphed, because this is discrete data.

10. Yes, the data points can be connected when graphed, because this is continuous data.

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4.2 Binomial Distributions

1. The answers are as follows:

a. The data is discrete, so it could be a binomial distribution.

b. The data is discrete, so it could be a binomial distribution.

c. The data is continuous, so it could not be a binomial distribution.

d. The data is continuous, so it could not be a binomial distribution.

2. The binomial distributions are a and b. Of these 2 distributions, b more closely approximates

a normal distribution.

3. The answers are as follows:

a. The data is continuous, so it could not be a binomial distribution.

b. The data is discrete, so it could be a binomial distribution.

c. The data is continuous, so it could not be a binomial distribution.

d. The data is discrete, so it could be a binomial distribution.

4. The binomial distributions are b and d. Of these 2 distributions, d more closely approximates

a normal distribution.

5. The probability of getting exactly 4 heads can be calculated as follows:

6. The probability of getting exactly 3 tails can be calculated as follows:

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7. The probability of getting exactly 6 heads can be calculated as follows:

8. The probability of getting exactly 2 tails can be calculated as follows:

9. South Hampton’s graph would more likely approximate a normal distribution, because it produced

more data, and the more data there is, the more likely a binomial distribution is to approximate a

normal distribution.

10. Graph b is most likely the one for North Liberty, and graph a is most likely the one for North

Hampton. This is because graph a more closely approximates a normal distribution.

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4.3 Binompdf Function

1. The probability that they will have exactly 2 boys is 0.3456, or 34.56%.

2. The probability that 4 will come up tails is 0.1611, or 16.11%.

3. The probability that the spinner lands on yellow 45 times is 0.0626, or 6.26%.

4. The probability of getting a 4 or greater 14 times is 0.1328, or 13.28%.

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5. The probability that Amy’s alarm clock goes off when it is supposed to on 6 days is 0.2573, or 25.73%.

6. The probability of randomly guessing the correct answer to 8 questions is 0.022, or 2.2%.

7. The probability that Bob choosing a diamond 5 times is 0.1651, or 16.51%.

8. The probability that 10 game cards are winners is 0.1319, or 13.19%.

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9. The probability that none of the people chose a 6 is 0.3487, or 34.87%.

10. The probability of getting your tax return audited exactly once in the next 5 years is 0.4069, or 40.69%.

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4.4 Binomcdf Function

1. The probability that Janet and David will have at least 2 boys is 0.8208, or 82.08%.

2. The probability that Janet and David will have at most 2 boys is 0.5248, or 52.48%.

3. The probability that it rains on 2 or more of the days is 0.5248, or 52.48%.

4. The probability of getting 15 or fewer heads is 0.2498, or 24.98%.

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5. The probability of getting at least 28 tails is 0.0083, or 0.83%.

6. The probability of getting selected for jury duty in at most 2 of the next 5 years is 0.9937, or 99.37%.

7. The probability of getting fewer than 7 clubs is 0.6074, or 60.74%.

8. The probability that Brady wins at least twice is 0.0932, or 9.32%.

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9. The probability of getting more than 106 even numbers is 0.1790, or 17.90%.

10. The probability that you get stopped by a red light 3 or fewer times is 0.5226, or 52.26%.

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4.5 Geometric Distributions

1. a and b are geometrically distributed

2. 0.031

3. 0.027

5. 0.034

6. 0.220

7. 0.167

8. 0.139; 0.116

9. 0.579

10. 0.070

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5.1 Mean

1. C

2. D The population mean is denoted by “mu,” and its symbol is µ.

3. B The mean is calculated by dividing the sum of all values by the number of values.

4. A 71.5 × 4 = 286 286 −(58 + 76 + 88) = 64

5. The means can be calculated as follows:

a.

b.

c.

6. 171.6 pounds 167.2 × 5 = 836 836 − (158.4 + 162.8 + 165.0 + 178.2) = 171.6

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7. The answers can be calculated as follows:

a. 12×5.1 = 61.2 feet

b. 8×4.8 = 38.4 feet

c. 61.2+38.4 = 99.6 feet. 99.5

20= 4.98 feet

9. The means can be calculated as follows:

a.

b.

c.

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10. The answers can be calculated as follows:

a. 31×4 = 124

b.

11. The means can be calculated as follows:

a.

b.

12. a. No Melanie’s answer is not correct. The temperature of 0◦C was recorded, but she did not include

it in the total.

b. She divided the sum of the numbers by 6, but she should have divided by 7. The mean daily

temperature should be −2 ◦C.

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13. The means can be calculated as follows:

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5.2 Ungrouped Data to Find the Mean

1. The mean can be calculated as follows:

2. The mean can be calculated as follows:

3. The frequency distribution table is as follows:

The mean can be calculated as follows:

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4. The frequency distribution table is as follows:

The mean can be calculated as follows:

5. The sum of the data values is 108.

6. There are 12 data values.

7. The mean of the data values is 9.

8. The mean of the data is:

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9. The mean of the data is:

10. The mean of the data is:

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5.3 Grouped Data to Find the Mean

1.

