Download Business Statistics: A Decision

Business Statistics:
A Decision-Making Approach
8th Edition
Chapter 5
Discrete Probability Distributions
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall
5-1
Chapter Goals
After completing this chapter, you should be able to:

Calculate and interpret the expected value of a discrete
probability distribution

Apply the binomial distribution to business problems

Compute probabilities for the Poisson and hypergeometric
distributions

Recognize when to apply discrete probability distributions
to decision making situations
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5-2
Random variable vs. Probability distribution
Experiment: Toss 2 Coins.
Let x = # heads.
4 possible outcomes
T
H
H
T
H
T
H
x Value
Probability
0
1/4 = 0.25
1
2/4 = 0.50
2
1/4 = 0.25
Probability
T
0.50
0.25
0
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1
2
x
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Random variable vs. Probability distribution

When the value of a variable is the outcome of
a statistical experiment, that variable is a random
variable.



Let the variable X represent the number of Heads that result
from the previous experiment.
The variable X can take on the values 0, 1, or 2. In this previous
example, X is a random variable
A probability distribution is a table or an equation that
links each outcome of a statistical experiment with its
probability of occurrence.

Just like the previous table
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5-4
Cumulative Variable vs.
Cumulative Probability Distribution


A cumulative probability refers to the
probability that the value of a random variable
falls within a specified range.
Like a probability distribution, a cumulative
probability distribution can be represented by
a table or an equation.

See the next slide
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5-5
Cumulative Probability Distribution
Cumulative
Number of heads: x Probability: P(X = x) Probability:
P(X < x)
0
0.25
0.25
1
0.50
0.75
2
0.25
1.00
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5-6
Discrete Random Variable

A discrete random variable is a variable that
can assume only a countable number of values
Many possible outcomes:

number of complaints per day

number of TV’s in a household

number of rings before the phone is answered
Only two possible outcomes:

gender: male or female

defective: yes or no

spreads peanut butter first vs. spreads jelly first
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5-7
Continuous Random Variable

A continuous random variable is a variable that
can assume any value on a continuum (can
assume an uncountable number of values)





thickness of an item
time required to complete a task
temperature of a solution
height, in inches
These can potentially take on any value,
depending only on the ability to measure
accurately.
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5-8
Discrete Probability Distribution

If a random variable is a discrete variable,
its probability distribution is called a discrete
probability distribution.
Number of heads
Probability
0
0.25
1
0.50
2
0.25
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5-9
Discrete Random Variable
Mean

Expected Value (or mean) of a discrete distribution
(Weighted Average)
E(x) = xP(x)

Example: Toss 2 coins,
x = # of heads,
compute expected value of x:
x
P(x)
0
0.25
1
0.50
2
0.25
E(x) = (0 x 0.25) + (1 x 0.50) + (2 x 0.25)
= 1.0
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5-10
Discrete Random Variable
Standard Deviation

Standard Deviation of a discrete distribution
σx 
 {x  E(x)}
2
P(x)
where:
E(x) = Expected value of the random variable
x = Values of the random variable
P(x) = Probability of the random variable having
the value of x
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5-11
Discrete Random Variable
Standard Deviation
(continued)

Example: Toss 2 coins, x = # heads,
compute standard deviation (recall E(x) = 1)
σx 
 {x  E(x)}
2
P(x)
σ x  (0  1) 2 (0.25)  (1  1) 2 (0.50)  (2  1) 2 (0.25)  0.50  .707
Possible number of heads
= 0, 1, or 2
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5-12
Probability Distributions
Ch. 5
Ch. 6
Discrete
Probability
Distributions
Continuous
Probability
Distributions
Binomial
Normal
Poisson
Uniform
Hypergeometric
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Exponential
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The Binomial Distribution

Characteristics of the Binomial Distribution:





A trial has only two possible outcomes – “success” or
“failure”
There is a fixed number, n (finite), of identical trials
The trials of the experiment are independent of each
other
The probability of a success, p, remains constant from
trial to trial
If p represents the probability of a success, then
(1-p) = q is the probability of a failure
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5-14
Binomial Distribution Examples

A manufacturing plant labels items as
either defective or acceptable

A firm bidding for a contract will either get
the contract or not

A marketing research firm receives survey
responses of “yes I will buy” or “no I will
not”

