Probability Distributions for Discrete Variables

PROBABILITY DISTRIBUTIONS

Part I: Probability Distributions for Discrete Variables

1

What is a Probability Distribution?

• Precisely, a probability distribution is • a relative frequency distribution of all • possible outcomes of an experiment

Under $20,000

$20,000 to <

$40,000

$40,000 to <

$60,000

$60,000 to <

$80,000

$80,000 to <

$100,000

Above $100,000

0

0.05

0.1

0.15

0.2

0.25

0.3

Income

Prob

abili

ty P

(x)

Income Relative frequency P(x)

Under $20,000 0.2380

$20,000 to < $40,000 0.2607

$40,000 to < $60,000 0.1941

$60,000 to < $80,000 0.1408

$80,000 to < $100,000 0.0907

Above $100,000 0.0757

1

Results = Positive (PO)

Results = Negative (NE) Total

Woman is pregnant (Y) 0.85 0.05 0.9Woman is not pregnant (N) 0.03 0.07 0.1Total 0.88 0.12 1

(a)

(b)Y & PO Y & NE N & PO N & NE

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Joint Events (i & j)

Prob

abili

ty P

(I &

j)

Examples of Probability Distribution

2

Different types of Probability Distribution

Probability distributions of Continuous Variables

Probability distributions of Discrete Variables

x

P(x)

80 80.5 81 81.5 82.0 82.5

Examples:

- Uniform distribution - Normal distribution

(b)

x

P(x)

0 1 2 3 4

Examples:

- Binomial distribution- Poisson distribution

(a)3

Random variable:

A variable is a characteristic that varies from one component of a population to another. It becomes a random variable when its values vary randomly or by chance

Discrete random variable:A variable that can take only integer values (e.g. 0, 1, and 2)

Continuous random variable:A variable that can take any value in a specified interval falling within its plausible range (e.g. 5, 5.6, 8.8, 10, and 12.4)

4

Working problem 5.1:

Decide whether the following random variables x are discrete or continuous. Explain your answer

1. Human height2. Human weight3. Temperature4. Number of graduating students5. Number of student in a stat class6. House area in square feet7. The number of eggs that a hen lays in a day8. The number of emergency room patients in one day in a hospital

5

The probability distribution of discrete variables

It is a relative frequency distribution, which lists each possible value of a discrete random variable and its associated probability. It should also satisfy the criteria:

6

Example: Does the Table below describe a probability distribution?

The probability distribution of discrete variables

probabilities p(x) corresponding to a random variable x

x P(x)0 0.151 0.22 0.053 0.254 0.255 0.36 0.1

Solution:•The probability values here are less than one, which satisfies the condition:

• However, the sum of probabilities

which is not equal one.

Because the sum requirement is not satisfied, we conclude that Table 5.1 does not describe a probability distribution. 7

Example: In an experiment in which two coins are tossed, what are the possible outcomes? If the outcome of interest is ‘head’, what are the probabilities of zero head, one head, and two heads? Construct a probability distribution illustrating this experiment.

Solution:Outcomes resulting from tossing two coins

Trial number Possible outcomes x = number of heads P(x)1 T, T 0 1/42 T, H 1 1/43 H, T 1 1/44 H, H 2 1/4

x = number of heads

P(x)

0 1/41 1/22 1/4

Probability values associated with the number of heads

8


(a) Does P(x) = x/6, where x can be 0, 1, or 5 determine a probability distribution? Explain why?

(b) Does P(x) = x/5, where x can be 0 or 1 , 2, or 3 determine a probability distribution? Explain why?

(c) Does P(x) = x/3, where x can be 0 or 1 , or 2 determine a probability distribution? Explain why?

Working problem 5.4: In a survey conducted by a Real-Estate broker, houses in a certain city were classified by the number of bathrooms available. The results of this survey are shown below. The first column represents the number of bathrooms, and the second column represents the number of houses. Construct a probability distribution representing this survey.

Number of bathrooms per house

Number of houses

1 4001.5 840

2 3302.5 920

3 3603.5 420

10

The mean and standard deviation of the probability distribution of discrete variables

The mean of a discrete random variable x (also called the expected value of x) is expressed by:

The standard deviation of a discrete variable x is expressed by:

11

Example: For the data of the Table below, calculate the mean of the discrete variable representing the number of bedrooms per house.

