Discrete Probability Distributions.
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This document discusses probability distributions and related concepts. It begins by defining key terms like probability distribution, random variable, discrete and continuous distributions. It then focuses on several specific discrete probability distributions - binomial, hypergeometric, and Poisson. For each, it provides the characteristics and formulas for calculating probabilities. Several examples are worked through to demonstrate calculating probabilities, means, variances and more for problems that fit each distribution.
![4
6-1919
Binomial Probability– Example:1
There are five flights daily from Pittsburgh via US
Airways into the Bradford, Pennsylvania, Regional
Airport. Suppose the probability that any flight
arrives late is 0.20.
(a) What is the probability that NONE of the flights
are late today?
(b) What is the probability that ONE of the flights are
late today?
3277.0)3277)(.1)(1(
)20.0-1()20.0(
)-1()()0(
0-50
05
-
C
CP xnx
xn
4096.0=)4096)(.20.0)(5(=
)20.0–1()20.0(C=
)π–1()π(C=)1(P
1–51
15
x–nx
xn
6-2020
For the example regarding the
number of late flights, recall
that =.20 and n = 5.
What is the average number
of late flights?
What is the variance of the
number of late flights?
Binomial Distribution – Mean and
Variance: Example
6-2121
Binomial Distribution – Mean and
Variance: Another Solution
6-22
Binomial Distribution - Example
EXAMPLE
Five percent of the worm gears
produced by an automatic, high-
speed Carter-Bell milling machine
are defective.
What is the probability that out of six
gears selected at random none will
be defective? Exactly one? Exactly
two? Exactly three? Exactly four?
Exactly five? Exactly six out of six?
Binomial – Shapes for Varying (n constant)
Binomial – Shapes for Varying n ( constant)
6-23
Harry Ohme is in charge of the electronics section of a
large department store. He has noticed that the probability
that a customer who is just browsing will buy something is
0.3. Suppose that 15 customers browse in the electronics
section each hour.
(a) What is the probability that at least one browsing
customer will buy something during a specified hour?
(b) What is the probability that at least four browsing
customers will buy something during a specified hour?
(c) What is the probability that no browsing customers will
buy anything during a specified hour?
(d) What is the probability that no more than four browsing
customers will buy something during a specified hour?
Binomial Probability (Levin and Rubin)
6-24
9953.0=)30.0–1()30.0(C-1=
)π–1()π(C=)1x(P)a(
0–510
015
x–nx
xn
≥
7032.0=
]1700.0+0916.0+0.0305+[0.0047–1
])30.–1()30(.C+)30.–1()30(.C+
)30.–1()30(.C+)30.–1()30(.C-[1=
P(3)]+P(2)+P(1)+[P(0)-1=
)π–1()π(C=)4x(P)b(
3–513
315
2–512
215
1–511
115
0–510
015
x–nx
xn≥
Binomial Probability (Levin and Rubin)](https://image.slidesharecdn.com/chap00614e-140409122840-phpapp02/85/Discrete-Probability-Distributions-4-320.jpg)
![5
6-25
0047.0=)30.0–1()30.0(C=
)π–1()π(C=)0=x(P)c(
0–510
015
x–nx
xn
5154.0=
]2186.0+1700.0+0916.0+0.0305+[0.0047=
])30.–1()30(.C+
)30.–1()30(.C+)30.–1()30(.C+
)30.–1()30(.C+)30.–1()30(.C[=
P(4)]+P(3)+P(2)+P(1)+[P(0)=)4x(P)d(
4–514
415
3–513
315
2–512
215
1–511
115
0–510
015
≤
Binomial Probability (Levin and Rubin)
6-26
The latest nationwide political poll indicates that for
Americans who are randomly selected, the probability
that they are conservative is 0.55, the probability that
they are liberal is 0.30, and the probability that they are
middle-of-the-road is 0.15. Assuming that these
probabilities are accurate, answer the following the
questions pertaining to a randomly chosen group of 10
Americans.