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The table is created by completing problems 2 – 4

2. The summation of the values of 𝑚𝑓 for each class can be calculated as follows:

32 + 150 + 360 + 180 + 48 = 770

3. The value of N is 100.

4. The mean weight for a baby can be calculated as follows:

The table is created by completing problems 5 – 7

5. The summation of the values of 𝑚𝑓 for each class can be calculated as follows:

210 + 320 + 800 + 600 + 630 = 2,560

6. The value of n is 50.

7. The mean age for a teacher can be calculated as follows:

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The table is created by completing problems 8 – 10

8. The summation of the values of 𝑚𝑓 for each class can be calculated as follows:

60,000 + 360,000 + 1,000,000 + 560,000 + 360,000 = 2,340,000

9. The value of N is 25.

10. The mean number of miles on a car can be calculated as follows:

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5.4 Median

1. B An outlier is a value that is very small or very large compared to the majority of values in a data

set.

2. B 10, 25, 38, 39, 39, 42, 71, 76 The median is the number in the 𝑛+1

2 position:

𝑛+1

2=

8+1

2=

9

2= 4.5.

The number below the 4.5 position is 39, and the number above the 4.5 position is 39. The median is 39+39

2=

78

2= 39.

3. D There are 23 seats, which is an odd number. The median is the number in the 𝑛+1

2 position:

𝑛+1

2=

23+1

2=

24

2= 12. The median seat is seat number 12.

4. C 55, 58, 62, 63, 68, 70, 71, 79, 81, 82 The median is the number in the 𝑛+1

2position:

𝑛+1

2=

10+1

2=

11

2= 5.5. The number below the 5.5 position is 68, and the number above the 5.5 position is 70. The

median is 68+70

2=

138

2= 69.

5. A 32, 34, 36, 38 The median is the number in the 𝑛+1

2 position:

𝑛+1

2=

4+1

2=

5

2= 2.5. The number

below the 2.5 position is 34, and the number above the 2.5 position is 36. The median is 34+36

2=

70

2= 35.

6. B The median is the number of bars sold by the student in the 𝑛+1

2 position:

𝑛+1

2=

30+1

2=

31

2=

15.5. The number below the 15.5 position is 21, and the number above the 15.5 position is 21. The

median is 21+21

2=

42

2= 21.

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7. The medians can be calculated as follows:

a.

The median is the number in the 𝑛+1

2 position:

𝑛+1

2=

10+1

2=

11

2= 5.5. The number below the 5.5

position is \$25,200, and the number above the 5.5 position is \$26,450. The median retail price for

the cars is 25,200+26,450

2=

51,650

2= \$25,825.

b.

The median is the number in the 𝑛+1

2 position:

𝑛+1

2=

10+1

2=

11

2= 5.5. The number below the 5.5

position is \$21,300, and the number above the 5.5 position is \$22,100. The median median dealer’s

cost for the cars is 21,300+22,100

2=

43,400

2= \$21,700.

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8. The median number of power outages can be calculated as follows:

The median is the number in the 𝑛+1

2 position:

𝑛+1

2=

12+1

2=

13

2= 6.5. The number below the 6.5

position is 3, and the number above the 6.5 position is 3. The median number of power outages is 3+3

2=

6

2= 3.

9. The median number of safety devices aboard the boats can be calculated as follows:

The median is the number in the 𝑛+1

2 position:

𝑛+1

2=

15+1

2=

16

2= 8. The number in the 8th position

is 8. The median number of safety devices aboard the boats is 8.

10. The median wage for the teacher’s assistant can be calculated as follows:

The median is the number in the 𝑛+1

2 position:

𝑛+1

2=

13+1

2=

14

2= 7. The number in the 7th position

is \$600. The median wage for the teacher’s assistant is \$600.

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5.5 Median of Large Sets of Data

1. To find the answer, enter the frequency table into L1 and L2 and use median command on the

MATH menu. The median age of the hockey players is 13.

2. To find the answer, enter the individual data values into L1 and use 1-Var Stats. The median score

for the 14 rolls of the die is 3.

3. To find the answer, enter the frequency table into L1 and L2 and use median command on the

MATH menu. The median score for the players is 5.

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4. To find the answer, enter the frequency table into L1 and L2 and use median command on the

MATH menu. The median score for Monday’s quiz is 3.

5. To find the answer, enter the frequency table into L1 and L2 and use median command on the

MATH menu. The median number of coins in a student’s pocket is 4.

6. To find the answer, enter the individual data values into L1 and use 1-Var Stats. The median

time of the participants in the race is 4.8 minutes.

