Discrete Probability Distributions - Ramayah
Transcript of Discrete Probability Distributions - Ramayah
Copyright © 2020 Pearson Education Ltd. All Rights Reserved.A L W A Y S L E A R N I N G Slide 1
Discrete Probability
Distributions
Chapter 5
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Objectives
In this chapter, you will learn:
◼ The properties of a probability distribution.
◼ How to calculate the expected value and varianceof a probability distribution.
◼ How to calculate probabilities from Binomial and Poisson distributions.
◼ How to use the binomial and Poisson distributions to solve business problems.
Definitions: Random Variables
Random Variables
Discrete
Random Variable
Continuous
Random VariableCh. 5
Chap 5-3
Ch. 6
Unpredictable quantity & it is determined
by the outcomes of an experiment.
produce outcomes
that come from a
counting process.
produce outcomes
that come from a
measurement.
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◼ Can only assume a countable number of values.
Examples:
◼ Roll a die twice
Let X be the number of times 4 occurs
(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).
Discrete Random Variables
Probability Distribution For
A Discrete Random Variable
▪ A mutually exclusive listing of all possible numerical
outcomes for that variable & a probability of
occurrence associated with each outcome.
Interruptions Per
Day In Computer
Network
Probability
0 0.35
1 0.25
2 0.20
3 0.10
4 0.05
5 0.05
Chap 5-5
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◼ Expected Value (or mean) of a discrete
variable (Weighted Average):
=
===N
i
ii xXPx1
)( E(X)
Interruptions Per Day In
Computer Network (xi)
Probability
P(X = xi) xiP(X = xi)
0 0.35 (0)(0.35) = 0.00
1 0.25 (1)(0.25) = 0.25
2 0.20 (2)(0.20) = 0.40
3 0.10 (3)(0.10) = 0.30
4 0.05 (4)(0.05) = 0.20
5 0.05 (5)(0.05) = 0.25
1.00 μ = E(X) = 1.40
Discrete Variables Expected Value (Measuring Center)
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◼ Variance of a discrete variable.
◼ Standard Deviation of a discrete variable.
where:
E(X) = Expected value of the discrete variable X
xi = the ith outcome of X
P(X=xi) = Probability of the ith occurrence of X
=
=−=N
1i
i
2
i )xP(XE(X)][xσ2
=
=−==N
1i
i
2
i )xP(XE(X)][xσσ 2
Discrete Variables Expected Value (Measuring Dispersion)
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Discrete Variables: Measuring Dispersion
(continued)
Interruptions Per
Day In Computer
Network (xi)
Probability
P(X = xi) [xi – E(X)]2 [xi – E(X)]2P(X = xi)
0 0.35 (0 – 1.4)2 = 1.96 (1.96)(0.35) = 0.686
1 0.25 (1 – 1.4)2 = 0.16 (0.16)(0.25) = 0.040
2 0.20 (2 – 1.4)2 = 0.36 (0.36)(0.20) = 0.072
3 0.10 (3 – 1.4)2 = 2.56 (2.56)(0.10) = 0.256
4 0.05 (4 – 1.4)2 = 6.76 (6.76)(0.05) = 0.338
5 0.05 (5 – 1.4)2 = 12.96 (12.96)(0.05) = 0.648
σ2 = 2.04, σ = 1.4283
=
=−=N
1i
i
2
i )xP(XE(X)][xσ
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Experiment: Toss 2 Coins. Let X = # heads.
X Value Probability
0 1/4 = 0.25
1 2/4 = 0.50
2 1/4 = 0.25
4 possible outcomes
T
T T
T H
Probability Distribution
0 1 2 X
0.50
0.25
0.00Pro
bab
ilit
y
H
H H
Chap 5-9
Example 2 of a Discrete Random
Variable Probability Distribution
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▪ Example 2: Toss 2 coins, X = number of heads,
compute expected value of X:
E(X) = (0)(0.25) + (1)(0.50) + (2)(0.25) = 1.0
N
Chap 5-10
μ =E(X ) = xiP(X = xi )i =1
X P(X=xi) xiP(X = xi)
0 0.25 (0)(0.25) = 0.00
1 0.50 (1)(0.50) = 0.50
2 0.25 (2)(0.25) = 0.50
1.00 μ = E(X) = 1.00
Discrete Random Variables Expected Value (Measuring Center)
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2σ =N
i i
i =1
[x -E(X )] P(X = x )
σ = (0−1)2(0.25)+ (1−1)2(0.50)+ (2−1)2(0.25) = 0.50 =0.707
Possible number of heads = 0, 1, or 2
Chap 5-11
▪ Example 2: Toss 2 coins, X = number of heads,
compute standard deviation (recall E(X) = 1)
X P(X=xi) [xi -E(X)]2 [xi -E(X)]2 P(X=xi)
0 0.25 (0-1)2 =1 0.25
1 0.50 (1-1)2 =0 0.00
2 0.25 (2-1)2 =1 0.25
Total 1.00 0.50
Discrete Random Variables Measuring Dispersion
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Exercise 1◼ You are trying to develop a strategy for investing in two
different stocks. The anticipated annual return for a $1,000
investment in each stock under four different economic
conditions has the following probability distribution:
a) Calculate the expected return for stock X and for stock Y.