The mean and standard deviation of the probability distribution of discrete variables

Calculations of mean of a discrete variable

Bedrooms per house Number of houses P(x) xP(x)1 200 0.1 0.12 310 0.155 0.313 330 0.165 0.4954 600 0.3 1.25 450 0.225 1.1256 110 0.055 0.33

20001 3.56

12

Example: For the data of the Table below, calculate the standard deviation of the discrete variable representing the number of bedrooms per house.

Number of rooms/house

Number of houses

P(x) xP(x) x2 x2P(x)1 200 0.1 0.1 1 0.12 310 0.155 0.31 4 0.623 330 0.165 0.495 9 1.4854 600 0.3 1.2 16 4.85 450 0.225 1.125 25 5.6256 110 0.055 0.33 36 1.98

2000 1 3.56

14.61

13

Working Problem: 5.5

In a survey conducted by a real-estate broker, houses in a certain city were classified by the number of full bathrooms available. The results of this survey are shown in the table below. If a house is picked from this area randomly:

- What is the chance that it will have 3 bathrooms?

- What is the chance that it will have less than 3 bathrooms?

Frequency of houses by the number of full bathrooms

Number of full bathrooms per house Number of houses

1 45

2 410

3 320

4 38

5 12

14

Working Problem: 5.6

In a survey conducted by a real-estate broker, houses in a certain city were classified by the number of full bathrooms available. The results of this survey are shown in the table below.

- Calculate the mean number of bathrooms per house

- Calculate the standard deviation of the number of bathrooms per house.

Frequency of houses by the number of full bathrooms

Number of full bathrooms per house Number of houses

1 45

2 410

3 320

4 38

5 12

15


A person plans to invest a $100,000. An investment consultant offered this person five investment packages the return of each will depend on whether the economy next year will be fair or good. The table below lists the anticipated return on each investment.

According to the investment consultant, next year’s economy has a 45% chance of being good and a 55% chance of being fair.

a. Determine the probability distribution of each random variable M, S, C, and Rb. Determine the expected value of each random variable.c. Which investment has the best expected payoff? Which has the worst?

16

The binomial probability distribution

OrParameters of theBinomial Distribution, n & P

The mean and variance of a binomial distribution are:

17

Basic characteristics of the binomial distribution are as follows:

•The binomial distribution is the most appropriate probability distribution for sampling from an infinitely large population, where p represents the fraction of defective or nonconforming items in the population, x represents the number of nonconforming items found in a random sample of n items. Examples of variables that are best approximated by a binomial distribution include success and failure in business, passing or failing a course, product defects, rejected items, and head or tail in tossing a coin.

•In a binomial experiment, there are n identical trials. In other words, the given experiment is repeated n times, where n is a positive integer.

•Each repetition of a binomial experiment is called a trial or a Bernoulli trial (after Jacob Bernoulli).

•Each trial has two and only two outcomes (success or failure, pass or fail, head or tail, etc).

•The probability of success is denoted by p and that of failure by q or (1-p).

•Binomial trials are independent. This means that the outcome of one trial does not affect the outcome of the next trial or any other trial.

18

The shape of the binomial distribution

19

Example: There are four flights daily from Atlantic City Airport, New Jersey, to Atlanta Airport, Georgia, via Air Tran Airline. Suppose the probability that any flight arrives on time is 0.70.

• Construct a binomial distribution of the random variable “number of on-time arrivals”. • What is the probability that none of the flights will arrive on time tomorrow? • What is the probability that exactly two of the flights will be on time?

Solution:If we are willing to assume that the variable arrival time follows a binomial distribution, we can construct this distribution using the binomial equation, and the two parameters n and p. In symbols, the x variable is described as x ~ b(n, p) or x ~ b(4, 0.7).

x= 0,1,2, …

x P(x)0 0.008101 0.075602 0.264603 0.411604 0.24010

20

x P(x)0 0.008101 0.075602 0.264603 0.411604 0.24010

0 0 0 0 00.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Binomial distribution (n = 4, p = 0.7)

Number of flights on time (x)

P(x

)

Binomial Distributions of flight on-time arrivals

The probability that none of the flights will arrive on time tomorrow = P(x= 0) = 0.0081

The probability that exactly two of the flights will be on time = P(x=2) = 0.26460 (about 26% chance that exactly two flights will be on time)

21

Example: Using the results of the previous Example, what is the expected value of arrival on time, and what is the extent of variation (variance)?

