discrete probability distribution data models
TRANSCRIPT
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Chapter 5
Some Important Discrete
Probability Distributions
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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 and
Poisson distributions
How to use the binomial and Poisson distributionsto solve business problems
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Introduction to ProbabilityDistributions
Random Variable
Represents a possible numerical value from
an uncertain event
Random
Variables
Discrete
Random Variable
Continuous
Random Variable
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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)
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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
0.25Probability
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Example 5.3: Number of Radios
Sold at Sound City in a Week
Business Statistics, A First Course (4e) 2008 Pearson Education Chap 5-6
Let x be the random variable of the number ofradios sold per week x has values x = 0, 1, 2, 3, 4, 5
Given: Frequency distribution of sales history
over past 100 weeks Let f be the number of weeks (of the past 100) during
which x number of radios were sold# Radios,x Frequency Relative Frequency
0 f(0) = 3 3/100 = 0.03
1 f(1) = 20 20/100 = 0.202 f(2) = 50 0.50
3 f(3) = 20 0.20
4 f(4) = 5 0.05
5 f(5) = 2 0.02
100 1.00
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Example 5.3 Continued
Interpret the relative frequencies as probabilities So for any valuex, f(x)/n =p(x)
Assuming that sales remain stable over time
Business Statistics, A First Course (4e) 2008 Pearson Education Chap 5-7
Number of Radios Sold at Sound City in a Week
Radios,x Probability,p(x)0 p(0) = 0.03
1 p(1) = 0.20
2 p(2) = 0.503 p(3) = 0.20
4 p(4) = 0.05
5 p(5) = 0.02
1.00
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Example 5.3 Continued
What is the chance that two radios will be
sold in a week?
p(x= 2) = 0.50
Business Statistics, A First Course (4e) 2008 Pearson Education Chap 5-8
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Example 5.3 Continued
What is the chance that fewer than 2 radios willbe sold in a week? p(x < 2) = p(x = 0 or x = 1)
= p(x = 0) + p(x = 1)= 0.03 + 0.20 = 0.23
What is the chance that three or more radioswill be sold in a week?
p(x 3) = p(x = 3, 4, or 5)= p(x = 3) + p(x = 4) + p(x = 5)= 0.20 + 0.05 + 0.02 = 0.27
Business Statistics, A First Course (4e) 2008 Pearson Education Chap 5-9
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Discrete Random VariableSummary Measures
Expected Value (or mean) of a discretedistribution (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
1i
ii )X(PXE(X)
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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 X
P(Xi) = Probability of the ith occurrence of X
Discrete Random VariableSummary Measures
N
1i
i
2
i
2 )P(XE(X)][X
(continued)
N
1ii
2
i
2 )P(XE(X)][X
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Example 5.3: Number of Radios
Sold at Sound City in a Week
Business Statistics, A First Course (4e) 2008 Pearson Education Chap 5-12
How many radios should be expected to be sold in aweek? Calculate the expected value of the number of radios sold, X
On average, expect to sell 2.1 radios per week
Radios,x Probability,p(x) x p(x)0 p(0) = 0.03 0 0.03 = 0.00
1 p(1) = 0.20 10.20 = 0.20
2 p(2) = 0.50 20.50 = 1.00
3 p(3) = 0.20 30.20 = 0.60
4 p(4) = 0.05 40.05 = 0.205 p(5) = 0.02 50.02 = 0.10
1.00 2.10
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Example 5.7: Number of Radios
Sold at Sound City in a Week
Business Statistics, A First Course (4e) 2008 Pearson Education Chap 5-13
Radios, x Probability,p(x) (x - X)2 p(x)
0 p(0) = 0.03 (02.1)2 (0.03) = 0.1323
1 p(1) = 0.20 (12.1)2 (0.20) = 0.2420
2 p(2) = 0.50 (2
2.1)2 (0.50) = 0.0050
3 p(3) = 0.20 (32.1)2 (0.20) = 0.1620
4 p(4) = 0.05 (42.1)2 (0.05) = 0.1805
5 p(5) = 0.02 (52.1)2 (0.02) = 0.1682
1.00 0.8900
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Example 5.7 Continued
Variance equals 0.8900
Standard deviation is the square root of the
variance Standard deviation equals 0.9434
Business Statistics, A First Course (4e) 2008 Pearson Education Chap 5-14
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Probability Distributions
Continuous
ProbabilityDistributions
Binomial
Poisson
Probability
Distributions
Discrete
ProbabilityDistributions
Normal
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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
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Binomial Probability Distribution(continued)
Observations are independent The outcome of one observation does not affect the
outcome of the other
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Possible Binomial DistributionSettings
A manufacturing plant labels items as
either defective or acceptable
A firm bidding for contracts will either get acontract 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
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Rule of Combinations
The number ofcombinations of selecting X
objects out of n objects is
X)!(nX!
n!Cxn
where:n! =(n)(n - 1)(n - 2) . . . (2)(1)
X! = (X)(X - 1)(X - 2) . . . (2)(1)
0! = 1 (by definition)
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P(X) = probability ofX successes in n trials,
with probability of success pon 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
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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
E l 5 10 I id f
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Example 5.10: Incidence of
Nausea after Treatment
Let x be the number of patients who will
experience nausea following treatment with
Phe-Mycin out of the 4 patients tested.
