Business Statistics: A Decision-Making Approach · 2019. 8. 23. · Business Statistics: A Decision-Making Approach, 6e © 2005 Prentice-Hall, Inc. Chap 5-1 Business Statistics: A

Business Statistics: A Decision-Making Approach, 6e © 2005 Prentice-Hall, Inc. Chap 5-1

Business Statistics: A Decision-Making Approach

6th Edition

Chapter 5Discrete and Continuous Probability Distributions


Chapter Goals

After completing this chapter, you should be able to:

Apply the binomial distribution to applied problems Compute probabilities for the Poisson and

hypergeometric distributions Find probabilities using a normal distribution table

and apply the normal distribution to business problems

Recognize when to apply the uniform and exponential distributions


Probability Distributions

ContinuousProbability

Distributions

Binomial

Hypergeometric

Poisson


DiscreteProbability

Distributions

Normal

Uniform

Exponential


A discrete random variable is a variable that can assume only a countable number of valuesMany possible outcomes: number of complaints per day number of TV’s in a household number of rings before the phone is answeredOnly two possible outcomes: gender: male or female defective: yes or no spreads peanut butter first vs. spreads jelly first

Discrete Probability Distributions


Continuous Probability Distributions

A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values) thickness of an item time required to complete a task temperature of a solution height, in inches

These can potentially take on any value, depending only on the ability to measure accurately.


The Binomial Distribution

Binomial

Hypergeometric

Poisson


DiscreteProbability

Distributions


The Binomial Distribution

Characteristics of the Binomial Distribution:

A trial has only two possible outcomes – “success” or “failure”

There is a fixed number, n, of identical trials The trials of the experiment are independent of each

other The probability of a success, p, remains constant from

trial to trial If p represents the probability of a success, then

(1-p) = q is the probability of a failure


Binomial Distribution Settings

A manufacturing plant labels items as either defective or acceptable

A firm bidding for a contract will either get the contract 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


Counting Rule for Combinations

A combination is an outcome of an experiment where x objects are selected from a group of n objects

)!xn(!x!nCnx −

=

where:n! =n(n - 1)(n - 2) . . . (2)(1)x! = x(x - 1)(x - 2) . . . (2)(1)0! = 1 (by definition)


P(x) = probability of x successes in n trials,with probability of success p on each trial

x = number of ‘successes’ in sample, (x = 0, 1, 2, ..., n)

p = probability of “success” per trialq = probability of “failure” = (1 – p)n = number of trials (sample size)

P(x)n

x ! n xp qx n x

!( )!

=−

−

Example: Flip a coin four times, let x = # heads:

n = 4

p = 0.5

q = (1 - .5) = .5

x = 0, 1, 2, 3, 4

Binomial Distribution Formula


n = 5 p = 0.1

n = 5 p = 0.5

Mean

0.2.4.6

0 1 2 3 4 5X

P(X)

.2

.4

.6

0 1 2 3 4 5X

P(X)

0

Binomial Distribution The shape of the binomial distribution depends on the

values of p and n

Here, n = 5 and p = .1

Here, n = 5 and p = .5


Binomial Distribution Characteristics

Mean

Variance and Standard Deviation

npE(x)μ ==

npqσ2 =

npqσ =Where n = sample size

p = probability of successq = (1 – p) = probability of failure


n = 5 p = 0.1

n = 5 p = 0.5

Mean

0.2.4.6

0 1 2 3 4 5X

P(X)

.2

.4

.6

0 1 2 3 4 5X

P(X)

0

0.5(5)(.1)npμ ===

0.6708.1)(5)(.1)(1npqσ

=

−==

2.5(5)(.5)npμ ===

1.118.5)(5)(.5)(1npqσ

=

−==

Binomial CharacteristicsExamples


Using Binomial Tablesn = 10

x p=.15 p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50

0123456789

10

0.19690.34740.27590.12980.04010.00850.00120.00010.00000.00000.0000

0.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.0000

0.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.0000

0.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.0000

0.01350.07250.17570.25220.23770.15360.06890.02120.00430.00050.0000

0.00600.04030.12090.21500.25080.20070.11150.04250.01060.00160.0001

0.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.0003

0.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010

109876543210

p=.85 p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 x

Examples: n = 10, p = .35, x = 3: P(x = 3|n =10, p = .35) = .2522

n = 10, p = .75, x = 2: P(x = 2|n =10, p = .75) = .0004


The Poisson Distribution

Binomial

Hypergeometric

Poisson


DiscreteProbability

Distributions


The Poisson Distribution

Characteristics of the Poisson Distribution: The outcomes of interest are rare relative to the

possible outcomes The average number of outcomes of interest per time

or space interval is λ The number of outcomes of interest are random, and

the occurrence of one outcome does not influence the chances of another outcome of interest

