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Chapter 5
Discrete Probability Distributions

Learning Objectives

1. Understand the concepts of a random variable and a probability distribution.

2. Be able to distinguish between discrete and continuous random variables.

3. Be able to compute and interpret the expected value, variance, and standard deviation for a discrete random variable.

4. Be able tocompute and work with probabilities involving a binomial probability distribution.

5. Be able to compute and work with probabilities involving a Poisson probability distribution.

6. Know when and how to use the hypergeometric probability distribution.

Solutions:

1. a. Head, Head (H,H)
Head, Tail (H,T)
Tail, Head (T,H)
Tail, Tail (T,T)

b. x =number of heads on two coin tosses

c.
|Outcome |Values of x |
|(H,H) |2 |
|(H,T) |1 |
|(T,H) |1 |
|(T,T) |0 |

d. Discrete. It may assume 3 values: 0, 1, and 2.

2. a. Let x = time (in minutes) to assemble the product.b. It may assume any positive value: x > 0.

c. Continuous

3. Let Y = position is offered
N = position is not offered

a. S = {(Y,Y,Y), (Y,Y,N), (Y,N,Y), (Y,N,N), (N,Y,Y), (N,Y,N), (N,N,Y), (N,N,N)}

b. Let N = number of offers made; N is a discrete random variable.

c.

|Experimental Outcome | (Y,Y,Y) | (Y,Y,N) |(Y,N,Y) | (Y,N,N) | (N,Y,Y) | (N,Y,N) | (N,N,Y) | (N,N,N) |
|Value of N |3 |2 |2 |1 |2 |1 |1 |0 |

4. x = 0, 1, 2, . . ., 12.

5. a. S = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}

b.

|Experimental Outcome |(1,1) |(1,2) |(1,3) |(2,1)|(2,2) |(2,3) |
|Number of Steps Required |2 |3 |4 |3 |4 |5 |

6. a. values: 0,1,2,...,20
discrete
b. values: 0,1,2,...
discrete
c. values: 0,1,2,...,50
discrete

d. values: 0 ≤ x ≤ 8
continuous

e. values: x > 0
continuous

7. a.f (x) ≥ 0 for all values of x.

Σ f (x) = 1 Therefore, it is a proper probability distribution.

b. Probability x = 30 is f (30) = .25

c. Probability x ≤ 25 is f (20) + f (25) = .20 + .15 = .35

d. Probability x > 30 is f (35) = .40

8. a.
|x |f (x) |
|1 |3/20 = .15 |
|2|5/20 = .25 |
|3 |8/20 = .40 |
|4 |4/20 = .20 |
| |Total 1.00 |

b.
[pic]
c. f (x) ≥ 0 for x = 1,2,3,4.

Σ f (x) = 1

9. a.
|x |f (x) |
|1 |15/462 |= 0.032 |
|2 |32/462 |= 0.069 |
|3 |84/462 |=0.182 |
|4 |300/462 |= 0.650 |
|5 |31/462 |= 0.067 |

b.
[pic]

c. All f (x) ≥ 0

Σ f (x) = 0.032 + 0.069 + 0.182 + 0.650 + 0.067 = 1.000

10. a.
|x |f(x) |
|1 |0.05 |
|2 |0.09 |
|3 |0.03 |
|4 |0.42 |
|5 |0.41|
| |1.00 |

b.
|x |f(x) |
|1 |0.04 |
|2 |0.10 |
|3 |0.12 |
|4 |0.46 |
|5 |0.28 |
| |1.00 |

c. P(4 or 5) = f (4) + f (5) = 0.42 + 0.41 = 0.83

d. Probability of very satisfied: 0.28...