Transcript Discrete Probability Distributions

Chapter 5
Discrete Probability Distributions
5.1 Random Variables
 Random Variable
 A random variable is a function that assigns a numerical value to each
outcome of an experiment.
 Example 5.1.1: We flip a coin two times and are interested in the number
of heads.
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The outcomes are HH, HT, TH and TT.
Let x be the “total number of heads observed”. Then x = 0, 1, or 2.
Therefore, the above outcomes can be assigned these numbers as follows.
Outcomes
Value of x (i.e., total number of heads)
TT
0
TH
1
HT
1
HH
2
The variable “total number of heads observed” in this experiment is called a
random variable.
5.1 Random Variables (cont.)
 Discrete Random Variable
 A random variable that can assume a finite number of possible values is called a
discrete random variable.
 That is, the possible values of a discrete random variable can be listed or
counted.
 Example 5.1.2: X is a discrete random variable if X represents the number of
people out of 200 who will make an airline reservation and then fail to show up.
 Example 5.1.3: You roll two dice, a red die and a blue die. Suppose X is the total
of the two dice. X is a discrete random variable because X has finite number of
possible values.
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Outcome
Red Die
Blue Die
Value of X
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1
1
2
2
1
2
3
3
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3
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6
6
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5.1 Random Variables (cont.)
 Continuous Random Variable
 A random variable that can take any value over some continuous range
of values is called a continuous random variable.
 Example 5.1.4: Let Y be the amount of rainfall during the month of
September. Y is a continuous random variable because the amount of
rainfall can be any nonnegative value.
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5.1 Random Variables (cont.)
 Probability Distribution
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A probability distribution is the list of all possible outcomes of a random
variable and their associated probabilities.
Example 5.1.5: Toss three fair coins and let X equal the number of tails
observed.
 The outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
Each outcome has equal probability.
 X = 0, 1, 2, and 3.
Value of X
Frequency (i.e., number of outcomes)
P(X)
0
1
0.125
1
3
0.375
2
3
0.375
3
1
0.125
8
1.000
P(X) is between 0 and 1 (inclusive)
∑P(X) = 1
5.2 Probability Distributions for Discrete Random Variables
 Three popular methods of describing probabilities associated with a
discrete random variable.
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List each value of X and its corresponding probability.
Use a histogram to convey the probabilities corresponding to the various
values of X.
Use a function that assigns a probability to each value of X.
5.2 Probability Distributions for Discrete Random Variables (cont.)
 List each value of X and its corresponding probability. Used in Example
5.1.5.
 Use a histogram to convey the probabilities corresponding to the various
values of X.
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Probability
3/8 –
2/8 –
1/8 –
–
0
1
2
x = number of heads
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5.2 Probability Distributions for Discrete Random Variables (cont.)
 Probability Mass Function (PMF)
 A PMF is a function that assigns a probability to each value of X.
 P(X = x) = some expression (usually containing x) that produces a
probability of observing x = P(x).
 P(x) is between 0 and 1 (inclusive) for each x
 ∑P(x) = 1
 Example 5.2.1: The function P(X=x) = x/30 for x = 0, 10, and 20 (and zero
elsewhere) is a probability mass function because:
 P(X=x) = x/30 assigns a probability to each value of x.
 ∑P(X=x) = 1
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Value of X
P(X = x) = x/30
0
0/30 = 0
10
10/30 = 1/3
20
20/30 = 2/3
Others
0
1.00
5.2 Probability Distributions for Discrete Random Variables (cont.)
 Probability Mass Function (PMF) (cont.)
 Example 5.2.2: A manager has four employees with 0, 1, 3, and 4 years of job
experience. The manager will assign two of the employees at random to a team.
Define X to be equal to the total number of years of job experience for the two
selected employees.
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What values can X assume?
X = {0+1, 0+3, 0+4 or 1+3, 1+4, 3+4} = {1, 3, 4, 5, 7}
What is the probability mass function of X?
P(X=i) = 1/6, where i = 1, 3, 5, and 7; and
P(X=j) = 2/6, where j = 4.
5.2 Probability Distributions for Discrete Random Variables (cont.)
 Mean of Discrete Random Variables
 The mean of a discrete random variable represents the average value of the
random variable if you were to observe this variable over an indefinite period of
time.