The probability of that an outcome of interest occurs in a given segment is the same for all segments


Poisson Distribution Formula

where:t = size of the segment of interest x = number of successes in segment of interestλ = expected number of successes in a segment of unit sizee = base of the natural logarithm system (2.71828...)

!xe)t()x(P

tx λ−λ=


Poisson Distribution Characteristics

Mean

Variance and Standard Deviation

λtμ =

λtσ2 =

λtσ =where λ = number of successes in a segment of unit size

t = the size of the segment of interest


Using Poisson Tables

X

λt

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

01234567

0.90480.09050.00450.00020.00000.00000.00000.0000

0.81870.16370.01640.00110.00010.00000.00000.0000

0.74080.22220.03330.00330.00030.00000.00000.0000

0.67030.26810.05360.00720.00070.00010.00000.0000

0.60650.30330.07580.01260.00160.00020.00000.0000

0.54880.32930.09880.01980.00300.00040.00000.0000

0.49660.34760.12170.02840.00500.00070.00010.0000

0.44930.35950.14380.03830.00770.00120.00020.0000

0.40660.36590.16470.04940.01110.00200.00030.0000

Example: Find P(x = 2) if λ = .05 and t = 100

.07582!

e(0.50)!xe)t()2x(P

0.502tx

==λ

==−λ−


Graph of Poisson Probabilities

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x)X

λt =0.50

01234567

0.60650.30330.07580.01260.00160.00020.00000.0000

P(x = 2) = .0758

Graphically:λ = .05 and t = 100

Chart2

0

1

2

3

4

5

6

7

x

P(x)

0.6065306597

0.3032653299

0.0758163325

0.0126360554

0.0015795069

0.0001579507

0.0000131626

0.0000009402

Histogram

X

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Number of Successes

P(X)

Histogram

0

0.6065306597

0.3032653299

0.0758163325

0.0126360554

0.0015795069

0.0001579507

0.0000131626

0.0000009402

0.0000000588

0.0000000033

0.0000000002

0

0

0

4.24643487230639E-16

1.4154782907688E-17

4.42336965865248E-19

1.30099107607426E-20

3.61386410020629E-22

9.51016868475338E-24

2.37754217118835E-25

Poisson2

Poisson Probabilities for Customer Arrivals

Data

Average/Expected number of successes:0.5

Poisson Probabilities Table

XP(X)P(=X)

00.6065310.6065310.0000000.3934691.000000

10.3032650.9097960.6065310.0902040.393469

20.0758160.9856120.9097960.0143880.090204

30.0126360.9982480.9856120.0017520.014388

40.0015800.9998280.9982480.0001720.001752

50.0001580.9999860.9998280.0000140.000172

60.0000130.9999990.9999860.0000010.000014

70.0000011.0000000.9999990.0000000.000001

80.0000001.0000001.0000000.0000000.000000

90.0000001.0000001.0000000.0000000.000000

100.0000001.0000001.0000000.0000000.000000

110.0000001.0000001.0000000.0000000.000000

120.0000001.0000001.0000000.0000000.000000

130.0000001.0000001.0000000.0000000.000000

140.0000001.0000001.0000000.0000000.000000

150.0000001.0000001.0000000.0000000.000000

160.0000001.0000001.0000000.0000000.000000

170.0000001.0000001.0000000.0000000.000000

180.0000001.0000001.0000000.0000000.000000

190.0000001.0000001.0000000.0000000.000000

200.0000001.0000001.0000000.0000000.000000

&A

Page &P

Poisson2

x

P(x)

Poisson


Data



XP(X)P(=X)