Solution:

n = 4, p = 0.7

22


There are three flights daily from Charlotte, NC , to New York, NY (JFK), via U.S. Air . Suppose the probability that any flight arrives late is 0.10.

- Construct a binomial probability distribution

- What is the expected value and the variance?

- What is the probability that none of the flights are late today?

-What is the probability that exactly one of the flights is late today?

- What is the probability that exactly two of the flights are late today?

23


For the experiment of tossing a coin 6 times, construct a binomial probability distribution.

- What is the expected value and the variance?

- What is the probability that 2 heads will turn up?

-What is the probability that five heads will turn up?

- What is the probability that no heads will turn up?

24

Working problem 5.10: A manufacturer of electromagnetic combined clutch-brake units used for automobiles finds that on average 1.2% of the units fails at high-speed accelerated test. This finding represents historical value based on routine inspection of 100 units daily. If we let x be the random variable representing the number of failing brake units,

- Construct a binomial distribution of this variable- Determine the expected value and the variance of failure of brakes

25

Working problem 5.11: A manufacturer of electromagnetic combined clutch-brake units used for automobiles finds that on average 1.2% of the units fails at high-speed accelerated test. This finding represents historical value based on routine inspection of 100 units daily. If we let x be the random variable representing the number of failing brake units:

- What is the probability that exactly one brake unit will fail? - What is the probability that less than two brake units will fail?- What is the probability that more than 4 brake units will fail?

26

Working Problem 5.12:

There are five bus trips from JFK Airport to Newark Airport. Suppose the probability that any bus arrives late is 0.20.

- Construct a binomial distribution describing the random event of bus late arrivals.

-Calculate the mean and the variance

27

Working Problem 5.13:

There are five bus trips from JFK Airport to Newark Airport. Suppose the probability that any bus arrives late is 0.20. What is the probability that none of the bus trips will be late today? What is the probability that one of the bus trips will be late today?

28

Example: You enter the class to find out that the teacher is given a previously unannounced quiz of 10 multiple-choice questions, each with one correct answer out of possible 4 answer choices (A, B, C, and D). Since you were not prepared for the quiz, you decided to make a random guess. • Does this situation reflect a binomial experiment? If the answer is ‘yes’, formulate the probability function• What are the values of the two parameters, n and p of the binomial distribution? • What is your chance to answer exactly 3 questions correctly?

Solution:This experiment fully satisfies the binomial model. The reasons are:• A fixed number of trials = n = 10• The trials are independent (the answer of one question is independent of the answer of another question) • Each trial is associated with two outcomes, correct or wrong• Each trial is associated with fixed probability p = 1/4, since the chance of a correct answer is one out of four. Thus, the probability remains constant

n = 10, p = 0.25

Your chance to answer exactly 3 questions correctly:

29

Example: In the previous example, find the probabilities of answering one, two, three,…., or ten questions correctly. Construct the Binomial Distribution.

Binomial distribution of multiple-choice quiz

x P(x) Cumulative probability0 0.05631 0.056311 0.18771 0.244032 0.28157 0.525593 0.25028 0.775884 0.14600 0.921875 0.05840 0.980276 0.01622 0.996497 0.00309 0.999588 0.00039 0.999979 0.00003 1.00000

10 0.00000 1.00000

0 1 2 3 4 5 6 7 8 9 100.00

0.05

0.10

0.15

0.20

0.25

0.30

x

P(x

)

Binomial Distributions at p = 0.25 and n = 10

2.500 expected value1.875 variance1.369 standard deviation

30

30+ 60+ 70+

Working problem 5.14

Based on a worldwide mortality statistics, there is about 60% chance that a person aged 30 will be alive at age 70. Suppose that five people aged 30 are selected at random.

- Find the probability that the number alive at age 70 will be exactly two - Find the probability that the number alive at age 70 will be at most two - Find the probability that the number alive at age 70 will be at least three - Find the probability that the number alive at age 70 will be none

- Determine the probability distribution of the number alive at age 70

- Determine the expected value and the variance of the number alive at age 70

31


A charity for deaf and blind offered 1000 tickets each of $5 to collect money. Only the first 4 draws win prices. The prices for the first 4 tickets are $200, $150, $100, and $50. If you buy one ticket, what would be your expected value of gain?