 The mean of a discrete random variable is written as µ and µ = ∑xP(x).
 Example 5.2.2: Toss three fair coins and let X equal the number of tails observed.
Find the mean number of tails observed.
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Value of X
P(X)
XP(X)
0
0.125
0(0.125) = 0
1
0.375
1(0.375) = 0.375
2
0.375
2(0.375) = 0.75
3
0.125
3(0.125) = 0.375
1.000
∑XP(X) = 1.500
5.2 Probability Distributions for Discrete Random Variables (cont.)
 Variance of Discrete Random Variables
 The variance of a discrete random variable, X, is a parameter describing the
variation of the corresponding population.
 The symbol used is 2.
 2 = ∑(x - µ)2 • P(x) = ∑x2P(x) - µ2
 Example 5.2.3: Toss three fair coins and let X equal the number of tails observed.
Find the variance for the number of tails observed.
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(X - µ)2P(X)
P(X)
XP(X)
0
0.125
0(0.125) = 0
1
0.375 1(0.375) = 0.375 1 - 1.5 = -0.5 (-0.5)2 = 0.25 0.25(0.375) = 0.09375
2
0.375
2 - 1.5 = 0.5
(0.5)2 = 0.25
0.25(0.375) = 0.09375
3
0.125 3(0.125) = 0.375 3 - 1.5 = 1.5
(1.5)2 = 2.25
2.25(0.125) = 0.28125
1.000
2(0.375) = 0.75
1.500
X-µ
(X - µ)2
Value of X
0 - 1.5 = -1.5 (-1.5)2 = 2.25 2.25(0.125) = 0.28125
∑(X - µ)2P(X) = 0.75
5.3 Binomial Random Variable
 Binomial Random Variable
 A discrete random variable that can assume one of two possible
outcomes in each trial of an experiment comprising of n independent
trials.
 An experiment in which each trial results in one of two mutually
exclusive outcomes is called a binomial experiment.
 Example 5.3.1: Flip a coin five times. Let X be the number of heads.
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A binomial experiment with 5 trials and each trial has two possible outcomes
– a head and a tail. Trials (flips) are independent.
X is a binomial random variable.
5.3 Binomial Random Variable (cont.)
 Characteristics of a binomial experiment
 The experiment consists of n repetitions, called trials.
 Each trial has two mutually exclusive possible outcomes, referred to as
success and failure.
 The n trials are independent.
 The probability for a success for each trial is denoted p; and remains the
same for each trial.
 The random variable x is the number of successes out of n trials.
 Example 5.3.2: Flip a coin five times. Let X be the number of heads. Is it
a binomial situation?
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n = 5.
Success = head, failure = tail.
The results on one flip do not affect the results on another flip.
p = the probability of flipping a head on a particular flip = ½.
X = the number of heads out of five flips.
5.3 Binomial Random Variable (cont.)
 Binomial Distribution
 The probability distribution of a binomial random variable is called a
binomial distribution.
 Probability Mass Function (PMF) of a binomial random variable is:
 n C x p x (1  p) nx for x  0, 1, 2,...., n
P( X  x)  
elsewhere
0
 (number of ways of getting X  x)(probabi lity of each one)
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Example 5.3.3: A lawyer estimates that 40% of the cases in which she
represented the defendant were won. If the lawyer is presently
representing 10 defendants, what is the probability that 5 of the cases
will be won?
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n = 10; x = 5; p = 0.4
P( X  5) n C x p x (1  p) n  x 10 C5 (0.4)5 (1  0.4)105  0.201
5.3 Binomial Random Variable (cont.)
 Using Binomial Table A.1 to Determine Probabilities
 The binomial PMFs have been tabulated in Table A.1 for various values
of n and p.
 If n = 4 and p = 0.3 and you wish to find the P(2) locate n = 4 and x = 2.
 Go across to p = 0.3 and you will find the corresponding probability
(after inserting the decimal in front of the number). This probability is
0.265.
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5.3 Binomial Random Variable (cont.)
 Shape of the Binomial Distribution
 Approximately bell-shaped (symmetric) if p is near ½ or if n is large.
 Skewed left for p > ½ and small n.
 Skewed right for p < ½ and small n.
