ck-12 basic probability and

CK-12 Basic Probability andStatistics, Solution Key

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Contents www.ck12.org

Contents

1 Independent and Dependent Event, Solution Key 11.1 Solution Key for Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 The Next Step ... Conditional Probability, Solution Key 82.1 Solution Key for Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Introduction to Discrete Random Variables, Solution Key 203.1 Solution Key for Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Probability Distributions, Solution Key 294.1 Solution Key for Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5 Measures of Central Tendency, Solution Key 395.1 Solution Key for Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

6 The Shape, Center and Spread of a Normal Distribution, Solution Key 486.1 Solution Key for Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

7 Organizing and Displaying Distributions of Data, Solution Key 577.1 Solution Key for Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

8 Organizing and Displaying Data for Comparison, Solution Key 698.1 Solution Key for Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

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www.ck12.org Chapter 1. Independent and Dependent Event, Solution Key

CHAPTER 1 Independent andDependent Event, Solution Key

Chapter Outline1.1 SOLUTION KEY FOR REVIEW QUESTIONS

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1.1. Solution Key for Review Questions www.ck12.org

1.1 Solution Key for Review Questions

1. (a) and (b) are independent. They cannot occur at the same time.2. (a) and (c) are independent. They cannot occur at the same time.3. (b) and (c) are dependent. They can occur at the same time.4. (a) and (c) are dependent. They can occur at the same time.5. P(A) = probability of getting a sum equal to 6.

P(A) = 536

6. P(A) = probability of getting a less than 6.

P(A) = 1036

P(A) = 518

7. P(A) = probability of getting a sum greater than 6.

P(A) = 2136

P(A) = 712

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Or: P(> 6) = 1− (P(6)+P(< 6))

P(> 6) = 1−( 5

36 +1036

)P(> 6) = 1− 15

36

P(> 6) = 3636 −

1536

P(> 6) = 2136

8. P(A) = probability of getting a sum of at least 6.

P(A) = 2636

P(A) = 1318

Or: P(sum = at least 6) = 1−P(< 6)

P(sum = at least 6) = 1− 1036

P(sum = at least 6) = 3636 −

1036

P(sum = at least 6) = 2636

P(sum = at least 6) = 1318

9. P(A) = probability of getting tails on a coin toss

P(A) = 12

P(B) = probability of getting a 5 on a die toss

P(B) = 16

P(A and B) = 12 ×

16

P(A∩B) = 112

10. The Venn diagram is given below:

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11. The Venn diagram is given below:

12. P(A) = probability of selecting a student who wants a uniform

P(A) = 0.45

P(B) = probability of selecting a student who does not want a uniform

P(B) = 0.55

P(A or B) = 0.45+0.55

P(A or B) = 1

13.

P(A) = probability of selecting a number that is a multiple of 4 = {4,8,12,16,20}

P(A) = 510

P(B) = probability of selecting a number that is a multiple of 5 = {10,20}

P(B) = 210

P(A and B) = 510 ×

210

P(A∩B) = 10100

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P(A∩B) = 110

14. P(A∪B) = P(A)+P(B)−P(A∩B)

P(A∪B) = 510 +

210 −

110

P(A∪B) = 610

P(A∪B) = 35

15. P(A) = probability of selecting a red jelly bean

P(A) = 1037

P(B) = probability of selecting a green jelly bean

P(B) = 1237

P(A or B) = 1037 +

1237

P(A∪B) = 2237

16. P(A) = probability of selecting a blue or green jelly bean

P(A) = P(Blue or Green)1037 +

1237

P(A) = 2737

P(B) = probability of selecting a red jelly bean on the second draw

P(B) = 1036

P(A and B) = 2737 ×

1036

P(A∩B) = 2701332

P(A∩B) = 1574

17. P(A) = probability of a student taking chemistry

P(A) = 1760

P(B) = probability of a student taking math

P(B) = 1360

P(A or B) = 1760 +

1360

P(A∪B) = 3060

P(A∪B) = 12

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18. (a) P(A) = probability of a student wanting candy

P(A) = 2875

P(B) = probability of a student wanting a pencil

P(B) = 2175

P(A or B) = 2875 +

2175

P(A∪B) = 4975

(b) P(C) = probability of wanting neither a candy bar nor a pencilP(C) = 18

7519. P(A) = probability of selecting a heart from a deck of cards

P(A) = 1352

P(B) = probability of selecting a spade from a deck of cards

P(B) = 1352

P(A and B) = 0 since a card cannot be both a heart and a spade.

P(A∩B) = 0

P(A∪B) = P(A)+P(B)−P(A∩B)

P(A∪B) = 1352 +

1352 −0

P(A∪B) = 2652

P(A∪B) = 12

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The events are mutually exclusive since P(A∩B) = 0.

20. P(A) = probability of selecting a heart from a deck of cards

P(A) = 1352

P(B) = probability of selecting a face card from a deck of cards

P(B) = 1252

P(A and B) = 1352 ×

1252

P(A∩B) = 1562704

P(A∩B) = 352

P(A∪B) = P(A)+P(B)−P(A∩B)

P(A∪B) = 1352 +

1252 −

352

P(A∪B) = 2252

P(A∪B) = 1126

The events are not mutually exclusive since P(A∩B) 6= 0.

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CHAPTER 2 The Next Step ...Conditional Probability, Solution Key


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www.ck12.org Chapter 2. The Next Step ... Conditional Probability, Solution Key


1. Answer:

a. Tree diagram:

• P(two red balls) = 37 ×

37 = 9

49• There are two ways that Thomas can pick a red ball on the second pick. He can pick a blue ball first

and a red ball second (P(blue then red) or P(blue and red)) or he can pick a red ball first and a red ballsecond (P(red then red) or P(red and red)). To find the total probability, we then must add these twoprobabilities together.

P(blue then red) or P(red then red) = P(blue and red)+P(red and red)

P(blue and red)+P(red and red) =47× 3

7+

949

P(blue and red)+P(red and red) =1249

+949



The probability of getting an eraser on the second pick is 37 or 42.8%.

2. Answer:

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a.b. There are two ways that Janet can pick an eraser on the second pick. She can pick a pencil first and an

eraser second (P(pencil then eraser) or P(pencil and eraser)) or she can pick an eraser first and an erasersecond (P(eraser then eraser) or P(eraser and eraser)). To find the total probability, we then must addthese two probabilities together.

P(pencil then eraser) or P(eraser then eraser) = P(pencil and eraser)+P(eraser and eraser)

P(pencil and eraser)+P(eraser and eraser) =23× 1

3+

13× 1

3

P(pencil and eraser)+P(eraser and eraser) =29+

19

P(pencil and eraser)+P(eraser and eraser) =39

P(pencil and eraser)+P(eraser and eraser) =13

The probability of getting an eraser on the second pick is 13 or 33.3%.

3. Answer:

a. Combinationb. Permutationc. Permutation

4. Use the formula nPr. 7P5 = 7 · 6 · 5 · 4 · 3 = 2520↑ starting with 7 multiply 5 factors (The number of ways toarrange 5 objects that are chosen from a set of 7 different objects.)

5.

4P2×5 P3

4P2 =4!

(4−2)! 5P3 =5!

(5−3)!

4P2 =4×3×2×1

2×1 5P3 =5×4×3×2×1

2×14P2 = 12 5P3 = 60

4P2× 5P3 = 12×60

4P2× 5P3 = 720

6. Use the formula nPr.. 5P4 = 5 ·4 ·3 ·2 = 120 ( 5 numbers to choose from, n = 5, want 4-digit numerals, r = 4)

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7. There are nine (9) letters in the word REFERENCE therefore n = 9. We want 9 letter arrangements; thereforewe are choosing nine (9) objects at a time. In this example r = 9. In this problem, we are using a group ofletters with letters that repeat. In the group of nine letters (REFERENCE), there are two R’s and four E’s. Sox1 = 2 and x2 = 4.

