1 1 slide 2009 university of minnesota-duluth, econ-2030 (dr. tadesse) chapter 5: probability...

1 1 Slide

Slide

2009 University of Minnesota-Duluth, Econ-2030 (Dr. Tadesse)

Chapter 5: Probability Distributions:

Discrete Probability Distributions

2 2 Slide

Slide


Learning Objectives

Identifying Types of Discrete Probability Distribution and their Respective Functional Representations

Calculating the Mean and Variance of a Discrete Random Variable with Each of the Different Discrete Probability Distribution.

3 3 Slide

Slide


Random Variable

4 4 Slide

Slide


Random Variables

Any variable that is used to represent the outcomes any experiment of interest to us is called Random Variable.

A random variable can assume (take) any value (Positive, negative, zero; finite, infinite; continuous and discrete values).

Depending upon the values they take, we can identify two types of random variables:

1. Discrete Variables2. Continuous Variables.

5 5 Slide

Slide


Both types of variables can assume either a finite number of values or an infinite sequence of values. Both types of variables can assume either a finite number of values or an infinite sequence of values.

6 6 Slide

Slide


Let x = number of TVs sold at a given store in one day. The number of TV units that can be sold in a given day is finite. It is also discrete: (0, 1, 2, 3, 4).

X can be considered as Discrete Random Variable

Let x = number of TVs sold at a given store in one day. The number of TV units that can be sold in a given day is finite. It is also discrete: (0, 1, 2, 3, 4).

X can be considered as Discrete Random Variable

Example: JSL Appliances

Discrete random variable with a finite number of values

7 7 Slide

Slide


Let Y = number of customers arriving in a store in one day. Y can take on the values 0, 1, 2, . . .

Let Y = number of customers arriving in a store in one day. Y can take on the values 0, 1, 2, . . .

Example: JSL Appliances

Discrete random variable with an infinite sequence of values

We can count the customers arriving. However, there is nofinite upper limit on the number that might arrive.

8 8 Slide

Slide


Random Variables

Question Random Variable x Type

Familysize

X = Number of dependents reported on tax return

Discrete

Distance fromhome to store

Y = Distance in miles from home to the store site

Continuous

Own dogor cat

Z = 1 if own no pet; = 2 if own dog(s) only; = 3 if own cat(s) only; = 4 if own dog(s) and cat(s)

Discrete

9 9 Slide

Slide


Probability Distributions

10 10 Slide

Slide


The probability distribution for a random variable is a distributionthat describes the values that the random variable of interest takes. The probability distribution for a random variable is a distributionthat describes the values that the random variable of interest takes.


The probability distribution is defined by a probability function, denoted by f(x)---the probability of the values of the random variable.

The probability distribution is defined by a probability function, denoted by f(x)---the probability of the values of the random variable.

11 11 Slide

Slide


For any probability function the following conditions must besatisfied:For any probability function the following conditions must besatisfied:


1. f(x) > 01. f(x) > 0

2. f(x) = 12. f(x) = 1

12 12 Slide

Slide



13 13 Slide

Slide


A Discrete Probability Distribution is a tabular, graphic orFunctional representation of a Random Variable with discrete outcomes that follows the principle of probability distribution

A Discrete Probability Distribution is a tabular, graphic orFunctional representation of a Random Variable with discrete outcomes that follows the principle of probability distribution


f(x) > 0f(x) > 0

f(x) = 1f(x) = 1

14 14 Slide

Slide


• Using past data on sales, a tabular representation of the probability distribution for TV sales was developed.

Number Units Sold of Days

0 80 1 50 2 40 3 10 4 20

200

x f(x) 0 .40 1 .25 2 .20 3 .05 4 .10 1.00

80/200

Discrete Probability Distributions--Example

16 16 Slide

Slide


1. Uniform Distribution

2. Binomial Distribution

3. Poisson Distribution

A Discrete Probability Distribution can assume one or more of the Following Distributions:

4. Hyper-Geometric Distribution

17 17 Slide

Slide


5.1) Discrete Uniform Probability Distribution

The probability distribution of a discrete probability distribution is given by the following formula. The probability distribution of a discrete probability distribution is given by the following formula.

f(x) = 1/nf(x) = 1/n

where:n = the number of values the random variable may assume

the values of the

random variable

are equally likely

18 18 Slide

Slide


where: f(x) = the probability of x successes in n trials n = the number of trials p = the probability of success on any one trial

( )!( ) (1 )