New job applicants either accept the offer
or reject it
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5-15
Counting Rule for Combinations

A combination is an outcome of an experiment
where x objects are selected from a group of n
objects
n!
C 
x! (n  x )!
n
x
where:
n
Cx = number of combinations of x objects selected from n objects
n! =n(n - 1)(n - 2) . . . (2)(1)
Order does not
x! = x(x - 1)(x - 2) . . . (2)(1)
matter
NOTE: 0! = 1 (by definition)
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i.e. ABC = CBA
only one outcome
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Binomial Distribution Formula
n!
x nx
P(x) 
p q
x ! (n  x ) !
P(x) = probability of x successes in n trials,
with probability of success p on each trial
x = number of successes in sample,
(x = 0, 1, 2, ..., n)
p = probability of “success” per trial
q = probability of “failure” = (1 – p)
n = number of trials (sample size)
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5-17
Binomial Distribution
Summary Measures


Mean
μ  E(x)  np
Variance and Standard Deviation
σ  npq
2
σ  npq
Where n = sample size
p = probability of success
q = (1 – p) = probability of failure
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5-18
Binomial Distribution

The shape of the binomial distribution depends on the
values of p and n
Mean

Here, n = 5 and p = 0.1
P(X)
.6
.4
.2
0
X
0
P(X)

Here, n = 5 and p = 0.5
.6
.4
.2
0
1
2
3
4
5
n = 5 p = 0.5
X
0
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n = 5 p = 0.1
1
2
3
4
5
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Binomial Characteristics
Examples
μ  np  (5)(0.1)  0.5
Mean
σ  npq  (5)(0.1)(1  .1)
 0.6708
P(X)
.6
.4
.2
0
X
0
μ  np  (5)(0.5)  2.5
σ  npq  (5)(0.5)(1  0.5)
 1.118
P(X)
.6
.4
.2
0
1
2
3
4
5
n = 5 p = 0.5
X
0
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n = 5 p = 0.1
1
2
3
4
5
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Binomial Distribution Example

Example: 35% of all voters support Proposition
A. If a random sample of 10 voters is polled,
what is the probability that exactly three of them
support the proposition?
i.e., find P(x = 3) if n = 10 and p = 0.35 :
n!
10!
x n x
P(x  3) 
p q 
(0.35) 3 (0.65) 7  0.2522
x! (n  x)!
3!7!
There is a 25.22% chance that 3 out of the 10 voters
will support Proposition A
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5-21
Using Binomial Tables
n = 10
x
p=.15
p=.20
p=.25
p=.30
p=.35
p=.40
p=.45
p=.50
0
1
2
3
4
5
6
7
8
9
10
0.1969
0.3474
0.2759
0.1298
0.0401
0.0085
0.0012
0.0001
0.0000
0.0000
0.0000
0.1074
0.2684
0.3020
0.2013
0.0881
0.0264
0.0055
0.0008
0.0001
0.0000
0.0000
0.0563
0.1877
0.2816
0.2503
0.1460
0.0584
0.0162
0.0031
0.0004
0.0000
0.0000
0.0282
0.1211
0.2335
0.2668
0.2001
0.1029
0.0368
0.0090
0.0014
0.0001
0.0000
0.0135
0.0725
0.1757
0.2522
0.2377
0.1536
0.0689
0.0212
0.0043
0.0005
0.0000
0.0060
0.0403
0.1209
0.2150
0.2508
0.2007
0.1115
0.0425
0.0106
0.0016
0.0001
0.0025
0.0207
0.0763
0.1665
0.2384
0.2340
0.1596
0.0746
0.0229
0.0042
0.0003
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
0.2051
0.1172
0.0439
0.0098
0.0010
10
9
8
7
6
5
4
3
2
1
0
p=.85
p=.80
p=.75
p=.70
p=.65
p=.60
p=.55
p=.50
x
Examples:
n = 10, p = 0.35, x = 3:
P(x = 3|n =10, p = 0.35) = 0.2522
n = 10, p = 0.75, x = 2:
P(x = 2|n =10, p = 0.75) = 0.0004
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall
5-22
Using PHStat

Select: Add-Ins / PHStat / Probability & Prob. Distributions / Binomial…
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5-23
Using PHStat