8. There are eleven (11) letters in the word MISSISSIPPI therefore n = 11. We want 11 letter arrangements;therefore we are choosing eleven (11) objects at a time. In this example r = 11. In this problem, we are usinga group of letters with letters that repeat. In the group of eleven letters (MISSISSIPPI), there are two P’s, fourI’s, and four S’s. So x1 = 2,x2 = 4 and x3 = 4.

9. There are eleven (11) letters in the word MATHEMATICS therefore n = 11. We want 11 letter arrangements;therefore we are choosing eleven (11) objects at a time. In this example r = 11. In this problem, we are usinga group of letters with letters that repeat. In the group of eleven letters (MATHEMATICS), there are two M’s,two A’s, and two T’s. So x1 = 2,x2 = 2 and x3 = 2.

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10. There are eight (8) cookies that we can order therefore n = 8. We want 8 different cookie orders; thereforewe are choosing eight (8) objects at a time. In this example r = 8. In this problem, we are using a groupof cookies with types of cookies that repeat. In the group of eight cookies (4 chocolate chip, 2 oatmealand 2 double chocolate cookies), there are four chocolate chips, two oatmeal, and two double chocolate. Sox1 = 4,x2 = 2 and x3 = 2.

11. There are fifteen (15) questions on the math test that therefore n = 15. We have a maximum of 15 differentanswer sheets possible; therefore we are choosing fifteen (15) objects at a time. In this example r = 15. Inthis problem, we are using a group of answer choices with types that repeat. In the group of answer choices(5 questions have the answer A, 4 have the answer B, 3 have the answer C, 2 have the answer D, and 1 has theanswer E), there are five A’s, four B’s, three C’s, and two D’s. So x1 = 5,x2 = 4,x3 = 3 and x4 = 2.

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12. 38 songs to choose from, and selecting 17

38C17 =38!

17!(38−17!)

38C17 =38!

17!(21!)

38C17 = 2.878×1010

13. 9 toppings to choose from and selecting 2

9C2 =9!

2!(9−2!)

9C2 =9!

2!(7!)

9C2 = 36

14. 17 randomly placed dots on a circle, choosing any two dots to draw lines

17C2 =17!

2!(17−2!)

17C2 =17!

2!(15!)

17C2 = 136

15. 13 people to choose from, choosing 4

13C4 =13!

4!(13−4!)

13C4 =13!

4!(9!)

13C4 = 715

16. There are 4 meats and you are selecting 1, therefore we have 4C1 There are 10 veggies available and you arechoosing 3, therefore we have 10C3

4C1 =4!

1!(4−1!) 10C3 =10!

3!(10−3!)

4C1 =4!

1!(3!) 10C3 =10!

3!(7!)

4C1 = 4 10C3 = 120

4C1× 10C3 = 4×120 = 480

17. There are 15 freshmen and 30 seniors; therefore there are 45 students in total. (a) Choosing 4 students to goto the State Math Championships, order is not important.

45C4 =45!

4!(45−4!)

45C4 =45!

4!(41!)

45C4 = 148995

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(b) Choosing 2 freshmen from a class of 15 freshmen, therefore 15C2 Choosing 2 seniors from a class of 30seniors, therefore 30C2

15C2 =15!

2!(15−2!) 30C2 =30!

2!(30−2!)

15C2 =15!

2!(13!) 30C2 =30!

2!(28!)

15C2 = 105 30C2 = 435

15C2× 30C2 = 105×435

15C2× 30C2 = 45675 possible arrangements

18. Answer:

P(R|B) = P(B AND R)P(B)

P(R|B) = 0.25×0.650.25

P(R|B) = 0.65

19. Step 1: List what you know

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First event = first die is a 2

P(2) =16

P(2 and sum is even) =16× 18

36

P(2 and sum is even) =18216


Step 2: Calculate the probability that the sum is even when the first card chosen is a 2.

P(sum is even|2) = P(2 and sum is even)P(2)

P(sum is even|2) =1

1216

P(sum is even|2) = 112× 6

1

P(sum is even|2) = 612


Step 3: Write your conclusion: Therefore the probability that the sum is even given that the first die that isrolled is a 2 is 50%



P(5) =16

P(5 and sum is even) =16× 18

36



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Step 2: Calculate the probability that the sum is even when the first card chosen is a 5.

P(sum is even|5) = P(5 and sum is even)P(5)

P(sum is even|5) =1

1216

P(sum is even|5) = 112× 6

1



Step 3: Write your conclusion: Therefore the probability that the sum is even given that the first die that isrolled is a 5 is 50%



P(5) =16

P(5 and sum is odd) =16× 18

36

P(5 and sum is odd) =18216

P(5 and sum is odd) =112

Step 2: Calculate the probability of that the sum is odd when the first card chosen is a 5.

P(sum is odd|5) = P(5 and sum is odd)P(5)

P(sum is odd|5) =1

1216

P(sum is odd|5) = 112× 6

1

P(sum is odd|5) = 612

P(sum is odd|5) = 12

Step 3: Write your conclusion: Therefore the probability that the sum is odd given that the first die that isrolled is a 5 is 50%

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First event = being dealt the king

P(King) =452

P(red card and King) =452× 26

51

P(red card and King) =1042652

P(red card and King) =26663

Step 2: Calculate the probability of choosing red card as a second card when a king is chosen as a first card.

P(red card|King) =P(King and red card)

P(King)

P(red card|King) =26

663452

P(red card|King) =26663× 52

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Step 3: Write your conclusion: Therefore the probability of selecting a red card as a second card when a kingis chosen as a first card is 51%.


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First event = being dealt the seven

P(seven) =452

P(black card and seven) =452× 26

51

P(black card and seven) =1042652

P(black card and seven) =26663

Step 2: Calculate the probability of choosing black card as a second card when a seven is chosen as a firstcard.

P(black card|seven) =P(seven and black card)

P(seven)

P(black card|seven) =26

663452

P(black card|seven) =26663× 52

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Step 3: Write your conclusion: Therefore the probability of selecting a black card as a second card when aseven is chosen as a first card is 51%.


First event = getting an allowance for doing chores

P(doing chores) = 0.55

P(doing chores and being good) = 0.25

Step 2: Calculate the probability of choosing being good given that they have received an allowance.

P(being good|doing chores) =P(doing chores and being good)

P(doing chores)

P(being good|doing chores) =0.250.55

P(being good|doing chores) = 0.45

Step 3: Write your conclusion: Therefore the probability of a child will choose to be good given that they aregiven an allowance for doing chores is 45%.


First event = student speaks French

P(French) = 0.45

P(French and English) = 0.15

Step 2: Calculate the probability of a student speaks English given that they can speak French.

P(English|French) =P(French and English)

P(French)

P(English|French) =0.150.45

P(English|French) = 0.33

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Step 3: Write your conclusion: Therefore the probability of a student speaks English given that they speakFrench is 33%.


First event = student takes ArtP(Art) = 0.65

P(Art and Statistics) = 0.10

Step 2: Calculate the probability of a student takes Statistics given that they take Art.

P(Statistics|Art) =P(Art and Statistics)

P(Art)

P(Statistics|Art) =0.100.65

P(Statistics|Art) = 0.15

Step 3: Write your conclusion: Therefore the probability of a student takes Statistics given that they are takingArt is 15%.