!( )!x n xn

f x p px n x

5.2) Binomial Distribution

The Probability distribution of a Binomial Distribution is given by the following function

19 19 Slide

Slide


The Probability of a Poisson Distribution is given by the following function

5.3) Poisson Distribution

f xex

x( )

!

where:f(x) = probability of x occurrences in an interval = mean number of occurrences in an interval e = 2.71828

20 20 Slide

Slide


The Probability of a Hyper-geometric Distribution is given by the following Function

5.4) Hypergeometric Distribution

n

N

xn

rN

x

r

xf )( for 0 < x < r

where: f(x) = probability of x successes in n trials n = number of trials N = number of elements in the population r = number of elements in the population

labeled success

21 21 Slide

Slide


5.1) Discrete Uniform Probability Distribution

The probability distribution of a discrete probability distribution is given by the following formula. The probability distribution of a discrete probability distribution is given by the following formula.

f(x) = 1/nf(x) = 1/n

where:n = the number of values the random variable may assume

the values of the

random variable

are equally likely

22 22 Slide

Slide


Expected Value (Mean) and Variance of Discrete Probability Distributions

23 23 Slide

Slide


Expected Value (Mean) for ……..Discrete Uniform Distribution

The expected value, or mean, of a random variable is a measure of its central location. The expected value, or mean, of a random variable is a measure of its central location.

E(x) = = xf(x)E(x) = = xf(x)

24 24 Slide

Slide


Variance and Standard Deviation of Discrete Uniform Probability Distribution

The variance summarizes the variability in the values of a random variable. The variance summarizes the variability in the values of a random variable.

The standard deviation, , is defined as the square root of the variance. The standard deviation, , is defined as the square root of the variance.

Var(x) = 2 = (x - )2f(x)Var(x) = 2 = (x - )2f(x)

25 25 Slide

Slide


• The likelihood of selling TV sets in any given day is considered equally likely (Uniform). The following table summarizes sales data on the past 200 days.

Number Units Sold of Days

0 80 1 50 2 40 3 10 4 20

200

x f(x) 0 .40 1 .25 2 .20 3 .05 4 .10

Example-TV Sales in a Given Store

Given this data what is the Average Number of TVs sold in a day?

x F(x) 0 .40 1 .65 2 .85 3 .90 4 1.00

26 26 Slide

Slide


Given the data, we can find the Expected Value (Mean Number) of TVs sold in a day as follows

expected number of TVs sold in a day

x f(x) xf(x) 0 .40 .00 1 .25 .25 2 .20 .40 3 .05 .15 4 .10 .40

E(x) = 1.20

Expected Value (MEAN) and Variance

E(x) = = xf(x)E(x) = = xf(x)

27 27 Slide

Slide


01234

-1.2-0.2 0.8 1.8 2.8

1.440.040.643.247.84

.40

.25

.20

.05

.10

.576

.010

.128

.162

.784

x - (x - )2 f(x) (x - )2f(x)x

TVs squaredStandard deviation of daily sales = 1.2884 TVs

Find the Variance and Standard Deviation of the Number of TV Sold in a given day.

Var(x) = 2 = (x - )2f(x)= 1.66Var(x) = 2 = (x - )2f(x)= 1.66

Var(x) = 2 = (x - )2f(x)Var(x) = 2 = (x - )2f(x)

28 28 Slide

Slide


5.2) The Binomial Distribution

29 29 Slide

Slide


5.2) Binomial Distribution

Four Properties of a Binomial Experiment

3. The probability of a success, denoted by p, does not change from trial to trial.3. The probability of a success, denoted by p, does not change from trial to trial.

4. The trials are independent.4. The trials are independent.

2. Only two outcomes, success and failure, are possible on each trial.2. Only two outcomes, success and failure, are possible on each trial.

1. The experiment consists of a sequence of n identical trials.1. The experiment consists of a sequence of n identical trials.

30 30 Slide

Slide



Typical Examples of a Binomial Experiment:

• Lottery: Win or Lose

• Election: A Candidate Wins or Loses• Gender of an Employee: is Male or Female• Flipping a coin: Heads or Tails

31 31 Slide

Slide



Our interest is in the number of successes occurring in the n trials. Our interest is in the number of successes occurring in the n trials.

let X denote the number of successes occurring in the n trials.let X denote the number of successes occurring in the n trials.

