Some Important Discrete Probability Distributions: Powerpoint To Accompany

3/08/2016
Chapter 5
Some
important
discrete
probability
distributions
PowerPoint to accompany:
Cover illustration: Raw Pixel/Shutterstock.com
Slide 1
Copyright 2016 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781486018956 / Berenson / Basic Business Statistics 4/E
Learning Objectives
After studying this Chapter you should have a

better understanding of:
The properties of a probability distribution
How to calculate the expected value and variance of a
probability distribution
How to calculate the probabilities from binomial
distribution
How to use the binomial distribution to solve business
problems
2 Copyright 2013 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781442549272/Berenson/Business Statistics /2e Slide 2
Introduction to Probability Distributions
Random variable
Represents a possible numerical value from an uncertain
event
Random
Variables
Ch. 5 Ch. 6
Discrete Continuous
Random Variable Random Variable
Slide 3
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Discrete Random Variables
Can only assume a countable number of values
Examples:
Roll a dice twice. Let X be the number of times 4
comes up; thus X could be 0, 1, or 2 times
Toss a coin five times. Let X be the number of

heads; thus X could = 0, 1, 2, 3, 4, or 5
Slide 4
Discrete Probability Distribution
Experiment: Toss 2 Coins. Let X = # heads
4 possible outcomes Probability Distribution
T T X Probability
0 1/4 = 0.25
T H 1 2/4 = 0.50
2 1/4 = 0.25
H T
Probability
0.50
0.25
H H
0 1 2 X
Slide 5
Discrete Random Variable Summary Measures
Expected value (or mean) of a discrete random variable

(weighted average)
N
= E(X) = X P( X )
i =1
i i
Toss 2 coins, X = # of heads, X P(X)

calculate expected value of X:
0 0.25
1 0.50
E(X) = (0 x 0.25) + (1 x 0.50) + (2 x 0.25) 2 0.25
= 1.0
Slide 6
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(continued)
Variance of a discrete random variable definition formula

N
2 = [X
i =1
i E(X)] 2 P(X i )
Variance of a discrete random variable alternative calculation

formula
N
= X P ( X ) E(X)
2 2 2
i i
i =1
where: E(X) = expected value of the discrete random variable X

Xi = the ith outcome of the discrete random variable X
P(Xi ) = probability of the ith occurrence of X
Slide 7
Example:
Toss two coins, X = # heads, calculate variance and standard
deviation
Recall from slide 6 that E(X) = 1
N
2 = i =1
2
X i P ( X i ) E(X) 2
2 = (02 * 0.25 + 12 * 0.5 + 2 2 * 0.25) (12 ) = 0.5
Standard deviation = 2 = 0 .5 = 0 .707
Slide 8
Applied Activity
Given the following distribution:
X P(X)
0 0.50
1 0.20
2 0.15
3 0.10
4 0.05
a. Calculate the expected value.

b. Calculate the standard deviation
Copyright 2013 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781442548473/Berenson/Basic Business Statistics/3e Slide 9
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The Binomial Probability Distribution
Probability
Distributions
Discrete
Probability
Distributions
Binomial
Poisson
Hypergeometric
Slide 10

4 essential properties of the binomial distribution
1 A fixed number of observations, or trials, n

e.g. 15 tosses of a coin; 10 light bulbs taken
from a warehouse
2 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 1p
Slide 11
3 Constant probability for each observation

e.g. probability of getting a tail is the same each
time we toss the coin
4 Observations are independent

the outcome of one observation does not affect
the outcome of the other
two sampling methods can be used to ensure
independence; either:
- selected from infinite population without
replacement; or
- selected from finite population with replacement
Slide 12
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Possible Binomial Scenarios
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
A new job applicant either accepts the offer or

rejects it
Slide 13
Rule of Combinations
The number of combinations of selecting X objects out of n

objects is:
n n!
= nC x =
X X! (n X)!
where:
n! =(n)(n - 1)(n - 2) . . . (2)(1)
X! = (X)(X - 1)(X - 2) . . . (2)(1)
0! = 1 (by definition)
Slide 14
The Binomial Distribution Formula
n!
P( X ) = p X (1 p)n X
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)
n = sample size (number of trials or observations)
p = probability of success
1-p = probability of failure
Slide 15
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Binomial Probability
Example:
A customer has a 50% probability of making a

purchase. Five customers enter the shop. What is the
probability of one customer making a purchase?
n!
Let x = # customer
P(X = 1) = p X (1 p) n X
X! (n X)!
purchases:
5!
= (0.5) 1 (1 0.5) 51
n=5 1! (5 1)!
p = 0.5
1 - p = (1 - 0.5) = 0.5 = (5)(0.5)(0 .5) 4
X=1
= 0.1562
Slide 16
Shape of the Binomial Distribution