b) Calculate the standard deviation for stock X and for stock
Y.
c) Would you invest in stock X or stock Y? Explain.
Probability Economic Condition
Returns
Stock X Stock Y
0.1 Recession -60 -130
0.2 Slow Growth 20 60
0.4 Moderate Growth 100 150
0.3 Fast Growth 160 200
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Continuous
Probability
Distributions
Binomial
Poisson
Probability
Distributions
Discrete
Probability
Distributions
Normal
Chap 5-13
Ch. 5 Ch. 6
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Introduction:
Binomial Probability Distribution
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◼ A fixed number of observations, n.◼ e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse.
◼ Each observation is classified into one of two mutually
exclusive & collectively exhaustive categories.
◼ e.g., head or tail in each toss of a coin; defective or not defective light
bulb.
◼ The probability of being classified as the event of
interest, π, is constant from observation to observation.◼ Probability of getting a tail is the same each time we toss the coin.
◼ Since the two categories are mutually exclusive and collectively exhaustive, when the probability of the event of interest is π, the probability of the event of interest not occurring is 1 – π.
◼ The value of any observation is independent of the value
of any other observation.
Binomial Probability Distribution
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◼ The probability of being classified as the event
of interest, π, is constant from observation to
observation.
◼ Probability of getting a tail is the same each time we toss the coin.
◼ Since the two categories are mutually exclusive and collectively exhaustive,
▪ P(event of interest) = π;
▪ P(event of interest not occurring) =1 - π
Binomial Probability Distribution
A manufacturing plant
labels items as either
defective or acceptable
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
A firm bidding for contracts will
either get a contract or not
Possible Applications for the Binomial Distribution
C
h
a
p
5
-
1
7
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P(X=x|n,π) = probability that X = x events of interest, given n and π
x = number of “events of interest” in sample,
(x = 0, 1, 2, ..., n)
n = sample size (number of trials
or observations)
π = probability of “event of interest”
1 – π = probability of not having an
event of interest
P(X=x |n,π)n
x! n xπ (1-π)
x n x!
( )!=
−
−
Example: Flip a coin four
times, let x = # heads:
n = 4
π = 0.5
1 - π = (1 - 0.5) = 0.5
X = 0, 1, 2, 3, 4
Binomial Distribution Formula
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Example: Binomial Distribution
Suppose you take a multiple choice test with 10
questions, and each question has 5 answer
choices (a, b, c, d, e), what is the probability
you get
a) exactly 4 questions correct just by guessing?
b) has less than 3 questions correct just by
guessing?
Chap 5-20
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0 1 2= 10 C (0.2)0 (0.8)10 + 10 C (0.2)1(0.8)9 + 10 C (0.2)2(0.8)8
The probability that the exactly 4 questions
correct just by guessing is 0.08808.
(b) P(X <3)=P(X =0)+P(X =1)+P(X =2)
=0.10737+0.26844+0.30199
=0.6778
The probability that has less than 3
questions correct just by guessing is 0.6778.
(a) P(X = 4) = 10C (0.2)4(1-0.2)6
4
Let X= answer correctly by guessing
X having a binomial distribution with
n=10, π=0.2, 1- π =0.8
= 0.08808
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0 1
0 1
2
P(X = 0) = 2C =
2 fairs dices are tossed together. If X represents the
number of times obtained even number faced up,
determine the probability distribution of X. Hence, plot
the graph of the distribution of X.
Let X = number of times obtained even number faced
up with n=2, π =3/6=1/2, 1-π=1-1/2=1/2.
X = {0,1,2}
1 1
1 1
1
1
2 2 4
1P(X =1) = 2C =
2 2 2
2 1
2 1
0
P(X = 2) = 2C1
4=
2 2
Example: Binomial Distribution Graph
Calculate the
probability
distribution of
the discrete
random variable.
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X 0 1 2
P(X=x) 14
12
14
Plot the graph of the binomial distribution.
The sum of all the probability in the binomial distribution is 1.
