IS 310 Business Statistics CSU Long Beach

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IS 310 – Business StatisticsIS 310 – Business Statistics

IS 310

Business Statistic

sCSU

Long Beach

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Chapter 5Chapter 5 Discrete Probability Distributions Discrete Probability Distributions

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Random VariablesRandom Variables Discrete Probability DistributionsDiscrete Probability Distributions Expected Value and VarianceExpected Value and Variance Binomial Probability DistributionBinomial Probability Distribution Poisson Probability DistributionPoisson Probability Distribution

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Random VariablesRandom Variables

A random variable is a variable that can take on A random variable is a variable that can take on values at random. Consider the following values at random. Consider the following experiments:experiments:

1.1. Asking 10 students if they watched a TV show last Asking 10 students if they watched a TV show last night (the number of students who watched the night (the number of students who watched the show is a random variable)show is a random variable)

2.2. Inspecting 20 items of a product to check quality of Inspecting 20 items of a product to check quality of the items (the number of defective items is a the items (the number of defective items is a random variable)random variable)

3.3. Tossing a coin five times (the number of heads Tossing a coin five times (the number of heads occurring is a random variable)occurring is a random variable)

4.4. Taking an exam with 100 questions (the number of Taking an exam with 100 questions (the number of correct answers is a random variable)correct answers is a random variable)

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Random Variables (Contd)Random Variables (Contd)

A random variable can be eitherA random variable can be either

Discrete or ContinuousDiscrete or Continuous

Discrete random variables take on certain specific values. Discrete random variables take on certain specific values.

Examples are the following: number of defective items in Examples are the following: number of defective items in an inspection (0, 1, 2, 3,….); number of correct answers an inspection (0, 1, 2, 3,….); number of correct answers in an exam (0, 1, 2, 3, …); number of heads obtained in in an exam (0, 1, 2, 3, …); number of heads obtained in tossing a coin five times (0, 1, 2, 3, 4, 5)tossing a coin five times (0, 1, 2, 3, 4, 5)

o---------o---------o---------o---------oo---------o---------o---------o---------o

The only values the discrete random variable can take on The only values the discrete random variable can take on are indicated by circlesare indicated by circles

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Random Variables ContdRandom Variables Contd

Continuous Random VariablesContinuous Random Variables

A continuous random variable can take on any values on a A continuous random variable can take on any values on a scale. Examples are distance traveled, time taken to go scale. Examples are distance traveled, time taken to go from one place to another, heights of individuals, from one place to another, heights of individuals, weights of individuals, temperature of cities, etc.weights of individuals, temperature of cities, etc.

o------------------------------------------oo------------------------------------------o

A continuous random variable can take on any value on A continuous random variable can take on any value on the above scalethe above scale

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Random VariablesRandom Variables

QuestionQuestion Random Variable Random Variable xx TypeType

FamilyFamilysizesize

xx = Number of dependents = Number of dependents reported on tax returnreported on tax return

DiscreteDiscrete

Distance fromDistance fromhome to storehome to store

xx = Distance in miles from = Distance in miles from home to the store sitehome to the store site

ContinuousContinuous

Own dogOwn dogor cator cat

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

DiscreteDiscrete

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The The probability distributionprobability distribution for a random variable for a random variable describes how probabilities are distributed overdescribes how probabilities are distributed over the values of the random variable.the values of the random variable.

The The probability distributionprobability distribution for a random variable for a random variable describes how probabilities are distributed overdescribes how probabilities are distributed over the values of the random variable.the values of the random variable.

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

Discrete Probability DistributionsDiscrete Probability Distributions

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Discrete Probability DistributionDiscrete Probability Distribution

Let’s consider the illustration in Section 5.2 Let’s consider the illustration in Section 5.2 (10-Page 190; 11-Page 197)(10-Page 190; 11-Page 197)

DiCarlo Motors in Saratoga, New York sold the DiCarlo Motors in Saratoga, New York sold the following number of cars over the past 300 following number of cars over the past 300 days:days:

0 cars on 54 days; 1 car on 117 days; 2 cars 0 cars on 54 days; 1 car on 117 days; 2 cars on 72 days; 3 cars on 42 days; 4 cars on 12 on 72 days; 3 cars on 42 days; 4 cars on 12 days; and 5 cars on 3 days.days; and 5 cars on 3 days.