00.90480.9048370.0000000.0951631.000000

10.09050.9953210.9048370.0046790.095163

20.00450.9998450.9953210.0001550.004679

30.00020.9999960.9998450.0000040.000155

40.00001.0000000.9999960.0000000.000004

50.00001.0000001.0000000.0000000.000000

60.00001.0000001.0000000.0000000.000000

70.00001.0000001.0000000.0000000.000000

&A

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Sheet1

Sheet2

Sheet3


Poisson Distribution Shape

The shape of the Poisson Distribution depends on the parameters λ and t:

0.00

0.05

0.10

0.15

0.20

0.25

1 2 3 4 5 6 7 8 9 10 11 12

x

P(x)

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x)

λt = 0.50 λt = 3.0

Chart3

0.0497870684

0.1493612051

0.2240418077

0.2240418077

0.1680313557

0.1008188134

0.0504094067

0.0216040315

0.0081015118

0.0027005039

0.0008101512

0.0002209503

x

P(x)

Histogram

X

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Number of Successes

P(X)

Histogram

0

0.0497870684

0.1493612051

0.2240418077

0.2240418077

0.1680313557

0.1008188134

0.0504094067

0.0216040315

0.0081015118

0.0027005039

0.0008101512

0.0002209503

0.0000552376

0.0000127471

0.0000027315

0.0000005463

0.0000001024

0.0000000181

0.000000003

0.0000000005

0.0000000001

Poisson2


Data

Average/Expected number of successes:3


XP(X)P(=X)

00.0497870.0497870.0000000.9502131.000000

10.1493610.1991480.0497870.8008520.950213

20.2240420.4231900.1991480.5768100.800852

30.2240420.6472320.4231900.3527680.576810

40.1680310.8152630.6472320.1847370.352768

50.1008190.9160820.8152630.0839180.184737

60.0504090.9664910.9160820.0335090.083918

70.0216040.9880950.9664910.0119050.033509

80.0081020.9961970.9880950.0038030.011905

90.0027010.9988980.9961970.0011020.003803

100.0008100.9997080.9988980.0002920.001102

110.0002210.9999290.9997080.0000710.000292

120.0000550.9999840.9999290.0000160.000071

130.0000130.9999970.9999840.0000030.000016

140.0000030.9999990.9999970.0000010.000003

150.0000011.0000000.9999990.0000000.000001

160.0000001.0000001.0000000.0000000.000000

170.0000001.0000001.0000000.0000000.000000

180.0000001.0000001.0000000.0000000.000000

190.0000001.0000001.0000000.0000000.000000

200.0000001.0000001.0000000.0000000.000000

&A

Page &P

Poisson2

x

P(x)

Poisson

x

P(x)

Sheet1


Data



XP(X)P(=X)

00.90480.9048370.0000000.0951631.000000

10.09050.9953210.9048370.0046790.095163

20.00450.9998450.9953210.0001550.004679

30.00020.9999960.9998450.0000040.000155

40.00001.0000000.9999960.0000000.000004

50.00001.0000001.0000000.0000000.000000

60.00001.0000001.0000000.0000000.000000

70.00001.0000001.0000000.0000000.000000

&A

Page &P

Sheet2

Sheet3

Chart2

0

1

2

3

4

5

6

7

x

P(x)

0.6065306597

0.3032653299

0.0758163325

0.0126360554

0.0015795069

0.0001579507

0.0000131626

0.0000009402

Histogram

X

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Number of Successes

P(X)

Histogram

0

0.6065306597

0.3032653299

0.0758163325

0.0126360554

0.0015795069

0.0001579507

0.0000131626

0.0000009402

0.0000000588

0.0000000033

0.0000000002

0

0

0

4.24643487230639E-16

1.4154782907688E-17

4.42336965865248E-19

1.30099107607426E-20

3.61386410020629E-22

9.51016868475338E-24

2.37754217118835E-25

Poisson2


Data



XP(X)P(=X)