Enter desired values in dialog box
Here: n = 10
p = 0.35
Output for x = 0
to x = 10 will be
generated by PHStat
Optional check boxes
for additional output
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5-24
PHStat Output
P(x = 3 | n = 10, p = 0.35) = 0.2522
P(x > 5 | n = 10, p = 0.35) = 0.0949
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5-25
The Poisson Distribution

Characteristics of the Poisson Distribution:

The outcomes of interest are rare relative to the
possible outcomes

The average number of outcomes of interest per time or
space interval is 

The number of outcomes of interest are random, and
the occurrence of one outcome does not influence the
chances of another outcome of interest

The probability that an outcome of interest occurs in a
given segment is the same for all segments
e.g. the number of customers per hour or the number of bags lost per flight
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Poisson Distribution
Summary Measures


Mean
μ  λt
Variance and Standard Deviation
σ  λt
2
σ  λt
where
 = number of successes in a segment of unit size
t = the size of the segment of interest
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5-27
Graph of Poisson Probabilities
0.70
Graphically:
0.60
 = .05 and t = 100
0
1
2
3
4
5
6
7
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
P(x)
X
t =
0.50
0.50
0.40
0.30
0.20
0.10
0.00
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0
1
2
3
4
5
6
7
x
P(x = 2) = 0.0758
5-28
Poisson Distribution Shape

The shape of the Poisson Distribution
depends on the parameters  and t:
t = 0.50
t = 3.0
0.70
0.25
0.60
0.20
0.40
P(x)
P(x)
0.50
0.30
0.15
0.10
0.20
0.05
0.10
0.00
0.00
0
1
2
3
4
5
x
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6
7
1
2
3
4
5
6
7
8
9
10
11
12
x
5-29
Poisson Distribution Formula
( t ) e
P( x ) 
x!
x
 t
where:
t = size of the segment of interest
x = number of successes in segment of interest
 = expected number of successes in a segment of unit size
e = base of the natural logarithm system (2.71828...)
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5-30
Using Poisson Tables
t
X
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
0
1
2
3
4
5
6
7
0.9048
0.0905
0.0045
0.0002
0.0000
0.0000
0.0000
0.0000
0.8187
0.1637
0.0164
0.0011
0.0001
0.0000
0.0000
0.0000
0.7408
0.2222
0.0333
0.0033
0.0003
0.0000
0.0000
0.0000
0.6703
0.2681
0.0536
0.0072
0.0007
0.0001
0.0000
0.0000
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
0.5488
0.3293
0.0988
0.0198
0.0030
0.0004
0.0000
0.0000
0.4966
0.3476
0.1217
0.0284
0.0050
0.0007
0.0001
0.0000
0.4493
0.3595
0.1438
0.0383
0.0077
0.0012
0.0002
0.0000
0.4066
0.3659
0.1647
0.0494
0.0111
0.0020
0.0003
0.0000
Example: Find P(x = 2) if  = 0.05 and t = 100
(t ) x e  t (0.50) 2 e 0.50
P( x  2) 

 0.0758
x!
2!
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5-31
The Hypergeometric Distribution

“n” trials in a sample taken from a finite population of
size N

Sample taken without replacement

Trials are dependent

The probability changes from trial to trial

Concerned with finding the probability of “x” successes
in the sample where there are “X” successes in the
population
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5-32
Hypergeometric Distribution
Formula
(Two possible outcomes per trial: success or failure)
P( x ) 
N X .
n x
N
n
C
C
X
x
C
Where
N = population size
X = number of successes in the population
n = sample size
x = number of successes in the sample
n – x = number of failures in the sample
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5-33
Hypergeometric Distribution
Example
■ Example: 3 Light bulbs were selected from 10. Of the
10 there were 4 defective. What is the probability that 2
of the 3 selected are defective?
N = 10
X=4
P(x  2) 
N X
n x
N
n
C
C
C
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X
x
n=3
x=2
6
1
4
2
C C
(6)(6)


 0.3
10
C3
120
5-34
Hypergeometric Distribution
in PHStat

Select: Add-Ins / PHStat / Probability & Prob. Distributions /
Hypergeometric …
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5-35
Hypergeometric Distribution
in PHStat
(continued)

Complete dialog box entries and get output …
N = 10
X=4
n=3
x=2
P(x = 2) = 0.3
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5-36
Chapter Summary

Reviewed key discrete distributions



Binomial
Poisson
Hypergeometric

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems
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5-37
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