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# Chapter 3-PART I PowerPoint PPT Presentation

Chapter 3-PART I. QUANTITATIVE METHODS FOR BUSINESS DISCREETE PROBABILITY DISTRIBUTION. John Kooti. Chapter 3, Part I Discrete Probability Distributions. Random Variables Discrete Random Variables Expected Value and Variance Binomial Probability Distribution

Chapter 3-PART I

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Chapter 3-PART I

QUANTITATIVE

METHODS FOR

DISCREETE PROBABILITY DISTRIBUTION

John Kooti

### Chapter 3, Part I Discrete Probability Distributions

• Random Variables

• Discrete Random Variables

• Expected Value and Variance

• Binomial Probability Distribution

• Poisson Probability Distribution

### Random Variables

• A random variableis a numerical description of the outcome of an experiment.

• A random variable can be classified as being either discrete or continuous depending on the numerical values it assumes.

• A discrete random variablemay assume either a finite number of values or an infinite sequence of values.

• A continuous random variablemay assume any numerical value in an interval or collection of intervals.

### Example: JSL Appliances

• Discrete random variable with a finite number of values:

Let x = number of TV sets sold at the store in one day

where x can take on 5 values (0, 1, 2, 3, 4)

• Discrete random variable with an infinite sequence of values:

Let x = number of customers arriving in one day

where x can take on the values 0, 1, 2, . . .

We can count the customers arriving, but there is no finite upper limit on the number that might arrive.

### Discrete Probability Distributions

• The probability distributionfor a random variable describes how probabilities are distributed over the values of the random variable.

• The probability distribution is defined by a probabilityfunction, denoted by f(x), which provides the probability for each value of the random variable.

• The required conditions for a discrete probability function are:

f(x) > 0

Sf(x) = 1

• We can describe a discrete probability distribution with a table, graph, or equation.

### Example: JSL Appliances

• Using past data on TV sales (below left), a tabular representation of the probability distribution for TV sales (below right) was developed.

Number

Units Soldof Daysxf(x)

0 80 0 .40

1 50 1 .25

2 40 2 .20

3 10 3 .05

420 4 .10

200 1.00

### Example: JSL Appliances

• A graphical representation of the probability distribution for TV sales in one day

.50

.40

Probability

.30

.20

.10

0 1 2 3 4

Values of Random Variable x (TV sales)

### Expected Value and Variance

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

• Expected value of a discrete random variable:

E(x) = m = Sxf(x)

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

• Variance of a discrete random variable:

Var(x) = s2 = S(x - m)2f(x)

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

### Example: JSL Appliances

• Expected Value of a Discrete Random Variable

xf(x)xf(x)

0.40 .00

1.25 .25

2.20 .40

3.05 .15

4.10 .40

1.20 = E(x)

The expected number of TV sets sold in a day is 1.2

### Example: JSL Appliances

• Variance and Standard Deviation

of a Discrete Random Variable

xx - m (x - m)2f(x)(x - m)2f(x)

_____ _________ ___________ _______ _______________

0-1.2 1.44.40 .576

1-0.2 0.04.25 .010

2 0.8 0.64.20 .128

3 1.8 3.24.05 .162

4 2.8 7.84.10 .784

1.660 = s 2

The variance of daily sales is 1.66 TV sets squared.

The standard deviation of sales is 1.29 TV sets.

### Example: JSL Appliances

• Formula Spreadsheet for Computing Expected Value and Variance

### Binomial Probability Distribution

• Properties of a Binomial Experiment

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

• Two outcomes, success and failure, are possible on each trial.

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

• The trials are independent.

### Example: Evans Electronics

• Binomial Probability Distribution

Evans is concerned about a low retention rate for employees. On the basis of past experience, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employees chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year.

Choosing 3 hourly employees a random, what is the probability that 1 of them will leave the company this year?

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

### Binomial Probability Distribution

• Binomial Probability Function

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

### Example: Evans Electronics

• Using the Binomial Probability Function

= (3)(0.1)(0.81)

= .243

### Example: Evans Electronics

• Using the Tables of Binomial Probabilities

### Example: Evans Electronics

• Using a Tree Diagram

Second

Worker

Third

Worker

Value

of x

First

Worker

Probab.