00.6065310.6065310.0000000.3934691.000000

10.3032650.9097960.6065310.0902040.393469

20.0758160.9856120.9097960.0143880.090204

30.0126360.9982480.9856120.0017520.014388

40.0015800.9998280.9982480.0001720.001752

50.0001580.9999860.9998280.0000140.000172

60.0000130.9999990.9999860.0000010.000014

70.0000011.0000000.9999990.0000000.000001

80.0000001.0000001.0000000.0000000.000000

90.0000001.0000001.0000000.0000000.000000

100.0000001.0000001.0000000.0000000.000000

110.0000001.0000001.0000000.0000000.000000

120.0000001.0000001.0000000.0000000.000000

130.0000001.0000001.0000000.0000000.000000

140.0000001.0000001.0000000.0000000.000000

150.0000001.0000001.0000000.0000000.000000

160.0000001.0000001.0000000.0000000.000000

170.0000001.0000001.0000000.0000000.000000

180.0000001.0000001.0000000.0000000.000000

190.0000001.0000001.0000000.0000000.000000

200.0000001.0000001.0000000.0000000.000000

&A

Page &P

Poisson2

x

P(x)

Poisson


Data



XP(X)P(=X)

00.90480.9048370.0000000.0951631.000000

10.09050.9953210.9048370.0046790.095163

20.00450.9998450.9953210.0001550.004679

30.00020.9999960.9998450.0000040.000155

40.00001.0000000.9999960.0000000.000004

50.00001.0000001.0000000.0000000.000000

60.00001.0000001.0000000.0000000.000000

70.00001.0000001.0000000.0000000.000000

&A

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Sheet1

Sheet2

Sheet3


The Hypergeometric Distribution

Binomial

Poisson


DiscreteProbability

Distributions

Hypergeometric


The Hypergeometric Distribution

“n” trials in a sample taken from a finite population of size N

Sample taken without replacement

Trials are dependent

Concerned with finding the probability of “x” successes in the sample where there are “X” successes in the population


Hypergeometric Distribution Formula

Nn

Xx

XNxn

CCC)x(P

−−=

.

WhereN = Population sizeX = number of successes in the populationn = sample sizex = number of successes in the sample

n – x = number of failures in the sample

(Two possible outcomes per trial)


Hypergeometric Distribution Formula

0.3120

(6)(6)C

CCC

CC2)P(x 103

42

61

Nn

Xx

XNxn =====−−

■ Example: 3 Light bulbs were selected from 10. Of the 10 there were 4 defective. What is the probability that 2 of the 3 selected are defective?

N = 10 n = 3X = 4 x = 2


The Normal Distribution


Distributions


Normal

Uniform

Exponential


The Normal Distribution

‘Bell Shaped’ Symmetrical Mean, Median and Mode

are EqualLocation is determined by the mean, μSpread is determined by the standard deviation, σ

The random variable has an infinite theoretical range: + ∞ to − ∞

Mean = Median = Mode

x

f(x)