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5.6 Mode

1. C

2. D

3. B

4. A

5. B

6. The modes are as follows:

a. 6

b. 16

7. There are two modes for the scores on the English quizzes. The modes are 6 and 7.

8. The mode of the amounts of time spent studying for the math test is 10 - 20 minutes.

9. The number of questions attempted most by the students was 42.

10. The mode of the numbers that were rolled is 3.

11. The mode of the numbers of games that students attended is 8.

12. The smallest possible value for m is 9.

13. The data set has two modes: 3 and 5. Each value appears 3 times. The distribution can be

described as bimodal.

14. Answers to this question will vary. Some acceptable responses would be:

• A business could use the mode to determine the most popular selling item.

• A sports team could use the mode to determine which player is most consistent.

• A teacher could use the mode to determine the length of a test by the number of questions

completed by the students on previous tests.

• A business could use the mode to determine the most popular sizes when ordering new stock.

15. The mode of the temperatures for the month of June is 74◦F.

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6.1 Standard Deviation of a Normal Distribution

1. “The typical measurement is approximately 65 kilograms, give or take 2 kilograms.”

2. “The typical measurement is approximately 25.0 miles per hour, give or take 0.5 miles per hour.”

3. “The typical measurement is approximately 375 feet, give or take 5 feet.”

4. The values within 1 standard deviation of the mean are from 16.8 − 0.7 years to 16.8 + 0.7 years, or

from 16.1 years to 17.5 years.

5. The values within 2 standard deviations of the mean are from 16.8 − (2 × 0.7) years to 16.8 + (2 × 0.7)

years, or from 15.4 years to 18.2 years.

6. The values within 3 standard deviations of the mean are from 16.8 − (3 × 0.7) years to 16.8 + (3 × 0.7)

years, or from 14.7 years to 18.9 years.

7. The graph should appear as follows:

8. The graph should appear as follows:

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9. The graph should appear as follows:

10. Answers will vary, but the actual percentages are 68% for the values within 1 standard deviation of

the mean, 95% for the values within 2 standard deviations of the mean, and 99.7% for the values

within 3 standard deviations of the mean.

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6.2 Variance of a Data Set

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6.3 Standard Deviation of a Data Set

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3. B

4. D

5. B

6. D

7. C

8. B

a. 68% of the volumes can be found between 7.4 oz and 7.6 oz.

b. 95% of the volumes can be found between 7.3 oz and 7.7 oz.

c. 99.7% of the volumes can be found between 7.2 oz and 7.8 oz.

a. 68% of the heights can be found between 51” and 61”.

b. 95% of the heights can be found between 46” and 66”.

c. 99.7% of the heights can be found between 41” and 71”.

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6.4 Applications of Variance and Standard Deviation

1. D

2. A

3. D

4. B

5. C

The variance is 4.0298883362 = 16.24.

6. D

The variance is 9.1487704092 = 83.7.

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7. A

8. B

a. The mean is 74.86.

b. The standard deviation is 14.63.

c. The variance is 14.635869562 = 214.21.

d. The normal distribution curve is shown below:

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a. The mean is 76.46.

b. The standard deviation is 14.01.

c. The variance is 14.014362882 = 196.40.

d. The normal distribution curve is shown below:

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6.5 Empirical Rule

1. The answer can be calculated as follows: 34% + 34% + 13.5% = 81.5%

2. The answer can be calculated as follows: 2.35% + 13.5% + 34% + 34% = 83.85%

3. The answer can be calculated as follows: 2.35% + 0.15% = 2.5%

4. The answer can be calculated as follows: 13.5% + 34% + 34% + 13.5% + 2.35% = 97.35%

5. The answer can be calculated as follows: 2.35% + 13.5% + 34% = 49.85%

6. The answer can be calculated as follows: 13.5% + 2.35% + 0.15% = 16%

7. The answer can be calculated as follows: 34% + 13.5% = 47.5%

8. a. The percentage of students waiting more than 11.5 minutes would be 68% + 13.5% + 2.35% +

0.15% = 84% of the students surveyed. 84% of 200 students = 0.84 × 200 = 168 students

b. The percentage of students waiting more than 18.5 minutes would be 13.5 + 2.35 + 0.15 =

16% of the students surveyed. 16% of 200 students = 0.16 × 200 = 32 students

c. The percentage of students waiting between 11.5 and 18.5 minutes would be 68% of the students

surveyed. 68% of 200 students = 0.68 × 200 = 136 students

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9. a. The percentage of babies weighing more than 7.3 lbs would be 13.5 + 2.35 + 0.15 = 16% of the

babies in the survey. 16% of 350 babies = 0.16 × 350 = 56 babies

b. The percentage of babies weighing more than 7.8 lbs would be 2.35 + 0.15 = 2.5% of the

babies in the survey. 2.5% of 350 babies = 0.025 × 350 = 9 babies

c. The percentage of babies weighing between 6.3 and 7.8 lbs minutes would be 68% + 13.5%

= 81.5% of the babies in the survey. 81.5% of 350 babies = 0.815 × 350 = 285 babies

You can use the data from the 1-Var Stats calculation to draw the normal distribution curve.

