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Chapter 4 – Discrete Probability Distribution Answer Key CK-12 Advanced Probability and Statistics Concepts 1 4.1 Discrete and Continuous Random Variables Answers 1. Discrete 2. Continuous 3. Continuous 4. Continuous 5. Discrete 6. Discrete 7. Continuous 8. Discrete 9. Discrete 10. Neither

4.1 Discrete and Continuous Random Variables

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Page 1: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 1

4.1 Discrete and Continuous Random Variables

Answers

1. Discrete

2. Continuous

3. Continuous

4. Continuous

5. Discrete

6. Discrete

7. Continuous

8. Discrete

9. Discrete

10. Neither

Page 2: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 2

4.2 Probability Distribution

Answers

1. a. x = -4, 0, 1, 3

b. X = 1

c. P(x > 0) = 0.9

d. P(x = -2) = 0

2. a.

X P(X)

2 1

3 2

4 3

5 4

6 5

7 6

8 5

9 4

10 3

11 2

12 1

b. 𝑃(𝑥 ≥ 8) = 15

36 𝑜𝑟

5

12

c. 𝑃(𝑥 < 8) = 21

36 𝑜𝑟

7

12

d. 𝑃(𝑥 𝑖𝑠 𝑜𝑑𝑑) = 18

36 𝑜𝑟

1

2 𝑃(𝑥 𝑖𝑠 𝑒𝑣𝑒𝑛) =

18

36 𝑜𝑟

1

2

e. 𝑃(𝑥 = 7) = 6

36 𝑜𝑟

1

6

Page 3: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 3

3. 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 1 𝑏𝑜𝑦) = 7

8

4. Suppose there are six numbers in a box: 1, 2, 3, 4, 5, 6.

a. Yes because what you pick on the second draw is 1, 2, 3, 4, 5, 6 as it is with the first

draw. The outcome of the second draw does not depend on the first draw.

b. No because what number you pick on the second draw depends on what you picked on

the first draw.

5. 𝑃(𝑑𝑟𝑎𝑤𝑖𝑛𝑔 𝑎 3 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑟𝑎𝑤) =3

12=

1

4 𝑜𝑟 0.25

6. 𝑃(𝑥 = 1) = 1

4 𝑜𝑟 0.25

7. Suppose a box has four slips of paper and on each slip are two numbers. The slips of paper

look like the following:

a. X would be the first number and Y would be the second number. So if the sequence you

needed to draw was 1, 3 (1st piece) this is not the same as 3,1 (the third piece of paper).

b. 𝑃(𝑥 ∙ 𝑦 = 3) = 2

4 𝑜𝑟

1

2

c. 𝑃(2𝑥 − 3𝑦 = 7) = 1

4 𝑜𝑟 0.25

8. True

9. Suppose two draws will be made at random with replacement from a box that has three slips

of paper, each with a number on it: 1, 2, and 3. Let represent the first draw and

represent the second draw.

a. 𝑃(𝑥1 = 1) = 1

3

b. 𝑃(𝑥1 = 1, 𝑥2 = 2) = 1

9

c. 𝑃(𝑥1 = 1) ∙ 𝑃(𝑥2 = 2) =3

9 𝑜𝑟

1

3

d. The events, choosing a number from 1, 2, and 3 then choosing a number again from 1, 2,

and 3 are independent here because the first draw was replaced.

e. No. Since removing slips with 1, 2, or 3 after the first draw, the second draw would be

dependent on what is remaining.

10. Random variable Y because P(3) = 0.5 and this is not possible with random variable X.

11. Suppose 2 1

af x

x

for 0,1,2,3x is a discrete probability distribution.

a. 5/9

b. 0.167

Page 4: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 4

4.3 Mean and Standard Deviation of Discrete Random

Variables

Answers

1. Consider the following probability distribution:

a. 𝜇 = 1.7

b. 𝜎2= 1.21

c. 𝜎 = 1.1

2. Expected Value = 1.29 or 1

3. Find the expected value. Expected Value = 3.6

4. Blank = .1

Expected Value = 4.2

5. Expected Value of S = 1

( ) ( )n

i

E x xp x

Suppose you have a discrete random variable X that has values and probabilities as shown in

the table below.

X P(X)

1 1/4

2 1/4

3 1/2

Then: 1 1 1

( ) 1 2 3 2.254 4 2

E x

Assume that the numbers drawn were 1, 2, 3, 3. Then the mean is calculated to be:

1 2 3 32.25

4

Page 5: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 5

6.

1 2 1 1 2 2

1 2

1 2

( ) ( )

1 1

6 6

36

E X X x p x x p x

x x

x x

7. Expected Value = 0.8

8. Blank = 3/11

Expected Value = 45/11

9.

a. Yes because the sum of the probabilities is 1.

b. Expected Value = 1.35. It means the average number of children would be 1.35. This is not likely since you cannot have 1.35 children.