μ

σ


By varying the parameters μ and σ, we obtain different normal distributions

Many Normal Distributions

Chart3

00

0.00221592422.95798147900153E-220.0000014867

0.002571320500.000002439

0.00297626620.00000000010.0000039613

0.00343638330.00000000010.0000063698

0.00395772580.00000000040.0000101409

0.00454678130.00000000090.0000159837

0.00521046740.00000000220.0000249425

0.00595612180.00000000530.0000385352

0.00679148460.00000001230.0000589431

0.00772467360.00000002820.0000892617

0.00876415020.00000006350.0001338302

0.00991867720.00000013990.0001986555

0.01119726510.00000030210.0002919469

0.012609110.00000063960.0004247803

0.01416351890.00000132730.0006119019

0.01586982590.00000269960.0008726827

0.01773729640.00000538230.0012322192

0.01977502080.00001051830.0017225689

0.0219917980.00002014830.0023840882

0.02439600930.00003783070.0032668191

0.02699548330.00006962490.0044318484

0.0297973530.00012560260.0059525324

0.03280790740.0002220990.0079154516

0.03603243720.00038495410.0104209348

0.03947507920.00065401160.0135829692

0.04313865940.00108912050.0175283005

0.04702453870.00177779050.0223945303

0.05113246230.00284445680.0283270377

0.05546041730.00446099980.0354745928

0.06000450030.00685771090.043983596

0.06475879780.01033332890.0539909665

0.06971528320.01526214050.0656158148

0.07486373280.02209554660.0789501583

0.08019166370.03135509790.0940490774

0.0856842960.04361397650.1109208347

0.09132454270.05946443790.1295175957

0.09709302750.07946997240.1497274656

0.10296813440.10410292120.171368592

0.10892608850.13367090270.194186055

0.11494107030.16823833270.217852177

0.12098536230.20755210430.2419707245

0.12702952820.25098249750.2660852499

0.13304262490.29749100760.2896915528

0.13899244310.34563551340.3122539334

0.14484577640.39361984480.3332246029

0.15056871610.43938954890.3520653268

0.15612696670.48076912580.3682701403

0.16148617980.51562922930.3813878155

0.16661230140.54206653460.391042694

0.17147192750.55857536080.3969525475

0.17603266340.56418958350.3989422804

0.18026348120.55857624590.3969525475

0.18413507020.54206825250.391042694

0.18762017350.51563168030.3813878155

0.19069390770.4807721730.3682701403

0.19333405840.439393030.3520653268

0.1955213470.3936235870.3332246029

0.19723966550.3456393470.3122539334

0.19847627370.29749477870.2896915528

0.1992219570.25098607670.2660852499

0.19947114020.20755539310.2419707245

0.1992219570.16824126510.217852177

0.19847627370.13367344430.194186055

0.19723966550.10410506560.171368592

0.1955213470.07947173530.1497274656

0.19333405840.05946585130.1295175957

0.19069390770.04361508220.1109208347

0.18762017350.03135594260.0940490774

0.18413507020.02209617680.0789501583

0.18026348120.01526260.0656158148

0.17603266340.01033365640.0539909665

0.17147192750.00685793910.043983596

0.16661230140.00446115530.0354745928

0.16148617980.00284456050.0283270377

0.15612696670.00177785810.0223945303

0.15056871610.00108916370.0175283005

0.14484577640.00065403850.0135829692

0.13899244310.00038497060.0104209348

0.13304262490.00022210890.0079154516

0.12702952820.00012560830.0059525324

0.12098536230.00006962820.0044318484

0.11494107030.00003783260.0032668191

0.10892608850.00002014930.0023840882

0.10296813440.00001051890.0017225689

0.09709302750.00000538260.0012322192

0.09132454270.00000269980.0008726827

0.0856842960.00000132730.0006119019

0.08019166370.00000063970.0004247803

0.07486373280.00000030220.0002919469

0.06971528320.00000013990.0001986555

0.06475879780.00000006350.0001338302

0.06000450030.00000002820.0000892617

0.05546041730.00000001230.0000589431

0.05113246230.00000000530.0000385352

0.04702453870.00000000220.0000249425

0.04313865940.00000000090.0000159837

0.03947507920.00000000040.0000101409

0.03603243720.00000000010.0000063698

0.03280790740.00000000010.0000039613

0.02979735300.000002439

0.026995483300.0000014867

0.024396009300.0000008972

Sheet1

xpopulationsampling populationsigman

-50.00221592422.95798147900153E-220.0000014867281

-4.90.002571320500.0000024390.70710678121

-4.80.00297626620.00000000010.000003961312

-4.70.00343638330.00000000010.0000063698

-4.60.00395772580.00000000040.0000101409

-4.50.00454678130.00000000090.0000159837

-4.40.00521046740.00000000220.0000249425

-4.30.00595612180.00000000530.0000385352

-4.20.00679148460.00000001230.0000589431

-4.10.00772467360.00000002820.0000892617

-40.00876415020.00000006350.0001338302

-3.90.00991867720.00000013990.0001986555

-3.80.01119726510.00000030210.0002919469

-3.70.012609110.00000063960.0004247803

-3.60.01416351890.00000132730.0006119019

-3.50.01586982590.00000269960.0008726827

-3.40.01773729640.00000538230.0012322192

-3.30.01977502080.00001051830.0017225689

-3.20.0219917980.00002014830.0023840882

-3.10.02439600930.00003783070.0032668191

-30.02699548330.00006962490.0044318484

-2.90.0297973530.00012560260.0059525324

-2.80.03280790740.0002220990.0079154516

-2.70.03603243720.00038495410.0104209348

-2.60.03947507920.00065401160.0135829692

-2.50.04313865940.00108912050.0175283005

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00.17603266340.56418958350.3989422804