The range of the differences in heights of the seedlings for the middle 68% of the data is 5.3

inches to 10.5 inches.

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7.1 Line Graphs

1. B The points associated with continuous data are joined, because all the fraction and decimal

values between two consecutive points are meaningful.

2. B The variable is quantitative, because it represents a number.

3. A The points that represent discrete data are not joined, because the values between two

consecutive points are not meaningful.

5. For more than 20 days of advertising on Walton’s Web Ads, Plan A would be the best plan to use.

6. From the graph, you can see that Plan A intersects Plan B at (20, 70). This means that for 20 days of

advertising on Walton’s Web Ads, both Plan A and Plan B would cost \$70.00. To advertise for more

than 10 days but less than 20 days, Plan B would be the best plan to use.

7. From the graph, you can see that Plan B intersects Plan C at (10, 50). This means that for 10 days of

advertising on Walton’s Web Ads, both Plan B and Plan C would cost \$50.00. To advertise for less

than 10 days, Plan C would be the best plan to use.

8. B The time to run depends upon the fitness level of the runner.

9. The fitness level of the runner would be on the x-axis.

10. The time it takes to run the 100 yard dash would be on the y-axis.

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7.2 Broken – Line Graphs

1. B A broken-line graph shows change over time with a series of straight lines that have no single

defined slope.

2. The total distance traveled on the bike ride was 24 + 24 = 48 miles.

3. The fastest speed traveled by the bike was 24−6

2.5−2=

18

0.5= 36 miles per hour.

4. The slowest speed traveled by the bike was 6−0

4−2.5=

6

1.5= 4 miles per hour.

5. The bicyclist stopped for 2 − 1.5 = 0.5 hours, or 30 minutes.

6. The return trip took 4−2 = 2 hours.

7. Bob is getting closer to the post office from 12:30 pm to 1:00 pm, from 2:30 pm to 3:30 pm, and

from 5:00 pm to 6:00 pm.

8. Bob is getting farther away from the post office from 1:00 pm to 2:30 pm and from 3:30 pm to 5:00

pm.

9. Bob traveled a total distance of 2 + 3 + 7 + 3 + 2 = 17 kilometers.

10. Bob’s average speed was approximately 17

5.5= 3.1 kilometers per hour.

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7.3 Scatter Plots

1. C The correlation of data on a scatter plot that consists of few points that are not bunched together is

considered to be weak.

2. C The term used to denote the relationship between 2 data sets is correlation.

3. a.

b.

c.

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4. a.

b.

5. There is a strong positive correlation between the prices of the 2 stocks.

6. If the price of stock A is \$12.00, the price of stock B can be expected to be around \$43.00.

7. If the price of stock B is \$47.75, the price of stock A can be expected to be around \$14.50.

8. There is a strong negative correlation between hours of exercise per week and resting heart rate.

9. If a 30-year-old male exercises 2 hours per week, he can be expected to have a resting heart rate of

10. If a 30-year-old male has a resting heart rate of 65 beats per minute, he can be expected to exercise

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7.4 Linear Regression Equations

1. D The line of best fit can be calculated with the TI-83 by using linear regression to provide an

equation for the straight line in the form y = ax + b.

2. The linear regression equation is approximately y = −13.64x + 81.

3. The correlation coefficient is −0.9782659134, while the coefficient of determination is

0.9570041973. The linear regression equation is a good fit for the data.

4. The approximate value of y when x = 3 can be calculated as follows:

y = −13.64x + 81

y = −13.64(3) + 81

y = −40.92 + 81

y = 40.08

5. The linear regression equation is approximately y = 0.41x + 15.1.

6. The correlation coefficient is 0.9982743732, while the coefficient of determination is

0.9965517241. The linear regression equation is a very good fit for the data.

7. The approximate value of y when x = 10 can be calculated as follows:

y = 0.41x + 15.1

y = 0.41(10) + 15.1

y = 4.1 + 15.1

y = 19.2

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8. The linear regression equation is approximately y = 1.11x− 10.1.

9. The correlation coefficient is 0.9171758199, while the coefficient of determination is 0.8412114846. The linear regression equation is a moderately good fit, but not a great fit, for the data.

10. The approximate value of x when y = 8 can be calculated as follows:

y = 1.11x− 10.1

8 = 1.11x− 10.1

8 + 10.1 = 1.11x− 10.1 + 10.1

18.1 = 1.11x

x = 16.3

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7.5 Pie Charts

1. D The number of students is 9 out of 25: 9

25= 0.36 = 36%.

2. The answers to this question will vary, but here is an example of one possible solution:

3. Answers will vary. One possible answer is that pie charts are helpful for interpreting data because

they show the size of each category of data in relation to the other categories.