(a) What is the probability that 4 are liberal ?
(b) What is the probability that none are conservative ?
(c) What is the probability that 2 are middle-of-the-road ?
(d) What is the probability that 8 are liberal ?
Binomial Probability (Levin and Rubin)
6-27
0003.0=)3.–1()3(.C=)π–1()π(C=)4(P)a( 4–014
410
x–nx
xn
2001.0=)55.–1()55(.C=)π–1()π(C=)10(P)b( 0–014
010
x–nx
xn
2759.0=)51.–1()15(.C=)π–1()π(C=)2(P)c( 2–014
210
x–nx
xn
00160.0=
00001.0+00014.0+00145.0=
)3.–1()3(.C
+)3.–1()3(.C+)3.–1()3(.C=
)10(P+)9(P+)8(P=)π–1()π(C=)8≥x(P)d(
01–0110
10
9–019
9
8–018
810
x–nx
xn
Binomial Probability (Levin and Rubin)
6-28
Problems Related to Meeting Conditions
of Binomial Distribution
In real life problems, the probability of
success or failure does not remain fixed
all the time
The trials may not be statistically
independent of each other
6-2929
Binomial – Shapes for Varying
(n constant)
6-3030
Binomial – Shapes for Varying n
( constant)](https://image.slidesharecdn.com/chap00614e-140409122840-phpapp02/85/Discrete-Probability-Distributions-5-320.jpg)
![7
6-37
More About the Poisson Probability
Distribution
•The Poisson probability distribution is always positively skewed and the
random variable has no specific upper limit.
•The Poisson distribution for the lost bags illustration, where µ=0.3, is highly
skewed.
•As µ becomes larger, the Poisson distribution becomes more symmetrical.
6-38
1465.=
!2
e4
=
!x
eµ
=)x(P
4–2u–x
The Lab Aid specializes in
caring for minor injuries,
colds, and flu. For the
evening hours of 6-10 PM
the mean number of arrivals
is 4.0 per hour. What is the
probability of 2 arrivals in an
hour?
Poisson Probability Distribution–
Example:2
6-39
The US Bureau of Printing and Engraving (BPE)
is responsible for printing this country’s paper
money. The BPE has an impressively small
frequency of printing errors; only 0.5 percent of
all bills are too flawed for circulation. What is the
probability that out of a batch of 1,000 bills
(a) None are too flawed for circulation?
(b) Ten are flawed for circulation?
(c) Fifteen are flawed for circulation?
Poisson Probability Distribution–
Example:3
6-40
Poisson Probability Distribution–
Example:3
01813.0=
!10
e×)5(
=)10(P
5–10
00016.0=
!15
e×)5(
=)15(P
5–15
Given us: n = 1000, π = 0.005
Thus: µ =nπ = (1000)(0.005) = 5
00674.0=
!0
e5
=
!x
eµ
=)0(P
5–0u–x
6-41
Concert Pianist Donna Prima has become quite
upset at the number of coughs occurring in the
audience just before she begins to play. On her
latest tour, Donna estimates that on average eight
coughs occur just before the start of her
performance. Ms. Prima has sworn to her
conductor that if she hears more than five coughs
at tonight’s performance, she will refuse to play.
What is the probability that she will not play
tonight?
Poisson Probability Distribution–
Example:4
6-42
Poisson Probability Distribution–
Example:4
8088.0=0.1912–1=
0.0916]+0.0573+0.0286+
0.0107+0.0027+[0.0003–1=
]
!5
e×)8(
+
!4
e×)8(
+
!3
e×)8(
+
!2
e×)8(
+
!1
e×)8(
+
!0
e×)8(
[–1=
8–58–48–3
8–28–18–0
Given us: µ = 8
)]5(P+)4(P+)3(P+)2(P+)1(P+)0(P[–1](https://image.slidesharecdn.com/chap00614e-140409122840-phpapp02/85/Discrete-Probability-Distributions-7-320.jpg)
Discrete Probability Distributions.