The shape of the binomial distribution depends on the values
of p and n
P(X) n = 5 p = 0.1
.6
.4
Here, n = 5 and p = 0.1 .2
0 X
0 1 2 3 4 5
P(X) n = 5 p = 0.5
.6
Here, n = 5 and p = 0.5 .4
.2
0 X
0 1 2 3 4 5
Slide 17
Characteristics of the Binomial Distribution
Mean
= E(x) = np
Variance and standard deviation
2 = np(1 - p)
= np(1 - p)
Where: n = sample size
p = probability of success
(1 p) = probability of failure
Slide 18
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Using the Binomial Table

n = 10
x p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50

0 0.1074 0.0563 0.0282 0.0135 0.0060 0.0025 0.0010 10
1 0.2684 0.1877 0.1211 0.0725 0.0403 0.0207 0.0098 9
2 0.3020 0.2816 0.2335 0.1757 0.1209 0.0763 0.0439 8
3 0.2013 0.2503 0.2668 0.2522 0.2150 0.1665 0.1172 7
4 0.0881 0.1460 0.2001 0.2377 0.2508 0.2384 0.2051 6
5 0.0264 0.0584 0.1029 0.1536 0.2007 0.2340 0.2461 5
6 0.0055 0.0162 0.0368 0.0689 0.1115 0.1596 0.2051 4
7 0.0008 0.0031 0.0090 0.0212 0.0425 0.0746 0.1172 3
8 0.0001 0.0004 0.0014 0.0043 0.0106 0.0229 0.0439 2
9 0.0000 0.0000 0.0001 0.0005 0.0016 0.0042 0.0098 1
10 0.0000 0.0000 0.0000 0.0000 0.0001 0.0003 0.0010 0
p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 x
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
Slide 19
A real estate agent knows (based on past

experience) that she sells 10% of houses within
the first week. She has five new houses to sell.
Find the probability that she will sell

exactly two houses within the week?
at least one house within the week?
Applied Activity
For n = 10 and p = 0.60, what is P(X= 7)?
For n = 5 and p = 0.80, what is P(X = 3)?
Suppose that you and two friends go to a McDonalds franchise,

which last month filled 90% of orders correctly.
1. What is the probability that:
all three orders will be filled correctly?
none of the three will be filled correctly?
at least two of the three will be filled correctly?
2. What is the mean and standard deviation of the number of
orders filled correctly?
7
3/08/2016
Chapter Summary
Addressed the probability of a discrete random variable

Discussed three important discrete probability
distributions
Calculation of probabilities for the Binomial distribution
22 Copyright 2013 Pearson Australia (a division of Pearson Australia Group Pty Ltd) 9781442549272/Berenson/Business Statistics /2e Slide 22
End of Chapter
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3/08/2016

Cover illustration: Raw Pixel/Shutterstock.com

After studying this Chapter you should have a

Introduction to Probability Distributions

Discrete Random Variables

Can only assume a countable number of values

Toss a coin five times. Let X be the number of

Discrete Probability Distribution

Experiment: Toss 2 Coins. Let X = # heads

4 possible outcomes Probability Distribution

Discrete Random Variable Summary Measures

Expected value (or mean) of a discrete random variable

Toss 2 coins, X = # of heads, X P(X)

Discrete Random Variable Summary Measures

Variance of a discrete random variable definition formula

Variance of a discrete random variable alternative calculation

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

Discrete Random Variable Summary Measures

2 = (02 * 0.25 + 12 * 0.5 + 2 2 * 0.25) (12 ) = 0.5

Standard deviation = 2 = 0 .5 = 0 .707

Given the following distribution:

a. Calculate the expected value.

The Binomial Probability Distribution

The Binomial Probability Distribution

1 A fixed number of observations, or trials, n

2 Two mutually exclusive and collectively

The Binomial Probability Distribution

3 Constant probability for each observation

4 Observations are independent

Possible Binomial Scenarios

A manufacturing plant labels items as either

A firm bidding for contracts will either get a contract

A marketing research firm receives survey responses

A new job applicant either accepts the offer or

The number of combinations of selecting X objects out of n

The Binomial Distribution Formula

P(X) = probability of X successes in n trials,

A customer has a 50% probability of making a

Shape of the Binomial Distribution

Characteristics of the Binomial Distribution

Using the Binomial Table

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

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

A real estate agent knows (based on past

Find the probability that she will sell

Suppose that you and two friends go to a McDonalds franchise,

Addressed the probability of a discrete random variable

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