The probability distribution is shown in Table The probability distribution is shown in Table 5.3 (10-Page 191; 11_Page 198).5.3 (10-Page 191; 11_Page 198).

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Discrete Probability DistributionDiscrete Probability Distribution

Table 5.3 Table 5.3

Number of cars soldNumber of cars sold ProbabilityProbability 0 54/300 = 0.180 54/300 = 0.18 1 117/300 = 0.391 117/300 = 0.39

2 72/300 = 0.242 72/300 = 0.24

3 42/300 = 0.143 42/300 = 0.14

4 12/300 = 0.044 12/300 = 0.04

5 3/300 = 0.015 3/300 = 0.01

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Sample ProblemSample Problem

Problem # 8 (10-Page 193; 11-Page 200)Problem # 8 (10-Page 193; 11-Page 200)

Number of operating rooms used over a 20-day period.Number of operating rooms used over a 20-day period.

Number of RoomsNumber of Rooms FrequencyFrequency Probability Probability

1 3 3/20 = 1 3 3/20 = 0.150.15

2 5 5/20 = 2 5 5/20 = 0.250.25

3 8 8/20 = 3 8 8/20 = 0.400.40

4 4 4/20 = 4 4 4/20 = 0.200.20

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Expected Value and VarianceExpected Value and Variance

The expected value of a random variable is obtained by The expected value of a random variable is obtained by multiplying each value of the random variable by its multiplying each value of the random variable by its probability and adding the resulting products.probability and adding the resulting products.

Let’s refer to the problem of car sales of DiCarlo Motors. Look Let’s refer to the problem of car sales of DiCarlo Motors. Look at Table 5.5 (10-Page 196) or Table 5.4 (11-Page 203).at Table 5.5 (10-Page 196) or Table 5.4 (11-Page 203).

No. of Cars Sold (xNo. of Cars Sold (x) ) Probability [f(xProbability [f(x)] )] x. f(x)x. f(x) 0 0.18 00 0.18 0 1 0.39 0.391 0.39 0.39 2 0.24 0.482 0.24 0.48 3 0.14 0.423 0.14 0.42 4 0.04 0.164 0.04 0.16

5 0.01 0.05 5 0.01 0.05

Expected Value of x = E(x) = 1.50 Expected Value of x = E(x) = 1.50

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What does Expected Value mean?What does Expected Value mean? Expected Value is the average value of the Expected Value is the average value of the

random variable over a long period of time.random variable over a long period of time.

Referring to DiCarlo Motors, the Expected Referring to DiCarlo Motors, the Expected Value of 1.5 means that DiCarlo can expect to Value of 1.5 means that DiCarlo can expect to sell, on the average, 1.5 cars per day over a sell, on the average, 1.5 cars per day over a long period of time.long period of time.

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The variance of a random variable is obtained The variance of a random variable is obtained by using formula 5.5 (10-Page 196; 11-Page by using formula 5.5 (10-Page 196; 11-Page 203). Calculations are shown in Table 5.6 (10-203). Calculations are shown in Table 5.6 (10-Page 197) or Table 5.5 (11-Page 204).Page 197) or Table 5.5 (11-Page 204).

The variance is calculated as 1.25 so the The variance is calculated as 1.25 so the standard deviation is √ 1.25 = 1.118.standard deviation is √ 1.25 = 1.118.

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Binomial Probability DistributionBinomial Probability Distribution

Two discrete probability distributions that we Two discrete probability distributions that we will study are:will study are:

Binomial Probability DistributionBinomial Probability Distribution

Poisson Probability DistributionPoisson Probability Distribution

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Binomial DistributionBinomial Distribution

Four Properties of a Binomial ExperimentFour Properties of a Binomial Experiment

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

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

2. Two outcomes, 2. Two outcomes, successsuccess and and failurefailure, are possible, are possible on each trial.on each trial.2. Two outcomes, 2. Two outcomes, successsuccess and and failurefailure, are possible, are possible on each trial.on each trial.

