Page 1
Chapter 4 β 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
Chapter 4 β 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. π(π₯ β₯ 8) = 15
36 ππ
5
12
c. π(π₯ < 8) = 21
36 ππ
7
12
d. π(π₯ ππ πππ) = 18
36 ππ
1
2 π(π₯ ππ ππ£ππ) =
18
36 ππ
1
2
e. π(π₯ = 7) = 6
36 ππ
1
6
Page 3
Chapter 4 β Discrete Probability Distribution Answer Key
CK-12 Advanced Probability and Statistics
Concepts 3
3. π(ππ‘ ππππ π‘ 1 πππ¦) = 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. π(ππππ€πππ π 3 ππ π‘βπ π πππππ ππππ€) =3
12=
1
4 ππ 0.25
6. π(π₯ = 1) = 1
4 ππ 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. π(π₯ β π¦ = 3) = 2
4 ππ
1
2
c. π(2π₯ β 3π¦ = 7) = 1
4 ππ 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. π(π₯1 = 1) = 1
3
b. π(π₯1 = 1, π₯2 = 2) = 1
9
c. π(π₯1 = 1) β π(π₯2 = 2) =3
9 ππ
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
Chapter 4 β 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. π = 1.7
b. π2= 1.21
c. π = 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
Chapter 4 β 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
Chapter 4 β 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 ΒΌ
2 ΒΌ
3 1/2
Page 7
Chapter 4 β 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) ΒΌ
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
Chapter 4 β 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
Chapter 4 β 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
Chapter 4 β 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
Chapter 4 β 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
Chapter 4 β 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
Chapter 4 β 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
Chapter 4 β 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 βnβ independent trials where the random
variable is the number of successes in βnβ trials.
7. (1 β p)(n β 1) represents the probability of failure for the number of trials up to the first
success. P = the probability of success and therefore 1 β p = the probability of failures.
βnβ 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