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CHAPTER 3 Introduction to DiscreteRandom Variables, Solution Key


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www.ck12.org Chapter 3. Introduction to Discrete Random Variables, Solution Key


1. The solutions are in the Table 3.1.

TABLE 3.1: Table for Problem 1

Probability Statement P(X)a. The probability of this event will never occur. e P(X) = 1.0b. The probability of this event is highly likely. d P(X) = 0.33c. The probability of this event is very likely. c P(X) = 0.67d. The probability of this event is somewhat likely. a P(X) = 0.00e. The probability of this event is certain. b P(X) = 0.95

2. The solutions are in the Table 3.2.

TABLE 3.2: Table for Problem 2

Probability Statement P(X)a. I bought a ticket for the State Lottery. The probabil-ity of a successful event (winning) is likely to be:

d P(X) = 0.80

b. I have a bag of equal numbers of red and green jellybeans. The probability of reaching into the bag andpicking out a red jelly bean is likely to be:

b P(X) = 0.50

c. My dad teaches math and my mom chemistry. Theprobability that I will be expected to study science ormath is likely to be:

c P(X) = 0.67

d. Our class has the highest test scores in the State MathExams. The probability that I have scored a great markis likely to be:

e P(X) = 1.0

e. The Chicago baseball team has won every game inthe season. The probability that the team will make itto the play offs is likely to be:

a P(X) = 0.01

3. Answers will vary depending on who is doing the problem and where they live.4. Answers will vary depending on who is doing the problem and where they live.5. P(scoring above 80%) = 0.20

Therefore, P(scoring below 80%) = 1−0.20 = 0.806. P(getting a job after university) = 0.85

Therefore, P(not getting a job after university) = 1−0.85 = 0.157. No it does not since ∑P(X) = 0.2+0.4+0.6+0.8 Or ∑P(X) = 2.08. No it does not since ∑P(X) = 0.202+0.174+0.096+0.078+0.055 Or ∑P(X) = 0.6059. Yes it does since ∑P(X) = 0.302+0.251+0.174+0.109+0.097+0.067 Or ∑P(X) = 1.0

10. Answer:

a. P(X) = 16 = 0.167

b. p = 0.167q = 1−0.167 = 0.833

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n = 10a = 1

P(X = a) = nCa× pa×q(n−a)

P(X = 1) = 10C1× p1×q(10−1)

P(X = 1) = 10C1× (0.167)1× (0.833)(10−1)

P(X = 1) = (10)× (0.167)× (0.193)

P(X = 1) = 0.322

Therefore, the probability of seeing a 6 when a die is rolled 10 times is 32.2%.11. Answer:

a. P(X) = 16 = 0.167

b. p = 0.167q = 1−0.167 = 0.833n = 15a = 1


P(X = 1) = 15C1× p1×q(15−1)

P(X = 1) = 15C1× (0.167)1× (0.833)(15−1)

P(X = 1) = (115)× (0.167)× (0.0774)

P(X = 1) = 0.194

Therefore, the probability of seeing a 6 when a die is rolled 15 times is 19.4%.12. Answer:

a. P(X) = 16 = 0.167

b. p = 0.167q = 1−0.167 = 0.833n = 15a = 7


P(X = 7) = 15C7× p7×q(15−7)

P(X = 7) = 15C7× (0.167)7× (0.833)(15−7)

P(X = 7) = (6435)× (3.62×10−6)× (0.231)

P(X = 7) = 0.00538

Therefore, the probability of seeing a 5 seven times (in 7 rolls) when a die is rolled 15 times is 0.538%.13. If we look at the chart below, we can see the number of times a five (5) shows up when rolling two die.

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(a) The probability of seeing at least one five (5) is:

P(X) =1136

= 0.306

(b) The probability of seeing at least one five (5) in exactly 5 of these 10 rolls p = 0.306q = 1−0.306 = 0.694n = 10 (10 trials)a = 5 (want to see at least one 5 in five of the ten rolls)


P(X = 5) = 10C5× p5×q(10−5)

P(X = 5) = 10C5× (0.306)5× (0.694)(10−5)

P(X = 5) = (252)× (0.00268)× (0.161)

P(X = 5) = 0.109

Therefore, the probability of rolling at least one 5 five times when two die are rolled 10 times is 10.9%.14. If we look at the chart below, we can see the number of times a five (5) shows up when rolling two die.

(a) The probability of seeing at least one five (5) is:

P(X) =1136

= 0.306

(b) The probability of seeing at least one five (5) in exactly 8 of these 15 rolls p = 0.306q = 1−0.306 = 0.694n = 15 (15 trials)a = 8 (want to see at least one 5 in eight of the fifteen rolls)


P(X = 8) = 15C8× p8×q(15−8)

P(X = 8) = 15C8× (0.306)8× (0.694)(15−8)

P(X = 8) = (6435)× (7.69×10−5)× (0.0775)

P(X = 8) = 0.384

Therefore, the probability of rolling at least one 5 eight times when two die are rolled 15 times is 3.84%.

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15. n = 7 (7 trials)p1 =

1352 = 0.25 (probability of drawing a heart)

p2 =1352 = 0.25 (probability of drawing a spade)

p3 =1352 = 0.25 (probability of drawing a club)

p4 =1352 = 0.25 (probability of drawing a diamond)

n1 = 2 (2 heart)n2 = 2 (2 spade)n3 = 2 (2 club)n4 = 3 (3 diamonds)k = 1 (for each trial there is only 1 possible outcome)

P(x = 7) =n!

n1!n2!n3! . . .nk!× (p1

n1× p2n2× p3

n3 . . . pknk)

P(x = 7) =7!

2!2!2!3!× (0.252×0.252×0.252×0.253)

P(x = 7) = 105× (0.0625)× (0.0625)× (0.0625)× (0.015625)

P(x = 7) = 4.00×10−4

Therefore, the probability of choosing two hearts, two spades, two clubs, and three diamonds is 0.0400%.16. n = 7 (7 trials)

p1 = 0.35 (35% probability the student watches NBX)p2 = 0.40 (40% probability the student watches FIX)p3 = 0.15 (15% probability the student watches TSA)p4 = 0.10 (10% probability the student watches MMA)n1 = 1 (1 student watching NBX)n2 = 2 (2 students watching FIX)n3 = 2 (2 students watching TSA)n3 = 3 (3 students watching MMA)k = 1 (for each trial there is only 1 possible outcome)

P(x = 7) =n!

n1!n2!n3! . . .nk!× (p1

n1× p2n2× p3

n3 . . . pknk)

P(x = 7) =7!

1!2!2!3!× (0.351×0.402×0.152×0.103)

P(x = 7) = 420× (0.35)× (0.16)× (0.015)× (0.015)

P(x = 7) = 5.29×10−4

Therefore, the probability from this survey is 0.0529%.17. Answers will vary but should include: Theoretical Probability –the calculated probability of an event Experi-

mental Probability –the actual probability of an event found through experimentation18. Answer:

a. What is the difference between theoretical and experimental probability? The difference between ex-perimental and theoretical probability is that theoretical is calculated and experimental is actual. Intheoretical probability, you are analyzing equally likely outcomes.

b. The more data that is collected, the closer the experimental probability gets to the theoretical probability.c. Say we are going to calculate the probability of getting heads when one coin is tossed 100 times. We

know that the probability of getting heads on the single coin toss is 0.5.

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P(H) = 0.5×0.5×0.5×0.5×0.5 . . .(100 times)

Or

P(H) = 0.5100

P(H) = 7.89×10−31

Now let’s say we are going to toss 100 coins just once. In order to get 100 heads, we do the followingprobability calculation. The total number of coins = 100 Each coin has the outcome of either a Head (H) orTail (T).

# o f f avorable choices =# possible letters in combination!

letter X!×letter Y !

# o f f avorable choices =100 letters!

100 head!×0 tails!# o f f avorable choices = 1

Denominator (Bottom) The number of possible outcomes = 2100 The number of possible outcomes = 1.27×1030 Now we just divide the numerator by the denominator.

P(100 heads) =1

1.27×1030

P(100 heads) = 7.89×10−31

So, yes tossing one coin 100 times is the same as tossing 100 coins one time.

