Statistics for business decision making and analysis robert stine and foster chapter 11

Chapter 11

Probability Models for
Counts

Copyright © 2011 Pearson Education, Inc.


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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11.1 Random Variables for Counts
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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11.1 Random Variables for Counts
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) = p
Var(B) = p(1-p)
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11.1 Random Variables for Counts
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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11.1 Random Variables for Counts
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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11.3 Properties of Binomial Random
Variables
Pharmaceutical Rep Example
E(Y) = np = (10)(0.40) = 4
We expect a rep to see 4 doctors in 10 visits.
Var(Y) = np(1 - p) = (1)(0.40)(0.60) = 2.4
SD(Y) = 1.55
A rep who has seen 8 doctors has performed
2.6 standard deviations above the mean.
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11.3 Properties of Binomial Random
Variables
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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11.3 Properties of Binomial Random
Variables
Binomial Probabilities
Binomial probability for y success in n trials

P ( Y = y ) = n C y p (1 − p )
y

n− y

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11.3 Properties of Binomial Random
Variables
Pharmaceutical Rep Example
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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11.3 Properties of Binomial Random
Variables
Probability Distribution for Rep Example

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11.3 Properties of Binomial Random
Variables
Pharmaceutical Rep Example
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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4M Example 11.1: FOCUS ON SALES
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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4M Example 11.1: FOCUS ON SALES

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) = 1.44
SD(X) = 1.2
The expected number is higher than the
observed number of 6.

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4M Example 11.1: FOCUS ON SALES
Mechanics – Probability Distribution

While 6 is not the most likely outcome,
it is still common.
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4M Example 11.1: FOCUS ON SALES
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
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


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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11.4 Poisson Model
The Poisson Probability Distribution

P( X = x ) = e

−λ

λ
x!
x

x = 0, 1, 2, ...

E(X) = λ
Var(X) = λ
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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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4M Example 11.2: DEFECTS IN
SEMICONDUCTORS
Method
The random variable is the number of
defects on a randomly selected wafer.
The Poisson model applies.

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