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Probability
Probability
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0
1
2
3
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5
6
7
8
9 10
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.1 –
x
0
1
2
(n = 10, p = .5)
(a)
5
6
7
8
9 10
8
9 10
x
.3 –
Probability
Probability
4
(n = 10, p = .8)
(b)
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3
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0
1
2
3
4
5
6
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(n = 10, p = .2)
(c)
8
9 10
x
.2 –
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0
1
2
3
4
5
6
7
(n = 20, p = .2)
(d)
20
x
5.3 Binomial Random Variable (cont.)
 Cumulative Binomial Probabilities
 Finding P(X ≤ k), that is, finding the total probability of all successes up
to and including k.
 P(X ≤ k) = P(X = 0) + P(X = 1) + P(X = 2) + ….. + P(X = k).
 P(X < k) = P(X = 0) + P(X = 1) + P(X = 2) + ….. + P(X = k - 1).
 P(X > k) = 1 - P(X ≤ k).
 P(X ≥ k) = 1 - P(X < k).
 Using Binomial Table A.2
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If n = 4 and p = 0.3 and you wish to find the P(x ≤ 2) locate n = 4 and x = 2.
Go across to p = 0.3 and you will find the corresponding probability (after
inserting the decimal in front of the number). This probability is 0.916.
5.3 Binomial Random Variable (cont.)
 Example 5.3.4: Suppose that 60% of all the employees of ABC
Company favor unionization. A poll of 20 employees is taken to
determine the number who favor unionization.
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Find the probability that at least 10 employees favor unionization.
p = 0.6, n = 20, P(x ≤ 10) = 0.245
[Using Table A.2]
Find the probability that more than 12 employees favor unionization.
P(x > 12) = 1 – P(x ≤ 12) = 1 – 0.584 (From Table A.2) = 0.416
5.3 Binomial Random Variable (cont.)
 Mean and Variance of a Binomial Random Variable
 µ = np
 2 = np(1 - p)
 Example 5.3.5: Suppose that 60% of all the employees of ABC Company
favor unionization. A poll of 20 employees is taken to determine the
number who favor unionization. Find the mean and standard deviation of
those who favor unionization.
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p = 0.6, n = 20, µ = np = 20(0.6) = 12.
2 = np(1 - p) = 20(0.6)(1 – 0.6) = 4.8,  = sqrt(4.8) = 2.19.
5.3 Binomial Random Variable (cont.)
 Finding Probabilities with Excel
 P(X ≤ 1) = P(X = 0) + P(X = 1). Assume n = 100 and p = 0.05.
 P(X = 0) is shown below.
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5.3 Binomial Random Variable (cont.)
 Finding Probabilities with Excel
 P(X = 1) is shown below.
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5.3 Binomial Random Variable (cont.)
 Finding Probabilities with Excel
 P(X ≤ 1) is shown below.
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5.4 Poisson Distribution
 The Poisson distribution is useful for counting the number of times a
particular event occurs over a specified period of time or over a
specified area.
 For example, arrivals of customers at a service facility follow Poisson
Distribution.
 Conditions for the Poisson Distribution:
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The number of occurrences in one measurement unit are independent
of the number of occurrences in any other nonoverlapping
measurement unit.
The expected number of occurrences in any given measurement unit
are proportional to the size of the measurement unit.
Events can not occur at exactly the same point in the measurement
unit.
5.4 Poisson Distribution (cont.)
 PMF of Poisson Distribution:
P( X  x) 
 xe 
x!
for x  0, 1, 2, 3, ......
 Mean and Variance of a Poisson Random Variable
 Mean of X = ∑xP(x) = µ = expected number of occurrences.
 Variance of X = 2 = ∑x2P(x) - µ2 = µ = expected number of occurrences.
 Example 5.4.1: A large bakery determined that the expected number of
delivery truck breakdowns per day is 1.5. Assume that the number of
breakdowns is independent from day to day.
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What is the probability that there will be exactly two breakdowns tomorrow?
µ = 1.5, P(X = 2) = 0.251.
What is the probability that there will be exactly two breakdowns during next two
days?
µ = 2(1.5) = 3, P(X = 2) = 0.224.
5.4 Poisson Distribution (cont.)
 Finding Poisson Probabilities with Statistical Software
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5.4 Poisson Distribution (cont.)
 Finding Poisson Probabilities with Statistical Software
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