In a multiple-choice exam of 20 questions, 5 possible answers were assigned for each question and only one answer is correct. If a student is answering these questions randomly, would this represent a binomial experiment? If the answer is ‘yes’, formulate the probability functionWhat are the values of the two parameters, n and p of the binomial distribution? Calculate the expected value and the variance.

- What is your chance to answer exactly 2 questions correctly?

- What is your chance to answer less than 3 questions correctly? - What is your chance to answer more than 15 questions correctly?

32


In a multiple-choice exam of 100 questions, 4 possible answers were assigned for each question and only one answer is correct. If a student is answering these questions randomly, how many questions would he/she be expected to answer?

Working problem 5.18:In a multiple-choice exam of 60 questions, 4 possible answers were assigned for each question and only one answer is correct. A student made a serious attempt at the first 40 questions. This student typically answer multiple-choice tests with a rate of success of about 80%. How many questions will this student be expected to answer correctly out of the first 40 questions?

If the same student then attempted to answer the remaining 20 questions randomly. Would you consider his attempts as a binomial experiment? If the answer is ‘yes”, what is the expected number of questions that this student will answer correctly out of these 20 questions? Estimate the total number of questions that this student and predict his letter grade?

Note: G ≥ 90 = A80 ≤ G < 90 = B70 ≤ G < 80 = C60 ≤ G < 70 = D

33

Example: If 15% of the bolts produced by a machine are defective, what is the probability that out of 3 bolts selected at random (a) 0, (b) 1, (c) less than 2, bolts will be defective?

Using Microsoft Excel® to Perform Binomial Distribution Analysis

Solution:

P = 0.15, 1- P = 0.85

0 1 2 30.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

Binomial distribution (n = 3, p = 0.15)

X

P(X

)

X P(X)0 0.614131 0.325132 0.057383 0.00338

34

1

23

4

Using Microsoft Excel® to Perform Binomial Distribution Analysis

• Go to Excel Spreadsheet• Click fx on the button bar• Select Statistical from

the “Or select a category” drop down list box• Select BINODIST from

the “Select a function” list • Click OK

P = 0.15, q = 0.85

Example:

?)2(?)1(?)0(

xPxPxP

35

• To answer Part (a) of this Example, type 0 in the Number_s window, and 3 in the Trials Window, 0.15 in the Probability_s window, and False in the Cumulative Window. • You should be able to see the value of the P(x = 0) as illustrated by the ellipse in the Figure, or you can Click Ok and the value will be presented

P = 0.15, q = 0.85Example: ?)2()(?)1()(?)0()( xPcxPbxPa

5

6

614.085.015.003

)0( 30

xP

Note:Under Cumulative, we use “FALSE” for getting the exact probability value (i.e. P(x=0) or P(x=1)), or “TRUE” for getting cumulative probability

36

939.0325.0614.0)1()0()2(

xPxPxP

325.085.015.013

)1( 21

xP

Note:Under Cumulative, we use “FALSE” for getting the exact probability value (i.e. P(x=0) or P(x=1)), or “TRUE” for getting cumulative probability

37

Working problem 5.19 (Use Excel to solve this problem):

A manufacturer of electromagnetic combined clutch-brake units used for automobiles finds that on average 1.2% of the units fails at high-speed accelerated test. This finding represents historical value based on routine inspection of 100 units daily. If we let x be the random variable representing the number of failing brake units:

- What is the probability that exactly one brake unit will fail?

- What is the probability that less than two brake units will fail?

- What is the probability that more than 4 brake units will fail?

38


You enter the class to find out that the teacher is given a previously unannounced quiz of eight multiple-choice questions, each with one correct answer out of possible 4 answer choices (A, B, C, and D). Since you were not prepared for the quiz, you decided to make a random guess.

- Use Excel function to determine your chance to answer exactly 3 questions correctly?

- Use Excel function to determine your chance to answer exactly 2 questions correctly?

- Use Excel function to determine your chance to answer no questions correctly?

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What is a Poisson Probability Distribution?

• A probability distribution used for modeling discrete variables with the concern being with the number of events occurring per a given time or space interval.

• The interval may represent any duration of time (e.g. per hour, per shift, per week, etc.) or any specified region (e.g. per unit area, per box, per shipment, etc.).