19.

Storing the data into L1

Graphing a histogram we see:

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P(2 heads) = 425 = 16%

20.

Storing the data into L1

Graphing a histogram we see:

P(4 heads) = 750 = 14%

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21.

# o f f avorable choices =10×9×8×7×6×5×4×3×2×1

(3×2×1)× (7×6×5×4×3×2×1)

# o f f avorable choices =3628800

(6)× (5040)


30240# o f f avorable choices = 120

The number of possible outcomes = 210 The number of possible outcomes = 1024 Now we just divide thenumerator by the denominator.

P(3 heads) =1201024

P(3 heads) = 0.117

P(3 heads) = 11.7%

22.

P(3 heads) =110

P(3 heads) = 0.10

P(3 heads) = 10%

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23.

# o f f avorable choices =12×11×10×9×8×7×6×5×4×3×2×1(8×7×6×5×4×3×2×1)× (4×3×2×1)


(40320)× (24)




P(8 heads) =4954096

P(8 heads) = 0.121

P(8 heads) = 12.1%

24.

# o f f avorable choices =14×12×12×11×10×9×8×7×6×5×4×3×2×1(7×6×5×4×3×2×1)× (7×6×5×4×3×2×1)

# o f f avorable choices =8.72×1010

(5040)× (5040)

# o f f avorable choices =8.72×1010



P(7 heads) =3432

16384P(7 heads) = 0.209

P(7 heads) = 20.9%

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www.ck12.org Chapter 4. Probability Distributions, Solution Key

CHAPTER 4 Probability Distributions,Solution Key


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1. The graphs or histograms alone can be elementary determined to be normal (standard) distributions, binomialdistributions, or exponential distributions based on shape.

a.

b.

c.

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d.

2. The graphs or histograms alone can be elementary determined to be normal (standard) distributions, binomialdistributions, or exponential distributions based on shape.

a.

b.

c.

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d.

3. There are four (4) trials, therefore n = 4. A success is when we two boys being born of the four children.Therefore a = 2. The probability of a success is 60% or 0.60 and thus p = 0.60. Therefore the probability ofa failure is 1−0.60 or 0.40. From this we know that q = 0.40.


P(3 girls) = 4C2× p2×q2

P(3 girls) = 4C2× (0.60)2× (0.40)2

P(3 girls) = (6)× (0.36)× (0.16)

P(3 girls) = 0.3456

Therefore, the probability of having exactly two (2) boys from the four (4) children is 34.6%. To usetechnology, you will involve the binompdf formula because you are looking at exactly two (2) boys fromfour (4) children.

4. There are four (4) trials, therefore n = 4. A success is when you see at least two (2) boys being born of thefour (4) children. Therefore a = 4, 3, and 2. The probability of a success is 60% or 0.60 and thus p = 0.60.Therefore the probability of a failure is 1−0.60 or 0.40. From this you know that q = 0.40.

P(4 boys) = 4C4× p4×q0

P(4 boys) = 4C4× (0.60)4× (0.40)0

P(4 boys) = (1)× (0.1296)× (1)

P(4 boys) = 0.1296

P(3 boys) = 4C3× p3×q1

P(3 boys) = 4C3× (0.60)3× (0.40)1

P(3 boys) = (4)× (0.216)× (0.40)

P(3 boys) = 0.3456

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P(2 boys) = 4C2× p2×q2

P(2 boys) = 4C2× (0.60)2× (0.40)3

P(2 boys) = (6)× (0.36)× (0.16)

P(2 boys) = 0.3456

The total probability for this example is:

P(X ≥ 2) = 0.1296+0.3456+0.3456

P(X ≥ 2) = 0.8208

Therefore, the probability of have at least two (2) boys in four (4) children is 82.1%. To use technology, youwill involve the binomcdf formula because you are looking at least two (2) boys in four (4) children.

5. There are four (4) trials, therefore n = 4. A success is when you see at most two (2) boys being born of thefour (4) children. Therefore a = 2, 1, and 0. The probability of a success is 60% or 0.60 and thus p = 0.60.Therefore the probability of a failure is 1−0.60 or 0.40. From this you know that q = 0.40.


P(2 boys) = 4C2× p2×q2

P(2 boys) = 4C2× (0.60)2× (0.40)3

P(2 boys) = (6)× (0.36)× (0.16)

P(2 boys) = 0.3456

P(1 boy) = 4C1× p1×q3

P(1 boy) = 4C1× (0.60)1× (0.40)3

P(1 boy) = (4)× (0.60)× (0.064)

P(1 boy) = 0.1536

P(0 boys) = 4C0× p0×q4

P(0 boys) = 4C0× (0.60)0× (0.40)4

P(0 boys) = (1)× (1)× (0.0256)

P(0 boys) = 0.0256

The total probability for this example is:

P(X ≤ 2) = 0.3156+0.1536+0.0256

P(X ≤ 2) = 0.5248

Therefore, the probability of have at least two (2) boys in four (4) children is 52.5%. To use technology, youwill involve the binomcdf formula because you are looking at most two (2) boys in four (4) children.

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6. 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 aquiz out of 25, the lowest score was 2 and the highest was 25. The mean score was 16.04.

7. 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 tellsthe manager that there are a number of very young people attending. When he looks at the survey results, hecan see that there are indeed children as young as 5 going to the movie.

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8. 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 groupgrowing well with a mean between 21 and 25 feet (upper part of the histogram) and one part of the park showsa group of trees that have a mean height between 6 and 10 feet (lower part of the histogram). In this part ofthe park, the park warden may need to look to see if there is a problem such as a beetle or other pest that isinvading the trees.

9. First you must plot the data either using pencil and paper or using the graphing calculator.

The resulting graph looks like the following.

At a glance it does look like an exponential curve but you really have to take a closer look by doing theexponential regression.

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Next graph this exponential equation onto our stat plot and see how close a match it is.

The equation, therefore, is y = 1.024(5.985x). The question asks “What is the count after 1 day?” In otherwords, “What is y when x = 24?”

y = 1.024(5.985x)

y = 1.024(5.98524)

y = 1.024(4.46×1018)

y = 4.57×1018 bacteria

Therefore there are 4.57× 1018 bacteria on the lunch room tabletop after 1 day. You can check this on ourcalculator as follows.

Your calculation might be a bit off because you rounded the values for a a and b in the equation y = abx, wherethe calculator did not.

10. First plot the data using pencil and paper or with the graphing calculator.

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The resulting graph looks like the following.

At a glance it does look like an exponential curve but you really have to take a closer look by doing theexponential regression.

Next graph this exponential equation onto our stat plot and see how close a match it is.

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The equation, therefore, is y = 8.11(0.5017x). The question asks “Why do you think his distance decreasedwith each jump?” Answers could vary here but a possible answer is that the grasshopper is reaching closer toa roadblock like a fence or a tree.