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

stationaritstationarityy

assumptioassumptionn

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Our interest is in the Our interest is in the number of successesnumber of successes occurring in the occurring in the nn trials. trials. Our interest is in the Our interest is in the number of successesnumber of successes occurring in the occurring in the nn trials. trials.

We let We let xx denote the number of successes denote the number of successes occurring in the occurring in the nn trials. trials. We let We let xx denote the number of successes denote the number of successes occurring in the occurring in the nn trials. trials.

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where:where: ff((xx) = the probability of ) = the probability of xx successes in successes in nn trials trials nn = the number of trials = the number of trials pp = the probability of success on any one trial = the probability of success on any one trial

( )!( ) (1 )

!( )!x n xn

f x p px n x

( )!( ) (1 )

!( )!x n xn

f x p px n x


Binomial Probability FunctionBinomial Probability Function

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( )!( ) (1 )

!( )!x n xn

f x p px n x

( )!( ) (1 )

!( )!x n xn

f x p px n x


Binomial Probability Binomial Probability FunctionFunction

Probability of a particularProbability of a particular sequence of trial outcomessequence of trial outcomes with with xx successes in successes in nn trials trials

Probability of a particularProbability of a particular sequence of trial outcomessequence of trial outcomes with with xx successes in successes in nn trials trials

Number of experimentalNumber of experimental outcomes providing exactlyoutcomes providing exactly

xx successes in successes in nn trials trials

Number of experimentalNumber of experimental outcomes providing exactlyoutcomes providing exactly

xx successes in successes in nn trials trials

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Example of Binomial DistributionExample of Binomial Distribution

Martin Clothing Store (10-Page 202; 11-Page 209))Martin Clothing Store (10-Page 202; 11-Page 209))

Given: The probability of a customer making a purchase is Given: The probability of a customer making a purchase is 0.3. Three customers walk into the store.0.3. Three customers walk into the store.

What is the probability that two of the three customers will What is the probability that two of the three customers will make a purchase?make a purchase?

This is an example of binomial distribution for the following This is an example of binomial distribution for the following reasons:reasons:

1. There are only two outcomes: making a purchase (success) 1. There are only two outcomes: making a purchase (success) or not making a purchase (failure).or not making a purchase (failure).

2. The probability of success is 0.3 . There are three trials 2. The probability of success is 0.3 . There are three trials (three customers) and we are trying to determine the (three customers) and we are trying to determine the probability of two successes.probability of two successes.

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Example of Binomial DistributionExample of Binomial Distribution

Martin Clothing Store ProblemMartin Clothing Store Problem

Let’s look at Figure 5.3 (10-Page 203; 11-Page Let’s look at Figure 5.3 (10-Page 203; 11-Page 210).210).

Formula 5.8 (10-Page 205; 11-Page 212) can be Formula 5.8 (10-Page 205; 11-Page 212) can be used to calculate the probability of two customers used to calculate the probability of two customers making a purchase.making a purchase.

n x n-xn x n-x P(x=2) = ( ) p (1 – p) = 0.189 P(x=2) = ( ) p (1 – p) = 0.189 xx

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Sample Problem on Binomial DistributionSample Problem on Binomial Distribution

Martin Clothing Store ProblemMartin Clothing Store Problem

Rather than using formula 5.8, we could use Rather than using formula 5.8, we could use Table 5 of Appendix B (10- Pages 930-937; 11-Table 5 of Appendix B (10- Pages 930-937; 11-Pages 989-997) to obtain directly the value of Pages 989-997) to obtain directly the value of any probability without any calculations.any probability without any calculations.

We need to know the values of p, x, and n to We need to know the values of p, x, and n to use Table 5.use Table 5.

For x=2, n=3, and p=0.3, the value of P(x=2) For x=2, n=3, and p=0.3, the value of P(x=2) = 0.189 from (10-Page 932; 11-Page 992).= 0.189 from (10-Page 932; 11-Page 992).