• In most applications, the Poisson distribution is often used to model the events of largely unlikely failure or rare product defects.

• The probability of an event described by a Poisson distribution is given by:

,....3,2,1,0!

P(x)x

xxe

where x is the number of target events, P(x) is the probability of x events in a given time or space interval, l is the expected average number of events per interval, and e = 2.7182818 is the base of the natural logarithm system.

• The mean and variance of the Poisson distribution are given by:

np

np

2

40

168.01234

0498.081!4

)3(!

4)P(x34

exex

Example: The mean number of accidents on the job per month at a certain company is 3. Model this variable by a Poisson distribution and determine the probability that in any given month four accidents will occur at this company?

3

32

The mean and variance of this distribution are:

Poisson Probability Distribution-Applications

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 160.00

0.05

0.10

0.15

0.20

0.25

Poisson distribution (l= 3)

X

P(X

)

0216.0!7

)3(!

)7(7

...............................

224.0!2

)3(!

)2(2

1494.0!1

)3(!

)1(1

0498.0!0

)3(!

)0(0

37

32

31

30

exePxat

exePxat

exePxat

exePxat

x

x

x

x

X P(X)0 0.049791 0.149362 0.224043 0.224044 0.168035 0.100826 0.050417 0.021608 0.008109 0.00270

10 0.0008111 0.0002212 0.0000613 0.0000114 0.0000015 0.0000016 0.0000017 0.00000

41

Example: A fast-food restaurant was trying to extend breakfast service from 10 am to 10:30 am to allow more customers, particularly those who wake up late, to be served breakfast. From past experience, the restaurant found that on average, the number of people showing up for breakfast service from 10:00 am to 10:30 am is 7.5 people. What advice would you give this restaurant?

To assist you in this task,

• The restaurant will allow the service to be extended if they can maintain more than 5 customers coming for breakfast service in this period.

- What is the chance that the restaurant will maintain more than 5 customers during the added period?

- What is the chance that the restaurant will maintain more than 10 customers during the added period?


7585.0]1094.00729.00389.00156.000414.0000553.0[1

]!5

5.7!4

5.7!3

5.7!2

5.7!1

5.7!0

5.7[1

)]5()4()3()1()0([1!

-15)P(x15)P(x

....)8()7()6(!

5)P(x

5.755.745.735.725.715.70

5

0

6

eeeeee

xPxPxPxPxPxe

or

xPxPxPxe

x

x

= 7.5

42

Example (Cont’d):


7585.05)P(x = 7.5

There is more than 75% chance that the restaurant will maintain more than 5 customers within the added half hour breakfast service.

The chance that the restaurant will maintain more than 10 customers during the added period:

138.00.862241)01P(x1)01P(x

0 2 4 6 8 10 12 14 16 18 20 22 240.000.020.040.060.080.100.120.140.16

Poisson distribution (l= 7.5)

X

P(X

)

There is a 75% chance the restaurant will maintain more than 5 customers and 14% chance it may even have more than 10 customers. 43

Example: Suppose that a new knitting machine used by a mill only fails to operate one time every 10 hours. This leads to a failure rate, , of 0.1failure per hour. What is the probability of two failures occurring randomly during the 10 hour period?


0 1 2 3 4 50.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00 Poisson distribution (l = 0.1)

X

P(X

)X P(X)

Cumulative probability

0 0.90484 0.904841 0.09048 0.995322 0.00452 0.999853 0.00015 1.000004 0.00000 1.000005 0.00000 1.00000

0045.0!2

1.0!

2)P(x1.02

e

xex

44

Working problem 5.21: The mean number of on-site accidental deaths per a period of five years in an electric-service company is 1.0. Assuming that a Poisson distribution can best represent this random variable, determine the probability that in any given five-year period, three on-site accidental deaths will occur? Construct the Poison distribution describing this variable. What is the mean and variance of this distribution?

Working problem 5.22: The mean number of car accidents in Peachtree street in Atlanta, Georgia is 3 per month. Assuming that a Poisson distribution can best represent this random variable, determine the probability that in any given month, 5 or more car accidents will occur? Construct the Poison distribution describing this variable. What is the mean and variance of this distribution?

45

Working Problem 5.23: Assume on-time flight per week is a rare situation for a particular airline. Suppose on average, this number is about 2.Construct a Poisson distribution of this random variable.