32 32 Slide

Slide


where:n = the number of trials p = the probability of success on any one trial f(x) = the probability of x successes in n trials

( )!( ) (1 )

!( )!x n xn

f x p px n x


Binomial Probability Function

33 33 Slide

Slide


( )!( ) (1 )

!( )!x n xn

f x p px n x


!!( )!

nx n x

( )(1 )x n xp p

Binomial Probability Function

Probability of a particular sequence of outcomes

with x successes

Probability of a particular sequence of outcomes

with x successes

Number of experimental outcomes providing exactly

x successes in n trials

Number of experimental outcomes providing exactly

x successes in n trials

34 34 Slide

Slide


Binomial Distribution

Example: Evans ElectronicsA local Electronics company is concerned about it low retention of employees. In recent years, management has seen an annual turnover of 10% in its hourly employees. Thus, for any hourly employee chosen at random, the company estimates that there is 0.1 probability that the person will leave the company in a year time.

35 35 Slide

Slide



Given the above information, if we randomly select 3 hourly employees, what is the probability that 1 of them will leave the company in one year ?

f xn

x n xp px n x( )

!!( )!

( )( )

1

1 23!(1) (0.1) (0.9) 3(.1)(.81) .243

1!(3 1)!f

Let: p = .10, n = 3, x = 1Solution:

37 37 Slide

Slide


Using Tables of Binomial Probabilities

n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50

3 0 .8574 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .12501 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

p


Page….383-388

38 38 Slide

Slide


Mean and Variance of A Binomial Distribution

(1 )np p

E(x) = = np

Var(x) = 2 = np(1 - p)

Expected Value

Variance

Standard Deviation

39 39 Slide

Slide


3(.1)(.9) .52 employees

E(x) = = 3(.1) = .3 employees out of 3

Var(x) = 2 = 3(.1)(.9) = .27

Expected Value

Variance

Standard Deviation

Given that p=0.1, for the 3 randomly selected hourly employees, what is the expected number and variance of workers who might leave the company this year?

40 40 Slide

Slide


5.3) Poisson Distribution

41 41 Slide

Slide


Poisson distribution refers to the probability distribution of a trial that involves cases of rare events that occur over a fixed time interval or within a specified region

6.4. Poisson Distribution

42 42 Slide

Slide


Examples….

• The number of errors a typist makes per page• The number of cars entering a service station per hour• The number of telephone calls received by a switchboard per

hour.

• The number of Bank Failures During a given Economic Recession.

• The number of housing foreclosures in a given city during a given year.

• The number of car accidents in one day on I-35 stretch from the Twin Cities to Duluth

6.4. Poisson Distribution

43 43 Slide

Slide


We use a Poisson distribution to estimate the number of occurrences of such discrete incidents over a specified interval of time or space

We use a Poisson distribution to estimate the number of occurrences of such discrete incidents over a specified interval of time or space

Thus a Poisson distributed random variable is discrete; Often times it assumes an infinite sequence of values (x = 0, 1, 2, . . . ).

Thus a Poisson distributed random variable is discrete; Often times it assumes an infinite sequence of values (x = 0, 1, 2, . . . ).

Poisson Distribution

44 44 Slide

Slide


The number of successes (events) that occur in a certain time interval is independent of the number of successes that occur in another time interval.

The probability of a success in a certain time interval is• the same for all time intervals of the same size,• proportional to the length of the interval.

The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller.

Properties of a Poisson Experiment

45 45 Slide

Slide


Poisson Probability Function


f xex

x( )

!

where:f(x) = probability of x occurrences in an interval = mean number of occurrences in an interval e = 2.71828

46 46 Slide

Slide


On Average 6 Patients per hour arrive at the emergency room of Mercy Hospital on

weekend evenings.

What is the probability of 4 arrivals in30 minutes on a weekend evening?

Poisson Distribution--Example

MERCY

47 47 Slide

Slide


Poisson Distribution-Example

Using the Poisson Probability Function

4 33 (2.71828)(4) .1680

4!f

MERCY

m= 6/hour = 3/half-hour, P(x = 4)?

f xex

x( )

!

48 48 Slide

Slide



Using Poisson Probability Tables

x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.00 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .04981 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .14942 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .22403 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .22404 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .16805 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .10086 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .05047 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .02168 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0081

MERCY

Page---390-395

50 50 Slide

Slide


Mean and Variance of Poisson Distribution

Another special property of the Poisson distribution is that the mean and variance are equal.

