Discrete Probability Distributions - Ramayah

35
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. ALWAYS LEARNING Slide 1 Discrete Probability Distributions Chapter 5

Transcript of Discrete Probability Distributions - Ramayah

Page 1: 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.

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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

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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

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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.