4. The parts of the pie chart are displayed with degrees, but they should be shown with percentages.

5. The number of students who have soccer as a favorite sport can be calculated as follows: 43.2

360= 0.12 (0.12)(550) = 66

6. The number of students who have hockey as a favorite sport can be calculated as follows: 28.8

360= 0.08 (0.08)(550) = 44

7. The number of students who have basketball as a favorite sport can be calculated as follows: 72.0

360= 0.2 (0.2)(550) = 110

8. The number of students who have baseball as a favorite sport can be calculated as follows: 100.8

360= 0.28 (0.28)(550) = 154

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9. The number of students who have football as a favorite sport can be calculated as follows: 115.2

360= 0.32 (0.32)(550) = 176

Chapter 7 – Organizing and Displaying Data Answer Key

CK-12 Basic Probability and Statistics Concepts 99

7.6 Stem – and – Leaf Plots

1. D The number of students is 7 out of 25: 7

25= 28%

From the stem and leaf plot, some information that is known is the following:

• The youngest Prime Minister was 39 when he was sworn into office.

• The oldest Prime Minister to be sworn into office was 74.

• The mode of the ages of the Prime Ministers when they were sworn into office was 46.

• The median age of the Prime Ministers when they were sworn into office was 54.5.

3. The mode of the data set is 56.

4. There are 41 data values, so the median is the value in the 21st position. Therefore, the median is 61.

5. The minimum of the data set is 10.

6. The maximum of the data set is 99.

7. The mode of the data set is 40.

8. There are 40 data values, so the median is the average of the values in the 20th and 21st positions.

Therefore, the median is 55+55

2= 55.

9. 24 of the data values are greater than 40.

10. The percentage of the data values that are less than 40 is 12

40=

3

10= 0.3 = 30%.

Chapter 7 – Organizing and Displaying Data Answer Key

CK-12 Basic Probability and Statistics Concepts 100

7.7 Bar Graphs

1. Joey has the most money in the bank.

2. Jean has the least money in the bank.

3. There are 4 bars that are below the line that represents \$800, so 4 people have less than \$800 in the

bank.

4. The tourists would most like to visit China.

5. The tourists would least like to visit France.

6. There are 2 bars that are above the line that represents 25 tourists, so 2 countries were the

most desired destination for more than 25 tourists.

7. The most popular superhero was Spider Man.

8. The least popular superhero was Iron Man.

9. 38 students had a favorite superhero of Wonder Woman.

10. The percentage of students who had a favorite superhero of Wonder Woman can be calculated as

follows: 38

10+44+16+38+12+30+18+32=

38

200= 0.19 = 19%.

Chapter 7 – Organizing and Displaying Data Answer Key

CK-12 Basic Probability and Statistics Concepts 101

7.8 Histograms

1. C, By definition, a distribution that has two peaks is bimodal.

2. There were 3 + 2 + 6 + 1 + 1 + 4 + 0 + 5 + 2 = 24 fish caught.

3. There were 4 + 0 + 5 + 2 = 11 fish caught that were over 35 cm in length.

4. There were 3 + 2 + 6 = 11 fish caught that were between 20 cm and 29 cm in length.

5. There is a blank space because no fish were caught that were between 38 cm and 41 cm.

6. There are 3 + 4 + 6 + 6 + 9 + 8 + 5 + 1 = 42 students in the class.

7. There are 5 + 1 = 6 students who have a height over 60 inches.

8. There are 6 + 9 + 8 + 5 = 28 students who have a height between 54 and 62 inches.

9. The distribution has only one peak, so the data is unimodal.

TABLE 7.2:

Bin Frequency

[50-60) 9

[60-70) 5

[70-80) 6

[80-90) 3

[90-100) 1

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7.9 Applications of Histograms

1. The bin size for the histogram is 10.

2. The bin with the highest frequency is [40-50).

3. The bins with the lowest frequency are [20-30) and [80-90).

4. The total number of data values represented by the histogram can be calculated as follows: 3 + 6 +

14 + 6 + 9 + 9 + 3 = 50

5. The percentage of the data values are in the bin [60-70) can be calculated as follows: 9

50= 0.18 =

18%

6. The bin size for the histogram is 5.

7. The bin with the highest frequency is [30-35).

8. The bin with the lowest frequency is [10-15).

9. The total number of data values represented by the histogram can be calculated as follows: 11 + 12 +

7 + 17 + 14 + 18 + 21 = 100

10. The percentage of data values that are in the bin [15-20) can be calculated as follows: 17

100= 0.17 = 17%

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7.10 Frequency Polygons

1. D By definition, a frequency polygon is a graph that uses lines to join the midpoints of the classes.

2. A Add the values and divide by 2: 14.5+23.5

2=

38

2= 19.

3. a. There were 5 + 9 + 14 + 18 + 12 + 8 + 4 = 70 players who played the sport.

b. The most common weight was between 82.5 kg and 87.5 kg.

c. Answers will vary. One possible answer is that the sport they were playing may have been rugby.

d. The weights of 55 kg and 105 kg are the weights that are one unit below and one unit above the

beginning and end points of the data set.

e. There are no weights recorded for 72.5 kg to 77.5 kg and for 77.5 kg to 82.5 kg.