10. a. No since the probability of getting a c is 0.

b. Expected Value = 3.4

11. Expected Value = 22.5 minutes

Page 6: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 6

4.4 Sums and Differences of Independent Random

Variables

Answers

1. 7

2. 2.42

3.

x P(x)

0 1/8

1 3/8

2 3/8

3 1/8

4. $2.40; $5.00

5. Expected Value = 2.28

Variance = 0.7397

6. True

7. Yes

8. Yes

9. a)

x P(x)

1 1/6

2 1/6

3 1/6

4 1/6

5 1/6

6 1/6

b)

y P(y)

1 ¼

2 ¼

3 1/2

Page 7: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 7

c)

Z P(z)

2 0.0417

3 0.0833

4 0.1667

5 0.1667

6 0.1667

7 0.1667

8 0.1250

9 0.0833

10. a) ¼

b)

Z P(z)

3 0.1667

6 0.3333

7 0. 1650

10 0.3333

11.

S P(S)

3 0.03

4 0.07

5 0. 10

6 0.13

7 0.17

8 0.17

9 0. 13

10 0.07

11 0.03

Page 8: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 8

4.5 Binomial Distribution and Probability

Answers

1.

x P(x)

0 0.4096

1 0.4096

2 0.1536

3 0.0256

4 0.0016

2. a)

x P(x)

0 0.328

1 0.4096

2 0.2048

3 0.0512

4 0.0064

5 0.0003

b)2

1

0.894

3. a) 0.591

b) 0.409

4. a) E(x) < 12

b) 2.45

c) 0.008

5. a) yes; n=250, p = 0.5

b) E(x) = 125

Page 9: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 9

6. a) Binomial distribution applies as there are two possible outcomes and each outcome is

independent of the other

b) Binomial distribution applies as this is the same as tossing a coin fifty times

c) Binomial distribution applies as you can draw out a yellow or a blue with the same

chances each time.

d) Binomial distribution does not apply as the result of each draw depends on the

previous draw(s).

e) Binomial distribution does not apply if you assume that the 20bolts were taken out at

once without replacement. Therefore there are no independent trials.

7. a) 0.117

b) 0.601

c) 0.399

8. a) 0.99996

b) 0.9997

9. a) i) 3; 1.22

ii) 1.8; 1.122

iii) 4.2; 1.122

Page 10: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 10

b) i)

ii)

iii)

c) As p increases to 0.5, the skewness to the higher values of x decreases and disappears at p

= 0.5. As p increases beyond 0.5, the histograms are skewed toward the lower values of x.

Page 11: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 11

10. 25; 3.54

11. 1.05; 0.945

12. a) Not a binomial experiment because there are more than two possible outcomes.

b) Not a binomial experiment because there is no mention of the number of trials.

13. a) ( 10) 0.176P x

b) ( 13) 0.0739P x

c) ( 8) 0.132P x

14. a) ( 30) 0.0820P x

b) ( 29) 0.214P x

c) ( 27) 0.216P x

d) ( 3) 0.435P x

15. a) ( 0) 0.573P x

b) ( 4) 0.0012P x

c) ( 5) 0.00008P x

d) QUESTION IS VAGUE

16. a) ( 7) 0.3823P x

b) ( 8) 0.9536P x

17. 0.008

18 a) ( ) 3.33; 1.05E x

b) ( ) 50; 5E x

c) ( ) 500; 19.36E x

d) ( ) 0.111; 0.314E x

e) ( ) 18; 2.68E x

19. 300

20. 0.624; 0.557

Page 12: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 12

4.6 Poisson Probability Distributions

Answers

1. m = 14.14

2. a) 0.001204

b) 0.01627

c) 0.9561

d) 0.9999

3. a) 0.0183

b) 0.5666

c) 0.2149

4. 4.372

5. a) 0.751

b) 0.818

6. 0.156

7. 0.224

8. 0.0842

9. 0.0050

10. 0

11. 0.224

12. 0.313

13. 0.224

Page 13: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 13

4.7 Geometric Probability Distributions

Answers

1. a) 0.0741

b) 0.6667

2. 0.216

3. 0.0812

4. a) the number of seniors who suspects to work full time in college

b) Geometric distribution

c)

x 1 2 3 4 5 6

P(x) 0.234 0.179 0.137 0.105 0.0806 0.0617

d) 4.27 ; therefore 4 or 5 people

e) 0.550

f)

Page 14: 4.1 Discrete and Continuous Random Variables

Chapter 4 – Discrete Probability Distribution Answer Key

CK-12 Advanced Probability and Statistics

Concepts 14

5. (i) The experiment consists of a sequence of independent trials.

(ii) Each trial results in one or two outcomes; successes (S) or failures (F)

(iii) The geometric random variable (X) is defined as the number of trials until the first S

is observed

(iv) The probability p(x) is the same for each trial.

6. Geometric probability distribution consists of a sequence of independent events where

the random variable is defined as the number of trials until a success is observed. A

Binomial probability distribution consists of “n” independent trials where the random

variable is the number of successes in “n” trials.

7. (1 – p)(n – 1) represents the probability of failure for the number of trials up to the first

success. P = the probability of success and therefore 1 – p = the probability of failures.

“n” represents the discrete random variable.

8. The expected value of a geometric random variable (x) is the mean which is the inverse

of the probability of successes for each trial (x).

9. 0.0348

10. 0.0767

11. a) 0.0504

b) 0.240

c) 0.117

d) 0.0579

e) 0.0824

f) 0.462

g) 0.303