Ten percent of all patients treated with Phe-
Mycin.
Find the probability that 0 of the 4 patients
treated will experience nausea Given: n = 4, p = 0.1, with x = 0
Then: q = 1 p = 1 0.1 = 0.9
Business Statistics, A First Course (4e) 2008 Pearson Education Chap 5-24
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Example 5.10 Continued
Business Statistics, A First Course (4e) 2008 Pearson Education Chap 5-25
.
%61.65exp
6561.09.01.01
9.01.0!04!0
!40
40
40
samplepossibleallof
innauseaeriencewouldpaitientssampledfourThe
xp
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Chap 5-26
Example 5.11: Incidence of Nausea
after Treatment
x = number of patients who will experience nausea following
treatment with Phe-Mycin out of the 4 patients tested
Find the probability that at least 3 of the 4 patients treated
will experience nausea
Setx = 3, n = 4,p = 0.1, so q = 1p = 10.1 = 0.9
Then:
0037.00001.0036.0
43
4or33
xpxp
xpxp
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Binomial DistributionCharacteristics
Mean
Variance and Standard Deviation
npE(x)
p)-np(12
p)-np(1 Where n = sample size
p = probability of success
(1 p) = probability of failure
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Back to Example 5.11
Chap 5-28
Of 4 randomly selected patients, how many
should be expected to experience nausea after
treatment?
Given: n = 4, p = 0.1
Then X = n p = 4 0.1 = 0.4
So expect 0.4 of the 4 patients to experience
nausea.
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Case Analysis
Case :1- Givenp = 0.5 and n = 5, P(X= 5) =
0.0312.
Case:2- Givenp = .6 and n = 5,(a) P(X= 5) = 0.0778
(b) P(X 3) = 0.6826
(c) P( X< 2) = 0.0870
(d) (a) P(X= 5) = 0.3277
(b) P(X 3) = 0.9421
(c) P( X< 2) = 0.0067
Business Statistics, A First Course (4e) 2008 Pearson Education Chap 5-29
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Case 4- Givenp = 0.90 and n = 3,
(a) P(X= 3) = = = 0.729
(b) P(X= 0) = = = 0.001(c) P(X 2) = P(X= 2) + P(X= 3) = + = 0.972
(d) E(X) = np = = 2.7
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The Poisson Distribution
Binomial
Poisson
Probability
Distributions
Discrete
ProbabilityDistributions
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The Poisson Distribution
Apply the Poisson Distribution when:
You wish to count the number of times an eventoccurs in a given area of opportunity
The probability that an event occurs in one area ofopportunity is the same for all areas of opportunity
The number of events that occur in one area ofopportunity is independent of the number of eventsthat occur in the other areas of opportunity
The average number of events per unit is (mu)
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Arrivals (e.g., customers, defects, accidents) must
be independent of each other.
Some examples of Poisson models in which
assumptions are sufficiently met are:
X= number of customers arriving at a bank
ATM in a given minute.
X= number of file server virus infections at
a data center during a 24-hour period.
Poisson Distribution
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Poisson Distribution Formula
where:
X = number of events in an area of opportunity
= expected number of eventse = base of the natural logarithm system (2.71828...)
!
)(
X
eXP
x
P i Di ib i
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Poisson DistributionCharacteristics
Mean
Variance and Standard Deviation
x
2
where = expected number of events
E l 5 13 ATC C t
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Example 5.13: ATC Center
Errors
An air traffic control (ATC) center has beenaveraging 20.8 errors per year and lately hasbeen making 3 errors per week
Let x be the number of errors made by the ATCcenter during one week
Given: = 20.8 errors per year
Then: = 0.4 errors per week There are 52 weeks per year so
for a week is:
= (20.8 errors/year)/(52 weeks/year)= 0.4 errors/week
Example 5 13: ATC Center
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Example 5.13: ATC Center
Errors Continued
Find the probability that 3 errors (x =3) will
occur in a week
Want p(x = 3) when = 0.4
Find the probability that no errors (x = 0) will
occur in a week Want p(x = 0) when = 0.4
0072.0!34.03
34.0
exp
6703.0!0
4.00
04.0
e
xp
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