discrete probability distribution data models

7/29/2019 Discrete Probability Distribution data models

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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


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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What is the chance that two radios will be

sold in a week?

p(x= 2) = 0.50



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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



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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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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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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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Variance equals 0.8900

Standard deviation is the square root of the

variance Standard deviation equals 0.9434



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Probability Distributions

Continuous

ProbabilityDistributions

Binomial

Poisson

Probability

Distributions

Discrete


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



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.

%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

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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



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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



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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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discrete probability distribution data models

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