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www.ck12.org Chapter 5. Measures of Central Tendency, Solution Key

CHAPTER 5 Measures of CentralTendency, Solution Key


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The Mean

Section A –Selected Response

1. C

x =∑x1 + x2 + x3 + . . .+ xn

n

x =10+39+71+42+39+76+38+25

8

x =340

8x = 42.5

2. B An outlier is a value that is very small or very large compared to the majority of values in a data set3. D The population mean is denoted by “mu” and its symbol is µ.4. B The mean is calculated by dividing the sum of all values by the number of values.5. A 71.5×4 = 286 286− (58+76+88) = 64

Section B –Long Answer Questions

1. (a) 34.33

x =∑x1 + x2 + x3 + . . .+ xn

n

x =20+14+54+16+38+64

6

x =2066

x = 34.33

(b) 44.57

x =∑x1 + x2 + x3 + . . .+ xn

n

x =22+51+64+76+29+22+48

7

x =3127

x = 44.57

(c) 62.8

x =∑x1 + x2 + x3 + . . .+ xn

n

x =40+61+95+79+9+50+80+63+109+42

10

x =62810

x = 62.8

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2. 171.6 pounds 167.2×5 = 836 836− (158.4+162.8+165.0+178.2) = 171.63. Answers:

a. 12×5.1 = 61.2b. 8×4.8 = 38.4c. 61.2+38.4 = 99.6 f eet 99.6

20 = 4.98 f eet

4. 31.0 advertisements

x =∑x1 + x2 + x3 + . . .+ xn

n

x =43+37+35+30+41+23+33+31+16+21

10

x =31010

x = 31.0

5. 7.7 pounds

TABLE 5.1:

Weight (Pounds) Number of Babies f Midpoint of Class m Product m f[3 - 5) 8 4 32[5 - 7) 25 6 150[7 - 9) 45 8 360[9 - 11) 18 10 180[11 - 13) 4 12 48

µ =∑m f

Nµ =

770100

µ =32+150+360+180+48

100µ = 7.7 pounds

The mean weight of each baby is 7.7 pounds.

6. (a) The mean mark for the boys is 72.

x =∑x1 + x2 + x3 + . . .+ xn

n

x =90+50+70+80+70

5

x =3605

x = 72

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(b) The mean mark for the girls is 50.

x =∑x1 + x2 + x3 + . . .+ xn

n

x =60+20+30+80+90+20

6

x =3006

x = 50

(c) The mean mark for all the students is 60.

x =∑x1 + x2 + x3 + . . .+ xn

n

x =90+50+70+80+70+60+20+30+80+90+20

11

x =66011

x = 60

7. Answer:

a. The sum of the four numbers is 124. 31×4 = 124b. The sum of the six numbers is 168. 28×6 = 168c. The mean of all the numbers is 29.2

x =∑x1 + x2 + x3 + . . .+ xn

n

x =124+168

10

x =29210

x = 29.2

8. (a) The mean weight of the nine dogs is 23.67 pounds.

x =∑x1 + x2 + x3 + . . .+ xn

n

x =22+19+26+18+29+33+20+16+30

9

x =213

9x = 23.67

(b) If the dog weighing 16 pounds and the one weighing 33 pounds are removed from the group, themean weight of the remaining dogs is 23.43 pounds.

x =∑x1 + x2 + x3 + . . .+ xn

n

x =22+19+26+18+29+20+30

7

x =1647

x = 23.43

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9. (a) No Melanie’s answer is not correct. The temperature of 0◦C was recorded but she did not include itin the total.

x =−7+0+−1+1+−4+−6+3

7

x =−14

7x =−2◦C

(b) She divided the sum of the numbers by 6 nut should have divided by 7. The mean daily temperatureshould be −2◦C.

10. Answer:

x =∑x1 + x2 + x3 + . . .+ xn

n

x =93+78+84+106+116+93+90+75+104+100+123+57

12

x =1119

12x = 93.25 (Honest Hoopers)

x =∑x1 + x2 + x3 + . . .+ xn

n

x =110+89+91+121+84+79+114+66+50+101+106+114

12

x =112512

x = 93.75 (Bouncy Baskets)

The Bouncy Baskets had the higher mean score.

Section C

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The Median


1. B 10, 25, 38, 39, 39, 42, 71, 76 The median is the number in the n+12 position. n+1

2 = 8+12 = 9

2 = 4.5 The numberbelow the 4.5 position is 39 and the number above the 4.5 position is 39. The median is 39+39

2 = 782 = 39

2. D There are 23 seats which is an odd number. The median is the number in the n+12 position. n+1

2 = 23+12 =

242 = 12 The median seat is seat number 12.

3. C 55, 58, 62, 63, 68, 70, 71, 79, 81, 82 The median is the number in the n+12 position. n+1

2 = 10+12 = 11

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

2 =1382 = 69

4. A 32, 34, 36, 38 The median is the number in the n+12 position. n+1

2 = 4+12 = 5

2 = 2.5 The number below the2.5 position is 34 and the number above the 2.5 position is 36. The median is 34+36

2 = 702 = 35

5. B The median is the number of bars sold by the student in the n+12 position.

n+12

=30+1

2=

312

= 15.5

The number below the 15.5 position is 21 and the number above the 15.5 position is 21. The median is21+21

2 = 422 = 21


1. Answer: (a) $25,825

$24,500 $14,280

$26,450 $16,725

$22,660 $22,660

$25,200 $24,500

$16,725 $25,200

$27,600 $26,450

$14,280 $27,600

$28,500 $28,500

$29,700 $29,700

$32,450 $32,450

• The median is the number in the n+12 position. n+1

2 = 10+12 = 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 is 25,200+26,4502 = 51,650

2 = $25,825

• The median retail price for the cars is $25,825.

(b) $21,700

• The median is the number in the n+12 position. n+1

2 = 10+12 = 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 is 21,300+22,1002 = 43,400

2 = $21,700

• The median dealer’s cost for the cars is $21,700.

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2. Three

4 5 3 4 2 1 0 3 2 7 2 3

0 1 2 2 2 3 3 3 4 4 5 7

The median is the number in the n+12 position. n+1

2 = 12+12 = 13

2 = 6.5 The number below the 6.5 position is3 and the number above the 6.5 position is 3. The median is 3+3

2 = 62 = 3 The median number of power

outages is 3.3. Eight

7 14 10 5 11 2 8 6 9 7 13 4 12 8 3

2 3 4 5 6 7 7 8 8 9 10 11 12 13 14


2 = 15+12 = 16

2 = 8 The number in the 8th position is 8. Themedian is 8. The median number of safety devices aboard the boats is 8.

4. $600

$700 $550 $760 $670 $500 $925 $600

$480 $390 $800 $850 $365 $525

$365 $390 $480 $500 $525 $550 $600 $670 $700 $760

$800 $850 $925


2 = 13+12 = 14

2 = 7 The number in the 7th position is $600.The median is $600. The median wage for the teacher’s assistant was $600.

5. Thirteen

6. Three

2,3,1,2,3,5,3,2,1,6,1,4,3,4

1,1,1,2,2,2,3,3,3,3,4,4,5,6


2 = 14+12 = 15

2 = 7.5 The number below the 7.5 position is 3and the number above the 7.5 position is 3. The median is 3+3

2 = 62 = 3 The median score for the roll of the

dice is 3.7. Five

Score 2 3 4 5 6 7 8

Number of Players 3 8 22 29 20 8 10

The median is the score of the player in the n+12 position. n+1

2 = 100+12 = 101


2 = 102 = 5 The median score

is 5.

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8. Three

Mark 0 1 2 3 4 5 6

Number of Students 2 6 7 10 5 3 7

The median is the score of the student in the n+12 position. n+1

2 = 40+12 = 41


2 = 62 = 3 The median score

on Monday’s test is 3.9. Four

Number of Coins 0 1 2 3 4 5 6 7

Number of Students 2 5 8 9 6 4 9 7

The median is the number of coins in the pocket of the student in the n+12 position.

n+12

=50+1

2=

512

= 25.5

The number below the 25.5 position is 4 and the number above the 25.5 position is 4. The median is 4+42 =

82 = 4 The median number of coins is 4.

10. 4.8 minutes

3.5 min 4.2 min 3.1 min 5.3 min 6.2 min 4.6 min





2 = 12+12 = 13

2 = 6.5 The number below the 6.5 position is4.6 and the number above the 6.5 position is 5.0. The median is 4.6+5.0

2 = 9.62 = 4.8 The median time for the

dirt bike race is 4.8 minutes.