4. The bin size of the histogram on which the frequency polygon is based is 10.

5. The point (40, 18) has the largest y-coordinate, so the bin that had the highest frequency was [35, 45).

6. Other than the points on the ends, the point (20, 4) has the smallest y-coordinate, so the bin that had

the lowest frequency was [15, 25).\

7. The point (50, 9) represents the bin [45, 55), so to find the number of data points that had a value

below 55, just add the y-coordinates of (20, 4), (30, 12), (40, 18), and (50, 9) to get 4 + 12 + 18 + 9 =

43. Since there were 4+12+18+9+7 = 50 data points in all, the percentage of the data that had a value

below 55 can be calculated as follows: 43

50= 0.86 = 86%.

8. 9 points would be connected to form the corresponding frequency polygon.

9. The points would be (15, 0), (25, 3), (35, 6), (45, 14), (55, 6), (65, 9), (75, 9), (85, 3), and (95, 0).

10. The frequency polygon would appear as follows:

Chapter 7 – Organizing and Displaying Data Answer Key

CK-12 Basic Probability and Statistics Concepts 104

7.11 Box – and – Whisker Plots

1. B The five–number summary consists of the minimum value, Q1, the median, Q3 and the maximum

value.

2. C The box contains 50% of the data, and each whisker contains 25% of the data.

3. B The horizontal lines on either side of the box of a box-and-whisker plot are called whiskers.

4. A If the median is located to the left of the center of the box, the distribution is positively skewed.

5. D The two horizontal lines of the box-and-whisker plot join Q1 and Q3.

6. The five-number summaries are as follows:

a. Min. Value → 60, Q1 → 68, Med → 72.5, Q3 → 77, Max. Value→ 83

b. Min. Value → 3, Q1 → 5.5, Med → 9, Q3 → 11.5, Max. Value → 15

7. The five-number summaries are as follows:

a. Min. Value → 4, Q1 → 8, Med → 13, Q3 → 16, Max. Value → 20. The median is located

to the right of the center of the box, which tells you that the distribution is negatively skewed.

b. Min. Value → 175, Q1 → 275, Med → 450, Q3 → 525, Max. Value → 625. The median is

located to the right of the center of the box, which tells you that the distribution is negatively

skewed.

8.

9.

The right whisker is longer than the left whisker, which indicates that the distribution is positively skewed.

Chapter 7 – Organizing and Displaying Data Answer Key

CK-12 Basic Probability and Statistics Concepts 105

10.

The left whisker is longer than the right whisker, which indicates that the distribution is negatively

skewed.

11.

The median is located to the right of the center of the box, which indicates that the distribution is

negatively skewed.

12.

The median is located to the right of the center of the box, which indicates that the distribution is

negatively skewed.

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CK-12 Basic Probability and Statistics Concepts 106

a.

b.

Chapter 7 – Organizing and Displaying Data Answer Key

CK-12 Basic Probability and Statistics Concepts 107

7.12 Applications of Box – and – Whisker Plots

1. The values for the five-number summary are as follows:

• Min. Value → 11, Q1 → 14, Med → 19.5, Q3 → 24, Max. Value → 26

• Answers will vary for the data set that could produce the box-and-whisker plot, but one possible data set is as follows: 11, 12, 14, 15, 19, 20, 21, 24, 25, 26

2. The values for the five-number summary are as follows:

• Min. Value → 70, Q1 → 72, Med → 80, Q3 → 89, Max. Value → 99

• Answers will vary for the data set that could produce the box-and-whisker plot, but one possible data set is as follows: 70, 72, 76, 80, 82, 89, 99

3. The values for the five-number summary are as follows:

• Min. Value → 2, Q1 → 6, Med → 11, Q3 → 12, Max. Value → 15

• Answers will vary for the data set that could produce the box-and-whisker plot, but one possible data set is as follows: 2, 3, 6, 7, 9, 11, 11, 11, 12, 13, 15

4. The values for the five-number summary are as follows:

• Min. Value → 44, Q1 → 49, Med → 50, Q3 → 50, Max. Value → 52

• Answers will vary for the data set that could produce the box-and-whisker plot, but one possible data set is as follows: 44, 49, 49, 49, 50, 50, 50, 50, 51, 52

5. The values for the five-number summary are as follows:

• Min. Value → 5, Q1 → 19.5, Med → 48.5, Q3 → 76.5, Max. Value → 94

• Answers will vary for the data set that could produce the box-and-whisker plot, but one possible data set is as follows: 5, 11, 16, 23, 30, 42, 55, 73, 74, 79, 85, 94

6. The values for the five-number summary are as follows:

• Min. Value → 9, Q1 → 19.5, Med → 21.5, Q3 → 27.5, Max. Value → 29

• Answers will vary for the data set that could produce the box-and-whisker plot, but one possible data set is as follows: 9, 19, 20, 21, 22, 26, 29, 29

