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Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-1

Business Statistics: A Decision-Making Approach

8th Edition

Chapter 5Discrete Probability Distributions


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


Experiment: Toss 2 Coins. Let x = # heads.

T

T

Random variable vs. Probability distribution

4 possible outcomes

T

T

H

H

H H 0 1 2 x

x Value Probability

0 1/4 = 0.25

1 2/4 = 0.50

2 1/4 = 0.25

0.50

0.25

Prob

abili

ty

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


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


Cumulative Probability Distribution


Number of heads: x Probability: P(X = x)Cumulative Probability: P(X < x)

0 0.25 0.25

1 0.50 0.75

2 0.25 1.00


A discrete random variable is a variable that can assume only a countable number of valuesMany 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

Discrete Random Variable


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.


If a random variable is a discrete variable, its probability distribution is called a discrete probability distribution.

Discrete Probability Distribution

Number of heads Probability

0 0.25

1 0.50

2 0.25


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:

E(x) = (0 x 0.25) + (1 x 0.50) + (2 x 0.25) = 1.0

x P(x)

0 0.25

1 0.50

2 0.25


Standard Deviation of a discrete distribution

where:E(x) = Expected value of the random variable x = Values of the random variableP(x) = Probability of the random variable having

the value of x

Discrete Random Variable Standard Deviation

P(x)E(x)}{xσ 2x


Example: Toss 2 coins, x = # heads, compute standard deviation (recall E(x) = 1)

Discrete Random Variable Standard Deviation

P(x)E(x)}{xσ 2x

.7070.50(0.25)1)(2(0.50)1)(1(0.25)1)(0σ 222x

(continued)

Possible number of heads = 0, 1, or 2


Probability Distributions

Continuous Probability

Distributions

Binomial

Hypergeometric

Poisson

Discrete Probability

Distributions

Normal

Uniform

Exponential

Ch. 5 Ch. 6


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


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


Counting Rule for Combinations

A combination is an outcome of an experiment where x objects are selected from a group of n objects

)!xn(!x!nCn

x

where:

Cx = number of combinations of x objects selected from n objectsn! =n(n - 1)(n - 2) . . . (2)(1)

x! = x(x - 1)(x - 2) . . . (2)(1)

n

NOTE: 0! = 1 (by definition)

Order does not matter

i.e. ABC = CBA only one outcome


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)

P(x)n

x ! n xp qx n x!

( ) !

Binomial Distribution Formula


Binomial Distribution Summary Measures

Mean

Variance and Standard Deviation

npE(x)μ

npqσ2

npqσ

Where n = sample sizep = probability of successq = (1 – p) = probability of failure


n = 5 p = 0.1

n = 5 p = 0.5

Mean

0.2.4.6

0 1 2 3 4 5X

P(X)

.2

.4

.6

0 1 2 3 4 5X

P(X)

0

Binomial Distribution

The shape of the binomial distribution depends on the values of p and n

Here, n = 5 and p = 0.1

Here, n = 5 and p = 0.5


n = 5 p = 0.1

n = 5 p = 0.5

Mean

0.2.4.6

0 1 2 3 4 5X

P(X)

.2

.4

.6

0 1 2 3 4 5X

P(X)

0

0.5(5)(0.1)npμ

0.6708.1)(5)(0.1)(1npqσ

2.5(5)(0.5)npμ

1.118.5)0(5)(0.5)(1npqσ

Binomial Characteristics

Examples


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 :

.25220(0.65)(0.35)3!7!10!qp

x)!(nx!n!3)P(x 73xnx

There is a 25.22% chance that 3 out of the 10 voters will support Proposition A


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

Using Binomial Tablesn = 10

x p=.15 p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50

012345678910

0.19690.34740.27590.12980.04010.00850.00120.00010.00000.00000.0000

0.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.0000

0.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.0000

0.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.0000

0.01350.07250.17570.25220.23770.15360.06890.02120.00430.00050.0000

0.00600.04030.12090.21500.25080.20070.11150.04250.01060.00160.0001

0.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.0003

0.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010

109876543210

p=.85 p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 x


Using PHStat

Select: Add-Ins / PHStat / Probability & Prob. Distributions / Binomial…


Using PHStat Enter desired values in dialog box

Here: n = 10p = 0.35

Output for x = 0 to x = 10 will be generated by PHStat

Optional check boxesfor additional output


P(x = 3 | n = 10, p = 0.35) = 0.2522

PHStat Output

P(x > 5 | n = 10, p = 0.35) = 0.0949


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


Poisson Distribution Summary Measures

Mean

Variance and Standard Deviation

λtμ

λtσ2

λtσ where = number of successes in a segment of unit size

t = the size of the segment of interest


Graph of Poisson Probabilities

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x)X

t =0.50

01234567

0.60650.30330.07580.01260.00160.00020.00000.0000

P(x = 2) = 0.0758

Graphically: = .05 and t = 100


Poisson Distribution Shape

The shape of the Poisson Distribution depends on the parameters and t:

0.00

0.05

0.10

0.15

0.20

0.25

1 2 3 4 5 6 7 8 9 10 11 12

x

P(x

)

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x)

t = 0.50 t = 3.0


Poisson Distribution Formula

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...)

!xe)t()x(P

tx


Using Poisson Tables

X

t

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

01234567

0.90480.09050.00450.00020.00000.00000.00000.0000

0.81870.16370.01640.00110.00010.00000.00000.0000

0.74080.22220.03330.00330.00030.00000.00000.0000

0.67030.26810.05360.00720.00070.00010.00000.0000

0.60650.30330.07580.01260.00160.00020.00000.0000

0.54880.32930.09880.01980.00300.00040.00000.0000

0.49660.34760.12170.02840.00500.00070.00010.0000

0.44930.35950.14380.03830.00770.00120.00020.0000

0.40660.36590.16470.04940.01110.00200.00030.0000

Example: Find P(x = 2) if = 0.05 and t = 100

.075802!e(0.50)

!)()2(

0.502

xetxP

tx


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


Hypergeometric Distribution Formula

Nn

Xx

XNxn

CCC)x(P

.

WhereN = population sizeX = number of successes in the populationn = sample sizex = number of successes in the sample

n – x = number of failures in the sample

(Two possible outcomes per trial: success or failure)


Hypergeometric Distribution Example

0.3120

(6)(6)C

CCC

CC2)P(x 103

42

61

Nn

Xx

XNxn

■ 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 n = 3 X = 4 x = 2


Hypergeometric Distribution in PHStat

Select: Add-Ins / PHStat / Probability & Prob. Distributions / Hypergeometric …


Hypergeometric Distribution in PHStat

Complete dialog box entries and get output …

N = 10 n = 3X = 4 x = 2

P(x = 2) = 0.3

(continued)


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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recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America.

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