The Mode


1. C2. D3. B4. A5. B

Section B –Long answer Questions

1. Answers:

a. The mode for this set of numbers is 6.b. The mode for this set of numbers is 16.

2. There are two modes for the scores on the English quizzes. The modes are 6 and 7.3. The modal number of minutes spent studying for the Math test are 10 - 20 minutes.4. The most common number of questions that students attempted to answer was 42.5. The modal score that was rolled was 3.6. The modal number of games that students attended was 8.

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7. The smallest value for m is 9.8. The data set has two modes 3 and 5. Each value appears 3 times. The distribution can be described as bimodal.9. 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.

10. The modal temperature for the month of June was 74◦F .

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www.ck12.org

CHAPTER 6 The Shape, Center andSpread of a Normal Distribution,

Solution KeyChapter Outline

6.1 SOLUTION KEY FOR REVIEW QUESTIONS

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www.ck12.org Chapter 6. The Shape, Center and Spread of a Normal Distribution, Solution Key


1. d2. a3. b4. d5. c6. b7. c8. d9. c

10. b

11.

(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__.

12.

a. 68% of the volumes can be found between ___51”__ and __61”__.b. 95% of the volumes can be found between ___46”__ and ___66”__.c. 99.7% of the volumes can be found between ___41”__ and ___71”__.

13. Answer:

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TABLE 6.1:

Data (x) Mean (µ) Data − Mean (x−µ)

Square of Data −Mean (x−µ)2

5 6.375 -1.375 1.89068 6.375 1.625 2.64069 6.375 2.625 6.890610 6.375 3.625 13.14064 6.375 -2.375 5.64063 6.375 -3.375 11.39067 6.375 0.625 0.39065 6.375 -1.375 1.8906

∑ 51 43.8748

σ =

√∑(x−µ)2

n

σ =

√43.8748

8σ =√

5.48435

σ = 2.342

Checking with the calculator you the data into the TI-84 calculator under [L1], then do 1-Var Stats using STAT[CALC].

Variance = 2.3422 = 5.485

14. Answer:

TABLE 6.2:



11 14.7 -3.7 13.6915 14.7 0.3 0.0916 14.7 1.3 1.6912 14.7 -2.7 7.2919 14.7 4.3 18.4917 14.7 2.3 5.2914 14.7 -0.7 0.4918 14.7 3.3 10.89

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TABLE 6.2: (continued)



15 14.7 0.3 0.0910 14.7 -4.7 22.09

∑ 147 80.1

σ =

√∑(x−µ)2

n

σ =

√80.110

σ =√

8.01

σ = 2.83

Checking with the calculator you the data into the TI-84 calculator under [L1], then do 1-Var Stats using STAT[CALC].

Variance = 2.832 = 8.01

15. The answers to the question can either be determined by hand (using pencil and paper) or by using the graphingcalculator. The graphing calculator is used to show the answers below. Enter the data into [L1] and do 1-VarStats under STAT [CALC].

a. The mean = 74.86b. The standard deviation is 14.63c. The variance is 14.632 = 214.04d. The normal distribution curve can be found below.

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16. The answers to the question can either be determined by hand (using pencil and paper) or by using the graphingcalculator. The graphing calculator is used to show the answers below. Enter the data into [L1] and do 1-VarStats under STAT [CALC].

a. The mean = 76.46b. The standard deviation is 14.01c. The variance is 14.012 = 196.28d. The normal distribution curve can be found below.

17. To view the keystrokes for drawing the normal distribution curve on your calculator, you could follow these.

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(a) How many students would you expect to wait more than 11.5 minutes?

The number of students waiting more than 11.5 minutes would be 68+ 13.6+ 2.15+ 0.13 = 83.88% of thestudents surveyed.

83.88% of 350 babies = 0.1588×350 = 55 babies

(b) How many students would you expect to wait more than 18.5 minutes?

The number of students waiting more than 18.5 minutes would be 13.6+2.15+0.13= 15.88% of the studentssurveyed.

15.88% of 200 students = 0.1588×200 = 31 students

(c) How many students would you expect to wait between 11.5 and 18.5 minutes?

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The number 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

18. To view the keystrokes for drawing the normal distribution curve on your calculator, you could follow these.

(a) How many babies would you expect to weigh more than 7.3 lbs?

The number of babies weighing more than 7.3 lbs would be 13.6+2.15+0.13 = 15.88% of the babies in thesurvey.

15.88% of 200 babies = 0.8388×200 = 167 students

(b) How many babies would you expect to weigh more than 7.8 lbs?

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The number of babies weighing more than 7.8 lbs would be 2.15+0.13 = 2.28% of the babies in the survey.

2.28% of 350 babies = 0.0228×350 = 8 babies

(c) How many babies would you expect to weigh between 6.3 and 7.8 lbs?

The number of babies weighing between 6.3 and 7.8 lbs minutes would be 68%+ 13.6% = 81.6% of thebabies in the survey.

81.6% of 350 babies = 0.816×350 = 285 babies

19. The answer to this question can either be determined by hand (using pencil and paper) or by using the graphingcalculator. The graphing calculator is used to show the answers below. Enter the data into [L1] and do 1-VarStats under STAT [CALC].

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

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To be in the top 0.13% in terms of time means the best time (or lowest time) therefore you have to look atthe lower end of the normal distribution curve. The time at the lower end of the normal distribution curve (orX−3σ) is 0.1 minutes or 6 seconds!

20. The answer to this question can either be determined by hand (using pencil and paper) or by using the graphingcalculator. The graphing calculator is used to show the answers below. Enter the data into [L1] and do 1-VarStats under STAT [CALC].

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

68% of the heights can be found

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www.ck12.org Chapter 7. Organizing and Displaying Distributions of Data, Solution Key

CHAPTER 7 Organizing and DisplayingDistributions of Data, Solution Key


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Line Graphs and Scatter Plots


1. B –The points associated with continuous data are joined because all the fraction and decimal values betweentwo consecutive points are meaningful.

2. C –The correlation of data on a scatter plot that consists of few points that are not bunched together isconsidered to be weak.

3. D –The line of best fit can be calculated by the TI83 by using linear regression to provide an equation for thestraight line in the form y = ax+b

4. B –The variable is quantitative because it represents a number.5. C –A term used to denote the relation between two data sets is ’correlation.’6. A –The points that represent discrete data are not joined because the values between two consecutive points

are not meaningful7. B –A broken line graph shows change over time with a series of straight lines that have no single defined

slope.


1. Answer:

TABLE 7.1:

Variable Quantitative Qualitative Discrete ContinuousMen’s favorite TVshows

x

Salaries of baseballplayers

x x

Number of childrenin a family

x x

Favorite color ofcars

x

Number of hoursworked weekly

x x

2. Answer:

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a.

b.

c.

3. Answer:

a. The fastest speed of the bus was 16 miles per hour.b. The bus was stopped four times.c. The bus was initially 2 miles from the bus depot.d. The total distance travelled by the bus was 38 miles.

4. Answer:

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a.b. The year 2007 would actually be the number 13 on the x−axis. The number of beetles sold in this year

would be approximately 98 to 100.c. The correlation of this graph is strong and positive.

5. From the graph, you can see that Plan B intersects Plan C at (10, 50). This means that for 10 days of advertisingon Walton’s Web Ads both plans would cost $50.00 To advertise for less than 10 days, Plan C would be thebest plan to use. From the graph, you can see that Plan A intersects Plan B at (20, 70). This means that for 20days of advertising on Walton’s Web Ads both plans would cost $70.00 To advertise for more than 10 daysbut less than 20 days, Plan B would be the best plan to use. To advertise for more than 20 days, Plan Awould be the best plan to use.

6. Answer: (a)

(b) From the graph, and using the TI83 to calculate a value, the fuel efficiency of a car travelling at a speed of47m/h would be approximately 35m.

