copyright © 2011 pearson education, inc. probability models for counts chapter 11

Copyright © 2011 Pearson Education, Inc.

Probability Models for Counts

Chapter 11

11.1 Random Variables for Counts

How many doctors should management expect a pharmaceutical rep to meet in a day if only 40% of visits reach a doctor? Is a rep who meets 8 or more doctors in a day doing exceptionally well?

Need a discrete random variable to model counts and provide a method for finding probabilities


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Bernoulli Random Variable

Bernoulli trials are random events with three characteristics:

Two possible outcomes (success, failure) Fixed probability of success (p) Independence


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Bernoulli Random Variable - Definition

A random variable B with two possible values, 1 = success and 0 = failure, as determined in a Bernoulli trial.

E(B) = pVar(B) = p(1-p)


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Counting Successes (Binomial)

Y, the sum of iid Bernoulli random variables, is a binomial random variable

Y = number of success in n Bernoulli trials (each trial with probability of success = p)

Defined by two parameters: n and p


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Binomial Distribution

Number of ‘successes’ in a sample of n observations (trials)

• Number of reds in 15 spins of roulette wheel

• Number of defective items in a batch of 5 items

• Number correct on a 33 question exam

• Number of customers who purchase out of 100 customers who enter store


Counting Successes (Binomial)

We can define the number of doctors seen by a pharmaceutical rep in 10 visits as a binomial random variable

This random variable, Y, is defined by n = 10 visits and p = 0.40 (40% success in reaching a doctor)


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11.2 Binomial Model

Assumptions

Using a binomial random variable to describe a real phenomenon

10% Condition: if trials are selected at random, it is OK to ignore dependence caused by sampling from a finite population if the selected trials make up less than 10% of the population


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11.3 Properties of Binomial Random Variables

Mean and Variance

E(Y) = np

Var(Y) = np(1 - p)


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Binomial Distribution Characteristics

.0

.5

1.0

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

n = 5 p = 0.1

n = 5 p = 0.5

( )E x n p Mean

Standard Deviation

(1 )n p p


Pharmaceutical Rep Example

E(Y) = np = (10)(0.40) = 4We expect a rep to see 4 doctors in 10 visits.

Var(Y) = np(1 - p) = (1)(0.40)(0.60) = 2.4SD(Y) = 1.55A rep who has seen 8 doctors has performed 2.6 standard deviations above the mean.


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Binomial Probabilities

Consist of two parts: The probability of a specific sequence of

Bernoulli trials with y success in n attempts The number of sequences that have y

successes in n attempts (binomial coefficient)


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Binomial Probabilities

Binomial probability for y success in n trials


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ynyyn ppCyYP 1

Binomial Probability Distribution Function

!( ) (1 )

! ( )!x n x x n xn n

p x p q p px x n x

p(x) = Probability of x ‘Successes’

n = Sample Size

p = Probability of ‘Success’

x = Number of ‘Successes’ in Sample

(x = 0, 1, 2, ..., n)

Binomial Probability Distribution Example

3 5 3

!( ) (1 )

!( )!

5!(3) .5 (1 .5)

3!(5 3)!

.3125

x n xnp x p p

x n x

p

Experiment: Toss 1 coin 5 times in a row. Note number of tails.

What’s the probability of 3 tails?

© 1984-1994 T/Maker Co.



P(Y = 8) = 10C8(0.4)8(0.6)2 = 0.011

The probability of seeing 8 doctors in 10 visits is only about 1%.


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Probability Distribution for Rep Example


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P(Y ≥ 8)= P(Y = 8) + P(Y = 9) + P(Y = 10)= 0.01062 + 0.00157 + 0.00010 = 0.01229

The probability of seeing 8 or more doctors in 10 visits is only slightly above 1%. This rep is doing exceptionally well!


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4M Example 11.1: FOCUS ON SALES

Motivation

A focus group with nine randomly chosen participants was shown a prototype of a new product and asked if they would buy it at a price of $99.95. Six of them said yes. The development team claimed that 80% of customers would buy the new product at that price. If the claim is correct, what results would we expect from the focus group?


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Method

Use the binomial model for this situation. Each focus group member has two possible responses: yes, no. We can use X ~ Bi(n = 9, p = 0.8) to represent the number of yes responses out of nine.


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Mechanics – Find E(X) and SD(X)

E(X) = np = (9)(0.8) = 7.2 Var(X) = np(1-p) = (9)(0.8)(0.2) = 1.44SD(X) = 1.2

The expected number is higher than the observed number of 6.


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Mechanics – Probability Distribution

While 6 is not the most likely outcome, it is still common.