- 1. 1 McGraw-Hill/Irwin Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved. Discrete Probability Distributions Chapter 6 6-2 1. Define the terms probability distribution and random variable. 2. Distinguish between discrete and continuous probability distributions. 3. Calculate the mean, variance, and standard deviation of a discrete probability distribution. 4. Describe the characteristics of and compute probabilities using the binomial probability distribution. 5. Describe the characteristics of and compute probabilities using the hypergeometric probability distribution. 6. Describe the characteristics of and compute probabilities using the Poisson probability distribution. GOALS 6-33 What is a Probability Distribution? Probability Distribution is a listing of all the outcomes of an experiment and the probability associated with each outcome It can be thought of a theoretical frequency distribution which describes how outcomes are expected to vary 6-44 Example of a Probability Distribution EXPERIMENT: Toss a coin three times. Observe the number of heads. The possible results are: zero heads, one head, two heads, and three heads. What is the probability distribution for the number of heads? 6-55 Probability Distribution of Number of Heads Observed in 3 Tosses of a Coin 6-66 Characteristics of a Probability Distribution The probability of a particular outcome is between 0 and 1 inclusive. The outcomes are mutually exclusive events. The list is exhaustive. The sum of the probabilities of the various events is 1
- 2. 2 6-7 Random Variables RANDOM VARIABLE A quantity resulting from an experiment that, by chance, can assume different values. EXAMPLES 1. The number of students in a class. 2. The number of children in a family. 3. The number of cars entering a carwash in a hour. 4. Number of home mortgages approved by Coastal Federal Bank last week. DISCRETE RANDOM VARIABLE A random variable that can assume only certain clearly separated values. It is usually the result of counting something. EXAMPLES 1. The length of each song on the latest Tim McGraw album. 2. The weight of each student in this class. 3. The temperature outside as you are reading this book. 4. The amount of money earned by each of the more than 750 players currently on Major League Baseball team rosters. CONTINUOUS RANDOM VARIABLE can assume an infinite number of values within a given range. It is usually the result of some type of measurement 6-88 The Mean of a Probability Distribution Mean The mean represents the central location of a probability distribution. The mean of a probability distribution is also referred to as its expected value. 6-99 The Variance, and Standard Deviation of a Discrete Probability Distribution Variance and Standard Deviation Measure the amount of spread in a distribution The computational steps are: 1. Subtract the mean from each value, and square this difference. 2. Multiply each squared difference by its probability. 3. Sum the resulting products to arrive at the variance. The SD is the positive square root of the variance. 6-10 Mean, Variance, and Standard Deviation of a Probability Distribution - Example John Ragsdale sells new cars for Pelican Ford. John usually sells the largest number of cars on Saturday. He has developed the following probability distribution for the number of cars he expects to sell on a particular Saturday. MEAN VARIANCE 136.1290.12 STANDARD DEVIATION 6-11 Expected Value Problem Mr. Byrd has decided to print either 25, 40, 55, or 70 thousand programs. Which number of programs will minimize the team’s expected losses? Program Sold 25,000 40,000 55,000 70,000 Probability 0.10 0.30 0.45 0.15 Harry Byrd is trying to decide how many programs to print for the team’s upcoming three-game series with the Oakland A’s. Each program costs 25 cents to print and sells for $1.25. Any programs unsold at the end of the series must be discarded. Mr. Byrd has estimated the following probability distribution for program sales: 6-12 Given: Per program cost = $0.25, Revenue = $1.25 and Profit = 1.25–0.25 =1.00 X P(X) 25000 40000 55000 70000 0.10 0.30 0.45 0.15 P(X) = 1.00 Probability Distribution Table Expected Value Problem