(c) From the calculator, the line of best fit has a linear equation that is approximately y =−.36x+52.6 wherey represents the fuel efficiency and x represents the speed. Using this equation

y =−.36x+52.6

29 =−.36x+52.6

29−52.6 =−.36x+52.6−52.6−23.6−.36

=−.36x−.36

65.5m/h = x

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The speed of the car would be approximately 65.6 miles per hour.7. Answers will vary. Students should include the beginning and end time of the trip as well as the intervals

when there was a stop. Reference could also be made to the speed of the vehicle, the total distance travelledand the length of time necessary to complete the trip.

8. (a)

m 5 14 2 10 16 4 18 2 8 11

n 6 13 4 10 15 7 16 5 8 12

(b)

m 13 3 18 9 20 15 6 10 21 4

n 7 14 9 16 7 13 10 13 3 19

Circle Graphs, Bar Graphs, Histograms and Stem-and-Leaf Plots


1. D –Count the number of all digits after the number 2 in the row that has 3 as its stem There is a total of 11digits.

2. B –Add the tops of the first three bars 3+2+6 = 113. C –By definition a distribution that has two peaks is bimodal.4. A –Add the values and divide by 2. 14.5+23.5

2 = 382 = 19

5. D –The number of students is 7 out of 25. 725 = 0.28 0.28(100%) = 28%

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

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7. D –By definition a frequency polygon is a graph that used lines to join the midpoints of the classes.8. A –The mode is the value that appears the most often and here it is 32.9. B –There are four bars that are below the line that represents $800.

10. D –The number of students is 9 out of 25. 925 = 0.36 0.36(100%) = 36%


1. 20 12 39 38 18 58 49 59 66 5023 32 43 53 67 35 29 13 42 5537 19 38 22 46 71 9 65 15 38

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

3. Ages of Canadian Prime Ministers when they were sworn into office.

• 52 74 60 39 65 46 55 66 54 51 70 47 69 47 57 46 4866 61 59 46 45

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From the stem and leaf plot some information that I know is:

• The youngest Prime Minister was 39 when he was sworn into office.• The oldest Prime Minister to be sworn into office was 74.• The modal age for the Prime Ministers was 46.• The median age for a Prime Minister was 54.5 years of age.

4. Answer:

TABLE 7.2:

Make of Car Number of CarsFord 8Honda 6Volkswagen 7Mazda 9

5. Answer:

a. There were 3+4+6+6+9+8+5+1 = 42 students in the class.b. There are 5+1 = 6 students who have a height over 60 inches.c. There are 6+9+8+5 = 28 students who have a height between 54 and 62 inches.d. The distribution has only one peak so the data is unimodal.

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6. Answer:

a. There are 5+9+14+18+12+8+4 = 70 players who played this sport.b. The most common weight was 85 kg.c. 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 above the beginning and

end points of the data set.e. There are no weights recorded for 75 kg and 80 kg.

7. Answers:

a. The parts of the circle graph are displayed with degrees and the parts should be shown with percentages.b. The number of students that participated in sports is 135

360 = 0.375 0.375(200) = 75c. The number of students who participated in CMT is 36

360 = 0.1 0.1(200) = 20d. The number of students who participated in volunteer work is 90

360 = 0.25 0.25(200) = 50

e.

8. 95 56 70 83 59 66 88 52 50 77 69 8054 75 68 78 51 64 55 67 74 57 73 53

TABLE 7.3:

Bin Frequency[50-60) 9[60-70) 5[70-80) 6[80-90) 3[90-100) 1

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9.10. The type of graph that tells me the most information is the histogram. I can quickly see the classes and how

many are in each one. As well, the shape of the histogram tells me about the distribution of the data set.(Answers will vary)

Box-and-Whisker Plots


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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 are called whiskers4. 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 join Q1 and Q3.


1. (a) Min. Value→ 60Q1→ 68 Med→ 72.5Q3→ 77 Max. Value→ 83

Q1 =67+69

2 = 1362 = 68 Median = 71+74

2 = 1452 = 72.5 Q3 =

75+792 = 154

2 = 77 (b) Min. Value→ 3Q1→ 5.5 Med→ 9Q3→ 11.5 Max. Value→ 15

Q1 =5+6

2 = 112 = 5.5 Median = 9 Q3 =

11+122 = 23

2 = 11.52. (a) Min. Value→ 4Q1→ 8 Med→ 13Q3→ 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→ 175Q1→275 Med→ 450Q3→ 525 Max. Value→ 625 The median is located to the right of the center of the boxwhich tells you that the distribution is negatively skewed.

3.Q1→ 35 Median = 44+50

2 = 942 = 47 Q3→ 64

4. 15 8 5 10 14 17 21 23 6 19 31 34 30 31 3 22 17 255 16

Q1→ 8 Median = 17+172 = 34

2 = 17 Q3→ 25

The right whisker is longer than the left whisker which indicates that the distribution is positivelyskewed.

5. 24 21 31 28 2927 22 27 30 3226 35 24 22 3430 28 24 32 2732 28 27 32 2320 32 28 32 34 The first step is to organize the data:

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Q1→ 24 Median = 28+282 = 56

2 = 28 Q3→ 32

6. The following are the screenshots show the steps necessary to produce the box-and- whisker plot usingthe TI83

Min. Value→ 44Q1→ 48.5 Med→ 51Q3→ 55 Max. Value→ 59

7.Q1→ 22 Median = 25 Q3→ 33

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The median is located to the left of the center of the box which indicates that the distribution is positivelyskewed.

8.Q1→ 31 Median = 35 Q3→ 38

The median is located to the right of the center of the box which indicates that the distribution isnegatively skewed.

9. The first step is to organize the data:

Q1→ 35.9 Median = 63.6+64.02 = 127.6

2 = 63.8 Q3→ 80.6

The median is located to the right of the center of the box which tells you that the distribution isnegatively skewed.

10. (a) Organize the given data set :

25, 33, 55, 32, 17, 19, 15, 18, 21

15, 17, 18, 19, 21, 25, 32, 33, 55

Determine the values for Q1 and Q3.

15, 17, 18 , 19, 21 , 25, 32, 33 , 55

Q1 =17+18

2=

352

= 17.5 Q3 =32+33

2=

552

= 27.5

Calculate the difference between Q1 and Q3. Q3−Q1 = 27.5−17.5= 10.0 Multiply this difference by 1.5.(10.0)(1.5) = 15.0 Compute the range: Q1− 15.0 = 17.5− 15.0 = 2.5 Q3 + 15.0 = 27.5+ 15.0 = 42.5Are there any data values below 2.5? No, there are no values below 2.5. Are there any values above 42.5?Yes, the value of 55 is above 42.5 and is therefore an outlier. (b) Organize the given data set:

149, 123, 126, 122, 129, 120

120, 122, 123, 126, 129, 149

Determine the values for Q1 and Q3. 120, 122 , 123, 126, 129 , 149

Q1 = 122 Q3 = 129

Calculate the difference between Q1 and Q3. Q3−Q1 = 129−122 = 7.0 Multiply this difference by 1.5.(7.0)(1.5) = 10.5 Compute the range: Q1− 10.5 = 122− 10.5 = 111.5 Q3 + 10.5 = 129+ 10.5 = 139.5Are there any data values below 111.5? No, there are no values below 111.5. Are there any values above139.5? Yes, the value of 149 is above 139.5 and is therefore an outlier.