7. The values for the five-number summary are as follows:

• Min. Value → 51, Q1 → 55.5, Med → 60, Q3 → 66.5, Max. Value → 69

• Answers will vary for the data set that could produce the box-and-whisker plot, but one possible data set is as follows: 51, 55, 56, 56, 60, 62, 65, 68, 69

Chapter 7 – Organizing and Displaying Data Answer Key

CK-12 Basic Probability and Statistics Concepts 108

8. The values for the five-number summary are as follows:

• Min. Value → 104, Q1 → 111, Med → 156, Q3 → 182, Max. Value → 193

• Answers will vary for the data set that could produce the box-and-whisker plot, but one possible data set is as follows: 104, 111, 146, 156, 167, 182, 193

9. The values for the five-number summary are as follows:

• Min. Value → 16, Q1 → 21, Med → 21, Q3 → 25, Max. Value → 28

• Answers will vary for the data set that could produce the box-and-whisker plot, but one possible data set is as follows: 16, 21, 21, 21, 21, 21, 21, 21, 23, 25, 25, 25, 26, 27, 28

10. The values for the five-number summary are as follows:

• Min. Value → 80, Q1 → 81.5, Med → 90, Q3 → 91.5, Max. Value → 93

• Answers will vary for the data set that could produce the box-and-whisker plot, but one possible data set is as follows: 80, 81, 82, 83, 90, 90, 91, 92, 93

Chapter 8 – Organizing and Displaying Data for Comparison Answer Key

CK-12 Basic Probability and Statistics Concepts 109

8.1 Basic Graph Types

1.

2.

3. Continuous data

4. Discrete data

5. Continuous data

6. Discrete data

7. Categorical (qualitative) data

8. Numerical (quantitative) data

9. Numerical (quantitative) data

10. Categorical (qualitative) data

Chapter 8 – Organizing and Displaying Data for Comparison Answer Key

CK-12 Basic Probability and Statistics Concepts 110

8.2 Double Line Graphs

1. The graph is wrong because it shows the time spent on games increasing and then decreasing for both

Michael and Scott multiple times. It should only show the time spent on games increasing or

staying the same.

2. Answers will vary, but they could include some of the following statements:

An 18-wheeler typically holds 100 gallons of fuel. The data in the graph shows that the tank needed

to be refilled at approximately 400 miles. Therefore, the 18-wheeler can go about 4 miles to the

gallon. A car can hold a variety of gallons of fuel. This car holds 20 gallons, which is usually typical

of a larger car or an SUV. From 100 to 200 miles, the car began with a full tank and then ran out of

gas, so the car can go about 5 miles per gallon.

3. Room temperature is approximately 20◦C.

4. Thomas’ oven was about 350◦C.

Chapter 8 – Organizing and Displaying Data for Comparison Answer Key

CK-12 Basic Probability and Statistics Concepts 111

6. Thomas let the cookies cool for about 15 minutes before he taste-tested them.

7. The cookies were approximately 30◦C when Thomas taste-tested them.

8. It took Jack about 5 minutes to boil the water.

9. Jack let the water cool for approximately 10 minutes before taste-testing his hot chocolate.

10. Jack’s hot chocolate was about 40◦C when he taste-tested it.

Chapter 8 – Organizing and Displaying Data for Comparison Answer Key

CK-12 Basic Probability and Statistics Concepts 112

8.3 Two-Sided Stem – and – Leaf Plors

1. The range for the number of home runs hit by the Mets was 0 to 51, while the range for the number

of home runs hit by the Phillies was 0 to 48.

2. The median for the number of home runs hit by the Mets was 22, while the median for the number

of home runs hit by the Phillies was 17.

3. The mode for the number of home runs hit by the Mets was 0, while the mode for the number of

home runs hit by the Phillies was 1.

4. The Mets had 13 players with 20 or more home runs, while the Phillies had 10 players with 20 or

more home runs. Therefore, the Mets had more players with 20 or more home runs.

5. The range for the highest scores for the girls was 92 to 204, while the range for the highest scores for

the boys was 105 to 195.

6. The median for the highest scores for the girls was 126+130

2= 128, while the median for the highest

scores for the boys was 144.

7. The mode for the highest scores for the girls was 125, while the mode for the highest scores for the

boys was 162.

8. The highest score for the girls was 204, while the highest score for the boys was 195. Therefore, a

girl had the highest score in the intramural bowling league.

Answers will vary, but the data suggests that there is a wider variation in the pulse rates for the group

of girls than for the group of boys. For the girls, the pulse rates ranged from 60 to 89, whereas for the

boys, the pulse rates ranged from 70 to 88. The median for the girls group is at 76, and the mode is

also at 76. For the group of boys, however, the median is at 85, and the mode is at 82. The boys seem

to have higher pulse rates.

