Business Statistics, A First Course (4e) © 2006 Prentice-Hall, Inc. Chap 5-1 Chapter 5 Some Important Discrete Probability Distributions Statistics for

Business Statistics, A First Course (4e) © 2006 Prentice-Hall, Inc. Chap 5-1

Chapter 5

Some Important Discrete Probability Distributions

Statistics for Managers4th Edition


Learning Objectives

In this chapter, you learn: The properties of a probability distribution To calculate the expected value, variance, and

standard deviation of a probability distribution To calculate probabilities from Binomial, Hypergeometric and Poisson distributions How to use the Binomial, Hypergeometric and

Poisson distributions to solve business problems


Introduction to Probability Distributions

Random Variable Represents a possible numerical value from

an uncertain event

Random

Variables

Discrete Random Variable

ContinuousRandom Variable

Ch. 5 Ch. 6


Discrete Random Variables

Can only assume a countable number of values

Examples:

Roll a die twiceLet X be the number of times 4 comes up (then X could be 0, 1, or 2 times)

Toss a coin 5 times. Let X be the number of heads

(then X = 0, 1, 2, 3, 4, or 5)


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

T

T

Discrete Probability Distribution

4 possible outcomes

T

T

H

H

H H

Probability Distribution

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

Pro

bab

ility


Discrete Random Variable Summary Measures

Expected Value (or mean) of a discrete distribution (Weighted Average)

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

N

1iii )X(PX E(X)


Variance of a discrete random variable

Standard Deviation of a discrete random variable

where:E(X) = Expected value of the discrete random variable X

Xi = the ith outcome of XP(Xi) = Probability of the ith occurrence of X


N

1ii

2i

2 )P(XE(X)][Xσ

(continued)

N

1ii

2i

2 )P(XE(X)][Xσσ


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


)P(XE(X)][Xσ i2

i

0.7070.50(0.25)1)(2(0.50)1)(1(0.25)1)(0σ 222

(continued)

Possible number of heads = 0, 1, or 2


Example of Expected ValueA company is considering a proposal to

develop a new product. The initial cash outlay would be $1 million, and development time would be three years. If successful, the firm anticipates that net profit (revenue minus initial cash outlay) over the 5 year life cycle of the product will be $1.5 million. If moderately successful, net profit will reach $1.2 million. If unsuccessful, the firm anticipates zero cash inflows. The firms assigns the following probabilities to the 5-year prospects for this product: successful, .60; moderately successful, .30; and unsuccessful, .10. What is the expected net profit? Ignore time value of money.


Calculation of Expected Value

To calculate the expected value

E(X) = Σ [(X) P(X)]

= (1.5)(.6) + (1.2)(.3) + (-1)(.1)

= + 1.16

Since the expected value Is positive this project passes the initial approval process

Statistics for Managers Using Microsoft Excel, 5e © 2008 Pearson Prentice-Hall, Inc. Chap 5-11

Investment ReturnsThe Mean

Consider the return per $1000 for two types of investments.

Economic

P(XiYi) Condition

Investment

Passive Fund X Aggressive Fund Y

0.2 Recession - $25 - $200

0.5 Stable Economy + $50 + $60

0.3 Expanding Economy + $100 + $350


Investment ReturnsThe Mean

E(X) = μX = (-25)(.2) +(50)(.5) + (100)(.3) = 50

E(Y) = μY = (-200)(.2) +(60)(.5) + (350)(.3) = 95

Interpretation: Fund X is averaging a $50.00 return and fund Y is averaging a $95.00 return per $1000 invested.


Investment ReturnsStandard Deviation

43.30

(.3)50)(100(.5)50)(50(.2)50)(-25σ 222X

71.193

)3(.)95350()5(.)9560()2(.)95200-(σ 222Y

Interpretation: Even though fund Y has a higher average return, it is subject to much more variability and the probability of loss is higher.


Investment ReturnsCovariance

8250

95)(.3)50)(350(100

95)(.5)50)(60(5095)(.2)200-50)((-25σXY

Interpretation: Since the covariance is large and positive, there is a positive relationship between the two investment funds, meaning that they will likely rise and fall together.


The Sum of Two Random Variables: Measures

Expected Value:

Variance:

Standard deviation:

XYYXYXYX 2σσσσ)Var( 222

)()()( YEXEYXE

2σσ YXYX


Portfolio Expected Return and Expected Risk

Investment portfolios usually contain several different funds (random variables)

The expected return and standard deviation of two funds together can now be calculated.

Investment Objective: Maximize return (mean) while minimizing risk (standard deviation).


Portfolio Expected Return and Expected Risk

Portfolio expected return (weighted average return):

Portfolio risk (weighted variability)

where w = portion of portfolio value in asset X

(1 - w) = portion of portfolio value in asset Y

)()1()(E(P) YEwXEw

XY2Y

22X

2P w)σ-2w(1σ)w1(σwσ


Portfolio Expect Return and Expected Risk

Recall: Investment X: E(X) = 50 σX = 43.30

Investment Y: E(Y) = 95 σY = 193.21

σXY = 8250Suppose 40% of the portfolio is in Investment X and 60% is in

Investment Y:

The portfolio return is between the values for investments X and Y considered individually.