- 3. 3 6-13 Conditional Loss Table Demand Possible Stock Options 25000 40000 55000 70000 25000 40000 55000 70000 $0 15000 30000 45000 $3750 0 15000 30000 $7500 3750 0 15000 $11250 7500 3750 0 Expected Value Problem 6-14 Expected loss table for stocking 25,000 programs Demand Condition Loss P(X) Expected Loss 25000 40000 55000 70000 $3750 0 15000 30000 0.10 0.30 0.45 0.15 $375 0 6,750 4,500 Total= $11,625 Expected loss table for stocking 40,000 programs Demand Condition Loss P(X) Expected Loss 25000 40000 55000 70000 $0 15,000 30,000 45,000 0.10 0.30 0.45 0.15 $0 4,500 13,500 6,750 Total= $24,750 Expected Value Problem 6-15 Expected loss table for stocking 55000 programs Demand Condition Loss P(X) Expected Loss 25000 40000 55000 70000 $7500 3750 0 15000 0.10 0.30 0.45 0.15 $ 750 1,125 0 2,250 Total= $ 4,125 Expected loss table for stocking 70000 programs Demand Condition Loss P(X) Expected Loss 25000 40000 55000 70000 $11250 7500 3750 0 0.10 0.30 0.45 0.15 1,125 2,250 1,687.5 0 Total = $5,062.5 Print55,000asthe lossisminimum Expected Value Problem 6-1616 Binomial Probability Distribution The Binominal Probability Distribution is a discrete probability distribution in which each trial or observation can assume only one of two states. Also called Bernoulli Process 6-1717 Binomial Probability Distribution Characteristics 1. There are only two possible outcomes on a particular trial of an experiment. 2. The probability of outcomes of trials remain fixed over time. 3. Each trial is independent of any other trial. 4. The outcomes are mutually exclusive. 5. The random variable is the result of counts. 6-1818 Binomial Probability Formula
- 4. 4 6-1919 Binomial Probability– Example:1 There are five flights daily from Pittsburgh via US Airways into the Bradford, Pennsylvania, Regional Airport. Suppose the probability that any flight arrives late is 0.20. (a) What is the probability that NONE of the flights are late today? (b) What is the probability that ONE of the flights are late today? 3277.0)3277)(.1)(1( )20.0-1()20.0( )-1()()0( 0-50 05 - C CP xnx xn 4096.0=)4096)(.20.0)(5(= )20.0–1()20.0(C= )π–1()π(C=)1(P 1–51 15 x–nx xn 6-2020 For the example regarding the number of late flights, recall that =.20 and n = 5. What is the average number of late flights? What is the variance of the number of late flights? Binomial Distribution – Mean and Variance: Example 6-2121 Binomial Distribution – Mean and Variance: Another Solution 6-22 Binomial Distribution - Example EXAMPLE Five percent of the worm gears produced by an automatic, high- speed Carter-Bell milling machine are defective. What is the probability that out of six gears selected at random none will be defective? Exactly one? Exactly two? Exactly three? Exactly four? Exactly five? Exactly six out of six? Binomial – Shapes for Varying (n constant) Binomial – Shapes for Varying n ( constant) 6-23 Harry Ohme is in charge of the electronics section of a large department store. He has noticed that the probability that a customer who is just browsing will buy something is 0.3. Suppose that 15 customers browse in the electronics section each hour. (a) What is the probability that at least one browsing customer will buy something during a specified hour? (b) What is the probability that at least four browsing customers will buy something during a specified hour? (c) What is the probability that no browsing customers will buy anything during a specified hour? (d) What is the probability that no more than four browsing customers will buy something during a specified hour? Binomial Probability (Levin and Rubin) 6-24 9953.0=)30.0–1()30.0(C-1= )π–1()π(C=)1x(P)a( 0–510 015 x–nx xn ≥ 7032.0= ]1700.0+0916.0+0.0305+[0.0047–1 ])30.–1()30(.C+)30.–1()30(.C+ )30.–1()30(.C+)30.–1()30(.C-[1= P(3)]+P(2)+P(1)+[P(0)-1= )π–1()π(C=)4x(P)b( 3–513 315 2–512 215 1–511 115 0–510 015 x–nx xn≥ Binomial Probability (Levin and Rubin)