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Sample ProblemsSample Problems


Given: p = 0.30 x = 3 (number of workers who Given: p = 0.30 x = 3 (number of workers who

take public take public transportation)transportation)

n = 10 (total number of workers in the sample)n = 10 (total number of workers in the sample)

a. f(3) = 0.2668 (From Table 5 in Appendix B)a. f(3) = 0.2668 (From Table 5 in Appendix B)

b. f(3 or more) = f(3) + f(4) + f(5) + f(6) + f(7) + f(8) b. f(3 or more) = f(3) + f(4) + f(5) + f(6) + f(7) + f(8) + f(9) + f(10) = 0.2668 + 0.2001 + 0.1029 + + f(9) + f(10) = 0.2668 + 0.2001 + 0.1029 + 0.0368 + 0.0090 + 0.0014 + 0.0001 + 0.0000 = 0.0368 + 0.0090 + 0.0014 + 0.0001 + 0.0000 = 0.620.62

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A Poisson distributed random variable is oftenA Poisson distributed random variable is often useful in estimating the number of occurrencesuseful in estimating the number of occurrences over a over a specified interval of time or spacespecified interval of time or space

A Poisson distributed random variable is oftenA Poisson distributed random variable is often useful in estimating the number of occurrencesuseful in estimating the number of occurrences over a over a specified interval of time or spacespecified interval of time or space

It is a discrete random variable that may assumeIt is a discrete random variable that may assume an an infinite sequence of valuesinfinite sequence of values (x = 0, 1, 2, . . . ). (x = 0, 1, 2, . . . ). It is a discrete random variable that may assumeIt is a discrete random variable that may assume an an infinite sequence of valuesinfinite sequence of values (x = 0, 1, 2, . . . ). (x = 0, 1, 2, . . . ).

Poisson DistributionPoisson Distribution

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Examples of a Poisson distributed random variable:Examples of a Poisson distributed random variable: Examples of a Poisson distributed random variable:Examples of a Poisson distributed random variable:

the number of knotholes in 14 linear feet ofthe number of knotholes in 14 linear feet of pine boardpine board the number of knotholes in 14 linear feet ofthe number of knotholes in 14 linear feet of pine boardpine board

the number of vehicles arriving at athe number of vehicles arriving at a toll booth in one hourtoll booth in one hour the number of vehicles arriving at athe number of vehicles arriving at a toll booth in one hourtoll booth in one hour


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Two Properties of a Poisson ExperimentTwo Properties of a Poisson Experiment

2.2. The occurrence or nonoccurrence in anyThe occurrence or nonoccurrence in any interval is independent of the occurrence orinterval is independent of the occurrence or nonoccurrence in any other interval.nonoccurrence in any other interval.

2.2. The occurrence or nonoccurrence in anyThe occurrence or nonoccurrence in any interval is independent of the occurrence orinterval is independent of the occurrence or nonoccurrence in any other interval.nonoccurrence in any other interval.

1.1. The probability of an occurrence is the sameThe probability of an occurrence is the same for any two intervals of equal length.for any two intervals of equal length.1.1. The probability of an occurrence is the sameThe probability of an occurrence is the same for any two intervals of equal length.for any two intervals of equal length.

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Poisson Probability FunctionPoisson Probability Function


f xe

x

x( )

!

f x

e

x

x( )

!

where:where:

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

= mean number of occurrences in an interval= mean number of occurrences in an interval

ee = 2.71828 = 2.71828

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Rather than using formula 5.11, one could use Rather than using formula 5.11, one could use Table 7 of Appendix B (10-Pages 939-944; 11-Table 7 of Appendix B (10-Pages 939-944; 11-Pages 999-1004) to calculate any probability. Pages 999-1004) to calculate any probability. We need to know the values of µ and x to use We need to know the values of µ and x to use Table 7.Table 7.

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Sample ProblemSample Problem


a. Given µ = 48 per hour = 4 per five-minutea. Given µ = 48 per hour = 4 per five-minute

f(3) = 0.1954 (From Table 7 in Appendix B)f(3) = 0.1954 (From Table 7 in Appendix B)

b.b. Given µ = 12 per 15-minuteGiven µ = 12 per 15-minute

f(10) = 0.1048 (From Table 7 in Appendix B)f(10) = 0.1048 (From Table 7 in Appendix B)

c.c. 4 calls f(0) = 0.01834 calls f(0) = 0.0183

d.d. f(0) = 0.0907 with µ = 2.4f(0) = 0.0907 with µ = 2.4

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End of Chapter 5End of Chapter 5

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IS 310 Business Statistics CSU Long Beach