L (.1)

3

.0010

Leaves (.1)

2

.0090

S (.9)

Leaves (.1)

L (.1)

.0090

2

Stays (.9)

.0810

1

S (.9)

L (.1)

.0090

2

Leaves (.1)

1

.0810

S (.9)

L (.1)

Stays (.9)

1

.0810

Stays (.9)

.7290

0

S (.9)

### Example: Evans Electronics

• Step 1: Select a cell in the worksheet where you

want the binomial probabilities to appear.

• Step 2: Select the Insert pull-down menu.

• Step 3: Choose the Function option.

• Step 4: When the Paste Function dialog box appears:

Choose Statistical from the FunctionCategory box.

Choose BINOMDIST from the FunctionName box.

Select OK.

continued

### Example: Evans Electronics

• Using an Excel Spreadsheet (continued)

• Step 5: When the BINOMDIST dialog box appears:

Enter 1 in the Number_s box (value of x).

Enter 3 in the Trials box (value of n).

Enter .1 in the Probability_s box (value of p).

Enter false in the Cumulative box.

[Note: At this point the desired binomial probability of .243 is automatically computed and appears in the right center of the dialog box.]

Select OK (and .243 will appear in the worksheet cell requested in Step 1).

### The Binomial Probability Distribution

• Expected Value

E(x) = m = np

• Variance

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

• Standard Deviation

• Example: Evans Electronics

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

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

### MINITAB APPLICATION

• Minitab Program

### Poisson Probability Distribution

• Properties of a Poisson Experiment

• The probability of an occurrence is the same for any two intervals of equal length.

• The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval.

### Poisson Probability Distribution

• Poisson Probability Function

where

f(x) = probability of x occurrences in an interval

m = mean number of occurrences in an interval

e = 2.71828

### Example: Mercy Hospital

Patients arrive at the emergency room of Mercy

Hospital at the average rate of 6 per hour on weekend

evenings. What is the probability of 4 arrivals in 30

minutes on a weekend evening?

• Using the Poisson Probability Function

m = 6/hour = 3/half-hour, x = 4

### Example: Mercy Hospital

• Using the Tables of Poisson Probabilities

### Poisson Probability Distribution

The Poisson probability distribution can be used as

an approximation of the binomial probability

distribution when p, the probability of success, is small

and n, the number of trials, is large.

• Approximation is good when p< .05 and n> 20

• Set m = np and use the Poisson tables.

### MINIAB APPLICATION

• Minitab Program

### Chapter 3, Part II Continuous Random Variables

• Continuous Random Variables

• Normal Probability Distribution

• Exponential Probability Distribution

### Continuous Probability Distributions

• A continuous random variablecan assume any value in an interval on the real line or in a collection of intervals.

• It is not possible to talk about the probability of the random variable assuming a particular value.

• Instead, we talk about the probability of the random variable assuming a value within a given interval.

• The probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be the area under the graphof the probability density functionbetween x1and x2.

### Uniform Probability Distribution

A random variable is uniformly distributedwhenever

the probability is proportional to the length of the

interval.

• Uniform Probability Density Function

f(x) = 1/(b - a) for a<x<b

= 0 elsewhere

• Expected Value of x

E(x) = (a + b)/2

• Variance of x

Var(x) = (b - a)2/12

where: a = smallest value the variable can assume

b = largest value the variable can assume

### Example: Slater's Buffet

Slater customers are charged for the amount of salad

they take. Sampling suggests that the amount of salad

taken is uniformly distributed between 5 ounces and 15

ounces.

• Probability Density Function

f(x) = 1/10 for 5 <x< 15

= 0 elsewhere

where

x = salad plate filling weight

### Example: Slater's Buffet

What is the probability that a customer will take

between 12 and 15 ounces of salad?

f(x)

P(12 <x< 15) = 1/10(3) = .3

1/10

x

5

15

10

12

### MINITAB APPLICATION

• Minitab Program

### The Normal Probability Distribution

• Graph of the Normal Probability Density Function

f(x)

x

m

### Normal Probability Distribution

• The Normal Curve

• The shape of the normal curve is often illustrated as a bell-shaped curve.