- 5. 5 6-25 0047.0=)30.0–1()30.0(C= )π–1()π(C=)0=x(P)c( 0–510 015 x–nx xn 5154.0= ]2186.0+1700.0+0916.0+0.0305+[0.0047= ])30.–1()30(.C+ )30.–1()30(.C+)30.–1()30(.C+ )30.–1()30(.C+)30.–1()30(.C[= P(4)]+P(3)+P(2)+P(1)+[P(0)=)4x(P)d( 4–514 415 3–513 315 2–512 215 1–511 115 0–510 015 ≤ Binomial Probability (Levin and Rubin) 6-26 The latest nationwide political poll indicates that for Americans who are randomly selected, the probability that they are conservative is 0.55, the probability that they are liberal is 0.30, and the probability that they are middle-of-the-road is 0.15. Assuming that these probabilities are accurate, answer the following the questions pertaining to a randomly chosen group of 10 Americans. (a) What is the probability that 4 are liberal ? (b) What is the probability that none are conservative ? (c) What is the probability that 2 are middle-of-the-road ? (d) What is the probability that 8 are liberal ? Binomial Probability (Levin and Rubin) 6-27 0003.0=)3.–1()3(.C=)π–1()π(C=)4(P)a( 4–014 410 x–nx xn 2001.0=)55.–1()55(.C=)π–1()π(C=)10(P)b( 0–014 010 x–nx xn 2759.0=)51.–1()15(.C=)π–1()π(C=)2(P)c( 2–014 210 x–nx xn 00160.0= 00001.0+00014.0+00145.0= )3.–1()3(.C +)3.–1()3(.C+)3.–1()3(.C= )10(P+)9(P+)8(P=)π–1()π(C=)8≥x(P)d( 01–0110 10 9–019 9 8–018 810 x–nx xn Binomial Probability (Levin and Rubin) 6-28 Problems Related to Meeting Conditions of Binomial Distribution In real life problems, the probability of success or failure does not remain fixed all the time The trials may not be statistically independent of each other 6-2929 Binomial – Shapes for Varying (n constant) 6-3030 Binomial – Shapes for Varying n ( constant)
- 6. 6 6-3131 Cumulative Binomial Probability Distribution A study in June 2003 by the Illinois Department of Transportation concluded that 76.2% of front seat occupants used seat belts. A sample of 12 vehicles is selected. What is the probability the front seat occupants in at least 7 of the 12 vehicles are wearing seat belts? 6-3232 Poisson Probability Distribution The Poisson probability distribution describes the number of times some event occurs during a specified interval. The interval may be time, distance, area, or volume. Assumptions of the Poisson Distribution (1) The random variable is the number of times some event occurs during a defined interval (2) The probability is proportional to the length of the interval. (3) The intervals don’t overlap and are independent. 6-3333 Poisson Probability Distribution The Poisson distribution can be described mathematically using the formula: 6-3434 Mean and Variance of Poisson Probability Distribution The mean number of successes can be determined in binomial situations by n, where n is the number of trials and the probability of a success. The variance of the Poisson distribution is also equal to n . 6-3535 Assume baggage is rarely lost by Northwest Airlines. Suppose a random sample of 1,000 flights shows a total of 300 bags were lost. Thus, the arithmetic mean number of lost bags per flight is 0.3 (300/1,000). If the number of lost bags per flight follows a Poisson distribution with u = 0.3, find the probability of not losing any bags. Poisson Probability Distribution– Example:1 6-3636 Poisson Probability Distribution- Table Assume baggage is rarely lost by Northwest Airlines. Suppose a random sample of 1,000 flights shows a total of 300 bags were lost. Thus, the arithmetic mean number of lost bags per flight is 0.3 (300/1,000). If the number of lost bags per flight follows a Poisson distribution with mean = 0.3, find the probability of not losing any bags.