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Message

The results of the focus group are in line with what we would expect to see if the development team’s claim is correct.


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11.4 Poisson Model

A Poisson Random Variable

Describes the number of events determined by a random process during an interval of time or space

Is not finite (possible values are infinite) Is defined by λ (lambda), the rate of events


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Poisson Process

1. Constant event probability

• Average of 60/hr is1/min for 60 1-minuteintervals

2. One event per interval• Don’t arrive together

3. Independent events• Arrival of 1 person does

not affect another’sarrival

© 1984-1994 T/Maker Co.

11.4 Poisson Model

The Poisson Probability Distribution

E(X) = λ

Var(X) = λ


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...,2,1,0!

xx

exXPx

Poisson Probability Distribution Function

p(x) = Probability of x given

= Expected (mean) number of ‘successes’

e = 2.71828 (base of natural logarithm)

x = Number of ‘successes’ per unit

p xx

( )!

x

e

-

Poisson Distribution Characteristics

.0

.2

.4

.6

.8

0 1 2 3 4 5

X

P(X)

.0

.1

.2

.3

X

P(X)

= 0.5

= 6

Mean

Standard Deviation

1

( )

( )N

i

E x

x p x

補上平均數和變異數的證明過程


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Binomial Distribution Thinking Challenge

You’re a telemarketer selling service contracts for Macy’s. You’ve sold 20 in your last 100 calls (p = .20). If you call 12 people tonight, what’s the probability ofA. No sales?B. Exactly 2 sales?C. At most 2 sales? D. At least 2 sales?

Binomial Distribution Solution*

n = 12, p = .20A. p(0) = .0687 B. p(2) = .2835C. p(at most 2) = p(0) + p(1) + p(2)

= .0687 + .2062 + .2835= .5584

D. p(at least 2) = p(2) + p(3)...+ p(12)= 1 – [p(0) + p(1)] = 1 – .0687 – .2062= .7251

11.4 Poisson Model

The Poisson Model

Uses a Poisson random variable to describe counts of data

Is appropriate for situations like• The number of calls arriving at the help desk in

a 10-minute interval• The number of imperfections per square meter

of glass panel


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4M Example 11.2: DEFECTS IN SEMICONDUCTORS

Motivation

A supplier claims that its wafers have 1 defect per 400 cm2. Each wafer is 20 cm in diameter, so the area is 314 cm2. What is the mean number of defects and the standard deviation?


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Method

The random variable is the number of defects on a randomly selected wafer. The Poisson model applies.


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Mechanics – Find λ

The assumed defect rate is 1 per 400 cm2.

Since a wafer has an area of 314 cm2, λ = 314/400 = 0.785E(X) = 0.785SD(X) = 0.886P(X = 0) = 0.456


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Poisson Distribution Example

Customers arrive at a rate of 72 per hour. What is the probability of 4 customers arriving in 3 minutes?

© 1995 Corel Corp.

Poisson Distribution Solution

72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval

-

4 -3.6

( )!

3.6(4) .1912

4!

x ep x

x

ep

Poisson Probability Table (Portion) x 0 … 3 4 … 9

.02 .980 … : : : : : : :

3.4 .033 … .558 .744 … .997 3.6 .027 … .515 .706 … .996 3.8 .022 … .473 .668 … .994 : : : : : : :

Cumulative Probabilities

p(x ≤ 4) – p(x ≤ 3) = .706 – .515 = .191

Thinking Challenge

You work in Quality Assurance for an investment firm. A clerk enters 75 words per minute with 6 errors per hour. What is the probability of 0 errors in a 255-word bond transaction?

© 1984-1994 T/Maker Co.

Poisson Distribution Solution: Finding *

75 words/min = (75 words/min)(60 min/hr) = 4500 words/hr

6 errors/hr = 6 errors/4500 words= .00133 errors/word

In a 255-word transaction (interval): = (.00133 errors/word )(255 words)

= .34 errors/255-word transaction

Poisson Distribution Solution: Finding p(0)*

-

0 -.34

( )!

.34(0) .7118

0!

x ep x

x

ep


Message

The chip maker can expect about 0.8 defects per wafer. About 46% of the wafers will be defect free.


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Best Practices

Ensure that you have Bernoulli trials if you are going to use the binomial model.

Use the binomial model to simplify the analysis of counts.

Use the Poisson model when the count accumulates during an interval.


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Best Practices (Continued)

Check the assumptions of a model.

Use a Poisson model to simplify counts of rare events.


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Pitfalls

Do not presume independence without checking.

Do not assume stable conditions routinely.


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copyright © 2011 pearson education, inc. probability models for counts chapter 11

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