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www.ck12.org Chapter 8. Organizing and Displaying Data for Comparison, Solution Key

CHAPTER 8 Organizing and DisplayingData for Comparison, Solution Key


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1. Answer:

TABLE 8.1:

Type of graph Type of Variablea. Histogram b discreteb. Stem-and-leaf plot d discretec. Broken line graph e discreted. Bar graph a continuouse. Pie chart c continuous

2. Answer:

TABLE 8.2:

Type of graph Type of Variablea. Broken line graph b qualitativeb. Bar graph a numericalc. Pie chart e categoricald. Stem-and-leaf plot c quantitativee. Histogram e numerical

3. Descriptions will vary but suggested answer is below.

4. Descriptions will vary but suggested answer is below.

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5. Answers will vary but could include some of the following statements An 18-wheeler typically holds 100gallons of fuel and can travel about 5 –6 miles per gallon. The data in the table shows that the tank needed tobe refilled at 500 miles (approximately) therefore the truck can go about 5 miles to the gallon. A car can holda variety of gallons of fuel. This car holds 20 gallons which is usually a larger car (or an SUV). From 100 to200 miles, the car began with a full tank and then ran out of gas so the car can go about 5 miles per gallon aswell.

6. Answers will vary but the data suggests that there is a wider variation in the pulse rates for the group of girlsthan for the group of boys. For the girls, the pulse rates ranged from 60 to 89 whereas the boys ranged from

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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 a higher pulse rate.

7. Answers will vary but the data suggests that there is a wider variation in the line-up time for the customersof Starbucks than for the customers at Just Us Coffee. For the Starbucks customers, the line-up time rangedfrom 8 to 27 minutes whereas the Just Us Coffee customers ranged from 10 to 29 minutes. The median forthe Starbucks customers’ line-up time is at 17 minutes and the mode is at 12 minutes. For the Just Us Coffeecustomers, however, the median is at 16 minutes and the mode is at 10 minutes. It would seem that the JustUs Coffee customers have a slightly better 9shorter) time in the line-up for their orders.

8. Answers will vary but the data suggests that there is a slightly wider variation in the heights for the group ofgirls than for the group of boys. For the girls, the heights ranged from 145 to 179 centimeters, whereas for theboys, the heights ranged from 153 to 185 centimeters. The median for the girls group is at 168.5 centimeters,and the mode is at 155 centimeters. For the group of boys, however, the median is at 169.5 centimeters, andthe mode is at 167 centimeters. The boys seem to be taller than the girls.

9. Answers will vary but the resulting chart should look similar to the following double bar graph.

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Eye color is an inherited trait. This means that the color of your eyes depends on both the mother and thefather of the child. Both the girls and boys groups have similar numbers for each of the four groups of eyecolors. If you look at the difference from boys and girls, you can see that there are more girls with green andbrown eye colors and less girls with hazel eyes. The number of girls with blue eyes equals the number of boyswith blue eyes (therefore no difference with regard to sex). If you were to look at sex difference for eye color,the biggest (most observable) difference can be found with green eyes. Green eyes are the least common ofall eye colors so this factor may be a significant difference but it is hard to say from this chart.

10. Answers will vary but the resulting chart should look similar to the following double bar graph. Robbie istrying to find out if any of the four food options would be desirable for the cafeteria menu. Using his surveyresults and plotting the yes versus no votes on a double bar graph, he can quickly see that fish burgers andbrown rice would be a good choice for new items on the menu. Although many students did vote for carrotsoup, the same number did not. Therefore it would be a gamble to place this item on the new menu. VegetarianPizza was voted more by the no votes than by the yes votes so Robbie has decided not to place this on themenu.

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11. Answers will vary but the resulting chart should look similar to the following double bar graph.

With the survey done by the guidance counselor, it was found that more girls were planning on going touniversity following high school and more boys were going to college, the military, or directly into some formof employment. It was also found that more girls were unsure than boys of what their plans were from thegroup who were unsure. Of the 186 students surveyed, 75 were attending university (or 50%) and 124 wereattending either university or college (which is 66.7%).

12. Answers will vary but the resulting chart should look similar to the following double box-and-whisker plot.

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Higher Level Study Times 5-Number Summary

Standard Level Study Times 5-Number Summary

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If you put all of this information into a table you would get the following result.

TABLE 8.3:

5-number summary Higher Level Study Times Standard Level Study TimesMinimum 5 2Lower Quartile (Q1) 6 6Median 10 7Upper Quartile (Q3) 16 12Maximum 19 20

Conclusions drawn may vary but some points made by students could include the following. Using the medians, 50%of the students spent 10 hours studying for the higher level exams and 50% of the students spent 7 hours studyingfor the standard level exams. This would seem reasonable since the higher level exams are on more material thanthe standard level exams. The range for the higher level exam study times was 19− 5 = 14 hours. The range forthe standard level exam study times was 20− 2 = 18 hours. Since the range for the higher level exam study timesis smaller, it means the study times are less spread out. This would then mean that the study times were morepredictable and reliable.

Also, the data in the standard level exam study times is not very even meaning that the box is not split into twoequal parts. This means that there are more scores in the lower part (between Q1 and Q3) than in the upper part.The same can be found for the higher level exam study times but not to the extreme as with the standard level examstudy times. Also the tails for the higher level exam study times and the standard level exam study times show thelower half is shorter than the upper half. This means that the lower half has less dispersed data than the upper half.For the higher level exam study times, the tails seem to be of more equal length meaning the data has a more equaldistribution, more so than the standard level exams.

13. Answers will vary but the resulting chart should look similar to the following double box-and-whisker plot.

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Math Scores 5-Number Summary

Verbal Scores 5-Number Summary

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TABLE 8.4:

5-number summary Math Scores Verbal ScoresMinimum 513 478Lower Quartile (Q1) 529 499Median 552 524Upper Quartile (Q3) 563 550Maximum 575 579

Conclusions drawn may vary but some points made by students could include the following. Using the medians,50% of the students scored on average 552 on the Math SAT portion of the exam and 50% of the students scored 524on the Verbal portion of the SATs. 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. The range for the Verbal scores was 579− 478 = 101points. Since the range for the Math scores is smaller, it means the math scores obtained by the students are lessspread out. This would then mean that the Math scores were more predictable and reliable.

Also, the data in the Verbal scores is very even meaning that the box is split into two equal parts. This means thatthere are equal scores in the lower part (between Q1 and Q3) as in the upper part. This is not the same as in theBox-and-whisker plot for the Math scores on the top where there are more scores in the lower half that in the upperhalf. Also the tails for the Math scores show the upper half is slightly shorter than the lower half. This means thatthe upper half has less dispersed data than the lower half. For the Verbal scores, the reverse is true. The upper tailseems to be slightly longer than the lower tail so the data in the lower half is less dispersed than in the lower half.

14. 2010 games: 5-number summary

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2006 games: 5-number summary


TABLE 8.5:

5-number summary 2010 Games 2006 GamesMinimum 6.4 7Lower Quartile (Q1) 7.7 7.7

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TABLE 8.5: (continued)

5-number summary 2010 Games 2006 GamesMedian 8.5 8Upper Quartile (Q3) 9.1 8.6Maximum 9.9 9.2

Conclusions drawn may vary but some points made by students could include the following. Using the medians,50% of the judge’s scores are at 8.5 for the 2010 Winter Games but in 2006, 50% of the scores were at 8.0. Therange for the scores in the 2010 Winter Games was 9.9− 6.4 = 3.5 points. The range for the scores in the 2006Winter Games was 9.2−7 = 2.2 points. Since the range for the 2006 Winter Games is smaller, it means the judgesscores are less spread out. This would then mean that the scoring was more predictable and reliable.

Also, the data in the 2006 Winter games is not very even meaning that the box is not split into two equal parts. Thismeans that there are more scores in the lower part (between Q1 and Q3) than in the upper part. The reverse can befound for the 2010 Winter games. The tails for the 2010 winter games show the lower half is longer than the upperhalf. This means that the upper half has less dispersed data than the lower half. For the 2006 Winter Games, the tailsseem to be of equal length meaning equal distribution of data.

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ck-12 basic probability and

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