- 7. 7 6-37 More About the Poisson Probability Distribution •The Poisson probability distribution is always positively skewed and the random variable has no specific upper limit. •The Poisson distribution for the lost bags illustration, where µ=0.3, is highly skewed. •As µ becomes larger, the Poisson distribution becomes more symmetrical. 6-38 1465.= !2 e4 = !x eµ =)x(P 4–2u–x The Lab Aid specializes in caring for minor injuries, colds, and flu. For the evening hours of 6-10 PM the mean number of arrivals is 4.0 per hour. What is the probability of 2 arrivals in an hour? Poisson Probability Distribution– Example:2 6-39 The US Bureau of Printing and Engraving (BPE) is responsible for printing this country’s paper money. The BPE has an impressively small frequency of printing errors; only 0.5 percent of all bills are too flawed for circulation. What is the probability that out of a batch of 1,000 bills (a) None are too flawed for circulation? (b) Ten are flawed for circulation? (c) Fifteen are flawed for circulation? Poisson Probability Distribution– Example:3 6-40 Poisson Probability Distribution– Example:3 01813.0= !10 e×)5( =)10(P 5–10 00016.0= !15 e×)5( =)15(P 5–15 Given us: n = 1000, π = 0.005 Thus: µ =nπ = (1000)(0.005) = 5 00674.0= !0 e5 = !x eµ =)0(P 5–0u–x 6-41 Concert Pianist Donna Prima has become quite upset at the number of coughs occurring in the audience just before she begins to play. On her latest tour, Donna estimates that on average eight coughs occur just before the start of her performance. Ms. Prima has sworn to her conductor that if she hears more than five coughs at tonight’s performance, she will refuse to play. What is the probability that she will not play tonight? Poisson Probability Distribution– Example:4 6-42 Poisson Probability Distribution– Example:4 8088.0=0.1912–1= 0.0916]+0.0573+0.0286+ 0.0107+0.0027+[0.0003–1= ] !5 e×)8( + !4 e×)8( + !3 e×)8( + !2 e×)8( + !1 e×)8( + !0 e×)8( [–1= 8–58–48–3 8–28–18–0 Given us: µ = 8 )]5(P+)4(P+)3(P+)2(P+)1(P+)0(P[–1
- 8. 8 6-43 Poisson Distribution as an Approximation of Binomial Distribution The Poisson Distribution is a good approximation of the Binomial Distribution when n ≥ 20 and π ≤ 0.05. 6-44 A hospital has 20 kidney dialysis machines and that the chance of any of them malfunctioning during any day is 0.02. What is the probability that exactly three machines will be out of service on the dame day? !x e×)πn( =)x(P )πn(–x 007.0= 6 e×)4.0( = !3 e×)02.0×20( =)3(P )4.0(–3 )02.0×20(–3 006.0= ).020-1()02.0(C=)3(P )π-1()π(C=)x(P 3-023 320 x-nx xn Approximation of Binomial Distribution: An Example Binomial DistributionPoisson Distribution 6-45 Hypergeometric Probability Distribution 1. An outcome on each trial of an experiment is classified into one of two mutually exclusive categories—a success or a failure. 2. The probability of success and failure changes from trial to trial. 3. The trials are not independent, meaning that the outcome of one trial affects the outcome of any other trial. Note: Use hypergeometric distribution if experiment is binomial, but sampling is without replacement from a finite population where n/N is more than 0.05 Formula: EXAMPLE PlayTime Toys, Inc., employs 50 people in the Assembly Department. Forty of the employees belong to a union and ten do not. Five employees are selected at random to form a committee to meet with management regarding shift starting times. What is the probability that four of the five selected for the committee belong to a union? Here’s what’s given: N = 50 (number of employees) S = 40 (number of union employees) x = 4 (number of union employees selected) n = 5 (number of employees selected) 6-4646 End of Chapter 6