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CK-12 Basic Probability and Statistics Concepts 113

Answers will vary, but the data suggests that the variations in the line-up times for the customers of

Starbucks and for the customers at Just Us Coffee are the same. For the Starbucks customers, the

line-up times ranged from 8 to 27 minutes, whereas for the Just Us Coffee customers, the line-up

times ranged from 10 to 29 minutes. The median for the Starbucks customers’ line-up times is at

17 minutes, and the mode is at 12 minutes. For the Just Us Coffee customers, however, the median

is at 16 minutes, and the mode is at 10 minutes. It would seem that the Just Us Coffee customers

have slightly shorter line-up times.

Chapter 8 – Organizing and Displaying Data for Comparison Answer Key

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8.4 Double Bar Graphs

1. More dogs were adopted than cats during the third and fourth quarters of last year.

2. More cats were adopted than dogs during the first and second quarters of last year.

3. In total, 15 + 12 + 18 + 21 = 66 dogs were adopted last year, and 21 + 24 + 15 + 18 = 78 cats were

4. Dealership A and Dealership B combined to sell the most cars on Saturday, when they sold 8 + 9 = 17

cars.

5. On Monday, Dealership B sold 4 − 1 = 3 more cars than Dealership A.

6. The biggest difference between cars sold by the 2 dealerships happened on Thursday, when

Dealership A sold 6 – 2 = 4 more cars than Dealership B.

7. 8 − 3 = 5 more of the non-college graduates primarily use television to follow the news than use the

newspaper.

follow the news. Only 2 + 1 = 3 people prefer this form of media. 9. The percentage of college graduates who primarily use the internet to follow the news can be

calculated as follows; 9

4+1+9+2+4=

9

20= 0.45 = 45%

Answers will vary, but here is a

possible answer: With the survey done

by the guidance counselor, it was

found that more girls were planning

on going to university following high

school, and more boys were planning

on going to college, the military, or

directly into some form of

employment. It was also found that

more girls were unsure of what their

plans were than boys. Of the 186

students surveyed, 75 (or 40%) were

planning on attending university, and

124 (or 66.7%) were planning on

attending either university or college.

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8.5 Double Box – and – Whisker Plots

Conclusions drawn may vary, but some points made by students could include the following:

Using the medians, 50% of the students spent at least 10 hours studying for the higher level exams,

and 50% of the students spent at least 7 hours studying for the standard level exams. This would

seem reasonable, since the higher level exams are on more material than the standard level exams.

The range for the higher level exam study times was 19 − 5 = 14 hours, while the range for the

standard level exam study times was 20 − 2 = 18 hours. Since the range for the higher level exam

study times is smaller, it means the study times are less spread out. This would then mean that the

study times are more predictable and reliable.

Also, the data in the standard level exam study times is not very even, meaning that the box is not split

into two equal parts. The same can be found for the higher level exam study times, but not to the

extreme as with the standard level exam study times. In addition, the tails for the higher level exam

study times and the standard level exam study times show the lower halves are shorter than the upper

halves. This means that the lower halves have less dispersed data than the upper halves. However,

for the higher level exam study times, the tails seem to be of more equal lengths, meaning the data

has a more equal distribution than the data for the standard level exams.

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2. The double box-and-whisker plots should appear as follows:

Conclusions drawn may vary, but some points made by students could include the following:

Using the medians, 50% of the students scored at least 552 on the math portion of the SAT exam,

and 50% of the students scored at least 524 on the verbal portion of the SAT exam. This is probably

expected, since the students were members of the AP Stats course. The range for the math scores

was 575 - 513 = 62 points, while the range for the verbal scores was 579 - 478 = 101 points. Since

the range for the math scores is smaller, it means the math scores obtained by the students are less

spread out. This would then mean that the math scores are more predictable and reliable.

Also, the data for the verbal scores is very even, meaning that the box is split into two equal parts.

This is not the same as in the box-and-whisker plot for the math scores on the top. In addition, the

tails for the math scores show the upper half is slightly shorter than the lower half. This means that

the upper half has less dispersed data than the lower half. For the verbal scores, the reverse is true.

The upper tail seems to be slightly longer than the lower tail, so the data in the lower half is less

dispersed than in the upper half.

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4. The double box-and-whisker plots should appear as follows:

5. The five-number summary for the 2006 games is as follows:

6. The five-number summary for the 2010 games is as follows:

7. Using the medians, 50% of the judge’s scores were at least 8 for the 2006 winter games, but in 2010,

50% of the scores were at least 8.5.

8. The range for the scores in the 2006 winter games was 9:2 - 7 = 2:2 points, while the range for the

scores in the 2010 winter games was 9:9 - 6:4 = 3:5 points.

9. Answers will vary, but the data in the 2006 winter games is not very even, meaning that the box is

not split into two equal parts. The reverse is true for the 2010 winter games. However, the tails for

the 2010 winter games show the lower half is longer than the upper half. This means that the upper

half has less dispersed data than the lower half. For the 2006 winter games, the tails seem to be of

equal lengths, meaning an equal distribution of data.

10. Answers will vary, but since the range for the 2006 winter games is smaller, it means the judges

scores are less spread out. This could then mean that the scoring was more predictable and reliable.

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