Another special property of the Poisson distribution is that the mean and variance are equal.

m = s 2

51 51 Slide

Slide


Poisson DistributionMERCY

Variance for Number of ArrivalsDuring 30-Minute Periods

m = s 2 = 3 m = s 2 = 3

52 52 Slide

Slide


5.4) Hyper-Geometric Distribution

53 53 Slide

Slide


Hyper-geometric Distribution

The hyper-geometric distribution is closely related to the binomial distribution. The hyper-geometric distribution is closely related to the binomial distribution.

However, for the hyper-geometric distribution: However, for the hyper-geometric distribution:

the trials are not independent, and the trials are not independent, and

the probability of success changes from trial to trial. the probability of success changes from trial to trial.

54 54 Slide

Slide


Hyper-geometric Probability Function


n

N

xn

rN

x

r

xf )( for 0 < x < r

where: f(x) = probability of x successes in n trials n = number of trials N = number of elements in the population r = number of elements labeled as success

55 55 Slide

Slide


Hyper-geometric Probability Function


( )

r N r

x n xf x

N

n

for 0 < x < r

number of waysn – x failures can be selectedfrom a total of N – r failures

in the populationnumber of ways

x successes can be selectedfrom a total of r successes

in the populationnumber of ways

a sample of size n can be selectedfrom a population of size N

56 56 Slide

Slide



( )r

E x nN

2( ) 11

r r N nVar x n

N N N

Mean

Variance

57 57 Slide

Slide


Hyper-geometric Distribution: Inspection

Electric fuses produced by a given company are packed in boxes. Each box carries 12 units of electric fuses. The role of the inspector in the company that manufactures the fuses is to make sure that all fuses in each box are in good condition.

Consider the following scenario: A worker inadvertently places 5 defective items in a box. As part of her job, the inspector randomly selects 3 fuses from the box that contains the defective fuses.

58 58 Slide

Slide



1. What is the probability that none of the three randomly selected fuses are defective?

2. What is the probability that the inspector finds only one of the three randomly selected fuses to be defective?

3. What is the probability that the inspector finds at least one of the three fuses defective?

59 59 Slide

Slide



1. What is the probability that none of the three randomly selected fuses are defective?

Solution:N=12; r=5; n=3; x=0;

P(x=0)?

1591.0220

35

!9!3!12!4!3!7

!5!0!5

)(

))(()0(

123

51203

50

xf

60 60 Slide

Slide



2. What is the probability that the inspector finds only one of the three randomly selected fuses to be defective?

Solution:N=12; r=5; n=3; x=1;

P(x=1)?

4773.0220

215

!9!3!12!5!2!7

!4!1!5

)(

))(()1(

123

51213

51

X

xf

61 61 Slide

Slide



2. What is the probability that the inspector finds at least one of the three fuses defective?

Solution:N=12; r=5; n=3; x 1

8409.0

1591.01

)0(1

?)1(

Xp

xf

62 62 Slide

Slide



Example: NevereadyBob has removed two dead batteries from a flashlight and inadvertently mingled them with the two good batteries that he intended to use as replacements. The four batteries look identical.

Then Bob randomly selects two of the four batteries. What is the probability that he selects the two good batteries?

ZAP

ZA

P

ZAPZAP

63 63 Slide

Slide



Using the Hyper-geometric Function

2 2 2! 2!

2 0 2!0! 0!2! 1( ) .167

4 4! 62 2!2!

r N r

x n xf x

N

n

N = 4 = number of batteries in total r = 2 = number of good batteries in total

x = 2 =number of good batteries to be selected n = 2 = number of batteries selected

167.06

1

!2!2

!4!2!0

!2

!0!2

!2

66 66 Slide

Slide



When the population size is large, a hyper-geometric distribution can be approximated by a binomial distribution with n trials and a probability of success p = (r/N).

When the population size is large, a hyper-geometric distribution can be approximated by a binomial distribution with n trials and a probability of success p = (r/N).

67 67 Slide

Slide



Consider a hyper-geometric distribution with n trials and let p = (r/n) denote the probability of a success on the first trial.

If the population size is large, the term (N – n)/(N – 1) approaches 1.

The expected value and variance can be written E(x) = np and Var(x) = np(1 – p).

Note that these are the expressions for the expected value and variance of a binomial distribution.

1 1 slide 2009 university of minnesota-duluth, econ-2030 (dr. tadesse) chapter 5: probability...

Documents

finite number of values

discrete values

tadesse random variable

number of customers

number of tvs

number of dependents

discrete variables

given store