• The highest point on the normal curve is at the mean, which is also the median and mode of the distribution.

• The normal curve is symmetric.

• The standard deviation determines the width of the curve.

• The total area under the curve is 1.

• Probabilities for the normal random variable are given by areas under the curve.

### Normal Probability Distribution

• Normal Probability Density Function

where

m = mean

s = standard deviation

p = 3.14159

e = 2.71828

### Standard Normal Probability Distribution

• A random variable that has a normal distribution with a mean of zero and a standard deviation of one is said to have a standard normal probability distribution.

• The letter z is commonly used to designate this normal random variable.

• Converting to the Standard Normal Distribution

• We can think of z as a measure of the number of standard deviations x is from m.

### Example: Pep Zone

Pep Zone sells auto parts and supplies including a

popular multi-grade motor oil. When the stock of this

oil drops to 20 gallons, a replenishment order is placed.

The store manager is concerned that sales are being

lost due to stockouts while waiting for an order. It has

been determined that leadtime demand is normally

distributed with a mean of 15 gallons and a standard

deviation of 6 gallons.

The manager would like to know the probability of a

stockout, P(x > 20).

### Example: Pep Zone

• Standard Normal

Distribution

z = (x - m)/s

= (20 - 15)/6

= .83

The Standard Normal table shows an area of .2967 for the region between the z = 0 line and the z = .83 line above. The shaded tail area is .5 - .2967 = .2033. The probability of a stockout is .2033.

Area = .2967

Area = .2033

Area = .5

z

0

.83

### Example: Pep Zone

• Using the Standard Normal Probability Table

### Example: Pep Zone

• Step 1: Select a cell in the worksheet where you want the normal probability to appear.

• Step 2: Select the Insert pull-down menu.

• Step 3: Choose the Function option.

• Step 4: When the Paste Function dialog box appears: Choose Statistical from the Function Category box. Choose NORMDIST from the Function Name box. Select OK.

continue

### Example: Pep Zone

• Using an Excel Spreadsheet (continued)

• Step 5: When the NORMDIST dialog box appears:

Enter 20 in the x box.

Enter 15 in the mean box.

Enter 6 in the standard deviation box.

Enter true in the cumulative box.

Select OK.

At this point, .7967 will appear in the cell selected in Step 1, indicating that the probability of demand during lead time being less than or equal to 20 gallons is .7967. The probability that demand will exceed 20 gallons is 1 - .7967 = .2033.

### Example: Pep Zone

If the manager of Pep Zone wants the probability of a

stockout to be no more than .05, what should the

reorder point be?

Let z.05 represent the z value cutting the tail area of .05.

Area = .05

Area = .5

Area = .45

z.05

0

### Example: Pep Zone

• Using the Standard Normal Probability Table

We now look-up the .4500 area in the Standard Normal Probability table to find the corresponding z.05 value.

z.05 = 1.645 is a reasonable estimate.

### Example: Pep Zone

The corresponding value of x is given by

x = m + z.05s

= 15 + 1.645(6)

= 24.87

A reorder point of 24.87 gallons will place the

probability of a stockout during leadtime at .05.

Perhaps Pep Zone should set the reorder point at 25 gallons to keep the probability under .05.

### Exponential Probability Distribution

• Exponential Probability Density Function

for x> 0, m > 0

wherem = mean

e = 2.71828

• Cumulative Exponential Distribution Function

wherex0 = some specific value of x

### Example: Al’s Carwash

The time between arrivals of cars at Al’s Carwash

follows an exponential probability distribution with a

mean time between arrivals of 3 minutes. Al would like

to know the probability that the time between two

successive arrivals will be 2 minutes or less.

P(x< 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866

### Example: Al’s Carwash

• Graph of the Probability Density Function

f(x)

.4

P(x< 2) = area = .4866

.3

.2

.1

x

1 2 3 4 5 6 7 8 9 10

### Relationship Between thePoisson and Exponential Distributions

• The continuous exponential probability distribution is related to the discrete Poisson distribution.

• The Poisson distribution provides an appropriate description of the number of occurrences per interval.

• The exponential distribution provides a description of the length of the interval between occurrences.