-What is the expected value and variance of the number of on-time flights per week? - What is the chance that the number of on-time flights per week will exceed 5?

46

Example: Suppose that historically a machine producing bolts yields a rate of 10% defective bolts. Find the probability that in a sample of 10 bolts chosen at random exactly two will be defective? Calculate this probability using Binomial and Poisson modeling of the variable.

Poisson & Binomial Probability Distribution-Applications

Using Binomial distribution:n = 10, p = 0.1

1937.0)90.0(10.0210

)2( 82

xnxqpxn

xP

0 1 2 3 4 5 6 7 8 9 100.000.050.100.150.200.250.300.350.400.45 Binomial distribution (n = 10, p = 0.1)

X

P(X

)

Using Poisson distribution:

11.010 np

1839.0!2

1!

)2(12

e

xxP ex

0 1 2 3 4 5 6 7 8 9 100.000.050.100.150.200.250.300.350.40

Poisson distribution (l = 1)

X

P(X

)

47

Using Microsoft Excel to Perform Poisson Distribution Analysis

Example: If the number of some type of metal defects is assumed to follow a Poisson distribution with a mean value 1 per 1000 square yard. • What is the probability of finding 2 defects?• What is the probability of finding at most 2 defects?• What is the probability of finding at least one defect?

184.0!2

1)2(!

)2(0

12

x xep

xXP e

Manual Calculations:

920.0184.0368.0368.0

!21

!11

!01)2()1()0(

!)2(

0

121110

x x eeeppp

xXP e

632.0368.01)0(1)1(1)1( xpxpxp

48

1

2

3

4

Using Microsoft Excel to Perform Poisson Distribution Analysis ( =1)

• Go to Excel Spreadsheet• Click fx on the button bar• Select Statistical from the “Or

select a category” drop down list box

• Select POISSON from the “Select a function” list

• Click OK

49

5

6184.0!2

1)2(!

)2(0

12

x xep

xXP e

• Type 2 at the X window, 1 in Mean window, and False in the Cumulative Window.

• You should be able to see the value of the P(x = 2) as illustrated by the circle in Figure 5.16. or you can Click Ok and the value will be presented

Using Microsoft Excel to Perform Poisson Distribution Analysis ( =1)

50

x x

eeex

XP e0

121110

!21

!11

!01

!)2(

632.0368.01)0(1)1(1)1(

xpxpxp

51

Example: Suppose that a manufacturer typically finds 2% defects in hard drive units produced for personal computers. Generate a Binomial distribution of nonconforming drives using a sample size of 400 drives.

Using Microsoft Excel to Perform Discrete Probability Distribution Simulation

n = 400, and p = 0.02

Binomial DistributionUSING EXCEL®

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1

2

3

Data Data Analysis Random Number Generation

53

4

5

Select one variable say 1000 random numbersSelect Binomial n = 400, p =0.02 select an output area

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6

1000 random numberseach representing number of defects in 400 inspected Samples

7

Perform Descriptive Statistics

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Average Number of Defects ≈ 8• This leads to average probability= 8/400 = 0.02 (the p value)• The Standard Deviation = 2.737• This is approximately equal tothe theoretical :

8.2)98.0)(02.0)(400( npq

Note

56

8

9

10

Perform Histogram

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Perform Histogram

58

To convert a histogram to a probabilityDistribution:1. Form a column labeled p(x) 2. Calculate p(x) for each value of xBy dividing the frequency of each x byThe total frequency (1000). 3. You can then make a graph usingExcel graph features

Note

Convert Histogram into a Probability Distribution

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Example: If a company producing notebook computers historically finds an average of 1.2 defects when inspecting random samples of constant size of 60 disks. Using Excel Spreadsheet, simulate the probability distribution of defects.

Using Microsoft Excel to Perform Discrete Probability Distribution Simulation

= 1.2

Poisson DistributionUSING EXCEL®

60

2

1

3

1000 random numberseach representing number of defects per trial - Microsoft Excel data

analysis- Random Number Generator- Generate, say 1000 numbers - Select Poisson, and = 1.2. - Press OK

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Perform Descriptive Statistics and Histogram

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Convert Histogram into a Probability Distribution

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Probability Distributions for Discrete Variables

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Transcript of Probability Distributions for Discrete Variables