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Binomial Distribution Characteristics
where
n = sample size
π = probability of the event of interest for any trial
(1 – π) = probability of no event of interest for any trial
X ~Binomial(n,π)
1. Mean, μ = E(X ) = nπ
2. Variance, σ2 = nπ(1-π)
3. Standard deviation, σ = nπ(1- π)
Chap 5-24
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Example
Let X = number of times obtained even number
faced up with n=2, π =3/6=1/2, 1-π=1-1/2=1/2.
X = {0,1,2}
Mean, nπ
= 2(0.5)
= 1
Standard deviation, n(1- )
= 2×0.5×0.5
= 0.707
Chap 5-25
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Exercise 2
◼ You and two friends decided to go to McDonald’s which
recently filled approximately 92.2% of the orders correctly.
What is the probability that:
a) all three orders will be filled correctly?
b) none of the three will be filled correctly?
c) at least 2 of the three will be filled correctly?
d) What are the mean and standard deviation of the
binomial distribution used in (a) through (c)? Interpret
these values.
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▪ Poisson distribution is used when you areinterested in the number of times an event occursin a given area of opportunity.
▪ An area of opportunity is a continuous unit orinterval of time, volume, or such area in whichmore than one occurrence of an event can occur.
The number of mosquitobites on a person
The number of scratches in a car’s paint
The Poisson Distribution: Definitions
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◼ Apply the Poisson Distribution when:
◼ You are interested in counting the number of times a particular event occurs in a given area of opportunity. An area of opportunity is defined by time, length, surface area, and so forth.
◼ The probability that an event occurs in a given area of opportunity is the same for all the areas of opportunity.
◼ The number of events that occur in one area of opportunity is independent of the number of events that occur in any other area of opportunity.
◼ The probability that two or more events will occur in an area of opportunity approaches zero as the area of opportunity becomes smaller.
◼ The average number of events per unit is (lambda).
The Poisson Distribution
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X ~Poisson(λ)
1. Mean, μ= λ
2. Variance, σ2 = λ
3. Standard deviation, σ = λ
e− x
x!
Chap 5-29
P(X = x | )=
where:
x = number of events in an area of opportunity
= expected number of events
e = base of the natural logarithm system (2.71828...)
Poisson Distribution Formula & Characteristics
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Example: Poisson Distribution
e− x
x!P(X = x | ) =
Suppose that we can expect some independent
event to occur ‘’ times over a specified time interval.
The probability of exactly ‘x’ occurrences is equal to:
Suppose a fast-food restaurant can expect 2
customers every 3 minutes, on average.
What is the probability that four or fewer patrons
will enter the restaurant in a 9 minute period?
Chap 5-30
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P(X 4)=P(X =0)+P(X =1)+P(X =2)+P(X =3)+P(X =4)
+ + + +e-6(6)0 e-6(6)1 e-6(6)2 e-6(6)3 e-6(6)4
=0! 1! 2! 3! 4!
=0.0025+0.0149+0.0446+0.0892+0.1339
=0.2851
The probability that four or fewer patrons will enter
the restaurant in a 9 minute period is 0.2851.
Let X= number of customers enter the restaurant
Given 3mins = 2 customers
9mins = 2 x 3 = 6 customers
e-λλx
P(X | λ) =x!
Chap 5-31
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X P(X=x)
0
1
0.6065
0.3033
2 0.0758
3
4
5
6
7
0.0126
0.0016
0.0002
0.0000
0.0000P(X = 2 | =0.50) = 0.0758
Graphically:
= 0.50
Find P(X = 2 | = 0.50)
P(X = 2 | λ = 0.50)
e-λλx
=x!
e-0.50(0.50)2
=2!
= 0.0758
Example
Graph of Poisson Probabilities
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Exercise 3
◼ The average number of calls received by the front
office receptionist o JLL services is 10 per hour.
Find the probability of receiving
a) Two calls per hour.
b) No more than two calls in an hour?
c) Two calls in 30 minutes.
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Exercise 4
◼ A toll-free phone number is available from 9 am to 9 pm for
your customers to register complaints about a product
purchased from your company. Past history indicates that a
mean of 0.8 calls is received per minute.
a) What properties must be true about the situation described here in
order to use the Poisson distribution to calculate probabilities
concerning the number of phone calls received in a one-minute
period?
Assuming that this situation matches the properties discussed
in (a), what is the probability that during a one-minute period
a) zero phone calls will be received?
b) three or more phone calls will be received?
c) What is the maximum number of phone calls that will be received in a
one-minute period 99.9% of the time?
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Chapter Summary
In this chapter we covered:
◼ The properties of a probability distribution.
◼ Computing the expected value and variance of a probability distribution.
◼ Computing probabilities from binomial and Poisson distributions.
◼ Using the binomial and Poisson distributions to solve business problems.