77)95()6(.)50(4.E(P)

04.133

8250)2(.4)(.6)((193.21))6(.(43.30)(.4)σ 2222P


Binomial Probability Distribution

A fixed number of observations, n e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse

Two mutually exclusive and collectively exhaustive categories e.g., head or tail in each toss of a coin; defective or not defective

light bulb Generally called “success” and “failure” Probability of success is p, probability of failure is 1 – p

Constant probability for each observation e.g., Probability of getting a tail is the same each time we toss

the coin


Binomial Probability Distribution(continued)

Observations are independent The outcome of one observation does not affect the

outcome of the other

Two sampling methods Infinite population without replacement Finite population with replacement


Possible Binomial Distribution Settings

A manufacturing plant labels items as either defective or acceptable

A firm bidding for contracts will either get a 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


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)

n = sample size (number of trials or observations)

p = probability of “success”

P(X)n

X ! n Xp (1-p)X n X!

( )!

Example: Flip a coin four times, let x = # heads:

n = 4

p = 0.5

1 - p = (1 - 0.5) = 0.5

X = 0, 1, 2, 3, 4

Binomial Distribution Formula


Example: Calculating a Binomial Probability

What is the probability of one success in five observations if the probability of success is .1?

X = 1, n = 5, and p = 0.1

0.32805

.9)(5)(0.1)(0

0.1)(1(0.1)1)!(51!

5!

p)(1pX)!(nX!

n!1)P(X

4

151

XnX


Binomial Distribution Characteristics

Mean

Variance and Standard Deviation

npE(x)μ

p)-np(1σ2

p)-np(1σ

Where n = sample size

p = probability of success

(1 – p) = probability of failure


Binomial Example

A doctor prescribes Nexium to 60% of his patients who have gastro-intestinal problems. What is the probability that out of the next 10 patients:

1. Six of them are given a prescription of Nexium

2. At least 6 are given a prescription of Nexium?


Sample size 10Probability of success 0.6

Binomial Probabilities TableX P(X) P(<=X) P(<X) P(>X) P(>=X)6 0.250822656 0.617719398 0.366896742 0.382280602 0.6331032587 0.214990848 0.832710246 0.617719398 0.167289754 0.3822806028 0.120932352 0.953642598 0.832710246 0.046357402 0.1672897549 0.040310784 0.993953382 0.953642598 0.006046618 0.046357402

10 0.006046618 1 0.993953382 0 0.006046618

Data


The Poisson Distribution

It is appropriate to use the Poisson distribution when:

You have an event that occurs randomly through time and space

You know the average number of successes you expect to observe over a given time frame or in a given space


Poisson Distribution Formula

where:

X = number of events in an area of opportunity

= expected number of events

e = base of the natural logarithm system (2.71828...)

!X

e)X(P

x


Poisson Example

On average there are three thread defects in a 10 yard bolt of fine wool fabric. What is the probability of finding no more than two thread defects in a randomly chosen 10 bolt lot?


Solution

Poisson Probabilities

DataAverage/Expected number of successes: 3

Poisson Probabilities TableX P(X) P(<=X) P(<X) P(>X) P(>=X)0 0.049787 0.049787 0.000000 0.950213 1.0000001 0.149361 0.199148 0.049787 0.800852 0.9502132 0.224042 0.423190 0.199148 0.576810 0.8008523 0.224042 0.647232 0.423190 0.352768 0.5768104 0.168031 0.815263 0.647232 0.184737 0.352768


Poisson Distribution Characteristics

Mean

Variance and Standard Deviation

λμ

λσ2

λσ

where = expected number of events


The Hypergeometric Distribution

“n” trials in a sample taken from a finite population of size N

Sample taken without replacement

Outcomes of trials are dependent

Concerned with finding the probability of “X” successes in the sample where there are “A” successes in the population


Hypergeometric Distribution Formula

n

N

Xn

AN

X

A

)X(P

WhereN = population sizeA = number of successes in the population

N – A = number of failures in the populationn = sample sizeX = number of successes in the sample

n – X = number of failures in the sample


Properties of the Hypergeometric Distribution

The mean of the hypergeometric distribution is

The standard deviation is

Where is called the “Finite Population Correction Factor” from sampling without replacement from a finite population

N

nAE(x)μ

1- N

n-N

N

A)-nA(Nσ

2

1- N

n-N


Using the Hypergeometric Distribution

■ Example: 3 different computers are checked from 10 in the department. 4 of the 10 computers have illegal software loaded. What is the probability that 2 of the 3 selected computers have illegal software loaded?

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

0.3120

(6)(6)

3

10

1

6

2

4

n

N

Xn

AN

X

A

2)P(X

The probability that 2 of the 3 selected computers have illegal software loaded is .30, or 30%.


Hypergeometric Distribution in PHStat

Complete dialog box entries and get output …

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

P(X = 2) = 0.3

(continued)


Chapter Summary

Addressed the probability of a discrete random variable

Discussed the Binomial distribution

Discussed the Poisson distribution

Discussed the Hypergeometric distribution

Documents

Business Statistics, A First Course (4e) © 2006 Prentice-Hall, Inc. Chap 5-1 Chapter 5 Some Important Discrete Probability Distributions Statistics for