0.10.18026348120.55857624590.3969525475

0.20.18413507020.54206825250.391042694

0.30.18762017350.51563168030.3813878155

0.40.19069390770.4807721730.3682701403

0.50.19333405840.439393030.3520653268

0.60.1955213470.3936235870.3332246029

0.70.19723966550.3456393470.3122539334

0.80.19847627370.29749477870.2896915528

0.90.1992219570.25098607670.2660852499

10.19947114020.20755539310.2419707245

1.10.1992219570.16824126510.217852177

1.20.19847627370.13367344430.194186055

1.30.19723966550.10410506560.171368592

1.40.1955213470.07947173530.1497274656

1.50.19333405840.05946585130.1295175957

1.60.19069390770.04361508220.1109208347

1.70.18762017350.03135594260.0940490774

1.80.18413507020.02209617680.0789501583

1.90.18026348120.01526260.0656158148

20.17603266340.01033365640.0539909665

2.10.17147192750.00685793910.043983596

2.20.16661230140.00446115530.0354745928

2.30.16148617980.00284456050.0283270377

2.40.15612696670.00177785810.0223945303

2.50.15056871610.00108916370.0175283005

2.60.14484577640.00065403850.0135829692

2.70.13899244310.00038497060.0104209348

2.80.13304262490.00022210890.0079154516

2.90.12702952820.00012560830.0059525324

30.12098536230.00006962820.0044318484

3.10.11494107030.00003783260.0032668191

3.20.10892608850.00002014930.0023840882

3.30.10296813440.00001051890.0017225689

3.40.09709302750.00000538260.0012322192

3.50.09132454270.00000269980.0008726827

3.60.0856842960.00000132730.0006119019

3.70.08019166370.00000063970.0004247803

3.80.07486373280.00000030220.0002919469

3.90.06971528320.00000013990.0001986555

40.06475879780.00000006350.0001338302

4.10.06000450030.00000002820.0000892617

4.20.05546041730.00000001230.0000589431

4.30.05113246230.00000000530.0000385352

4.40.04702453870.00000000220.0000249425

4.50.04313865940.00000000090.0000159837

4.60.03947507920.00000000040.0000101409

4.70.03603243720.00000000010.0000063698

4.80.03280790740.00000000010.0000039613

4.90.02979735300.000002439

50.026995483300.0000014867

5.10.024396009300.0000008972

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The Normal Distribution Shape

x

f(x)

μ

σ

Changing μ shifts the distribution left or right.

Changing σ increases or decreases the spread.


Finding Normal Probabilities

Probability is the area under thecurve!

a b x

f(x) P a x b( )≤ ≤

Probability is measured by the area under the curve


f(x)

xμ

Probability as Area Under the Curve

0.50.5

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

1.0)xP( =∞


Empirical Rules

μ ± 1σ encloses about 68% of x’s

f(x)

xμ μ+1σμ−1σ

What can we say about the distribution of values around the mean? There are some general rules:

σσ

68.26%


The Empirical Rule

μ ± 2σ covers about 95% of x’s

μ ± 3σ covers about 99.7% of x’s

xμ2σ 2σ

xμ3σ 3σ

95.44% 99.72%

(continued)


Importance of the Rule

If a value is about 2 or more standarddeviations away from the mean in a normaldistribution, then it is far from the mean

The chance that a value that far or farther away from the mean is highly unlikely, giventhat particular mean and standard deviation


The Standard Normal Distribution Also known as the “z” distribution Mean is defined to be 0 Standard Deviation is 1

z

f(z)

0

1

Values above the mean have positive z-values, values below the mean have negative z-values


The Standard Normal

Any normal distribution (with any mean and standard deviation combination) can be transformed into the standard normaldistribution (z)

Need to transform x units into z units


Translation to the Standard Normal Distribution

Translate from x to the standard normal (the “z” distribution) by subtracting the mean of x and dividing by its standard deviation:

σμxz −=


Example

If x is distributed normally with mean of 100and standard deviation of 50, the z value for x = 250 is

This says that x = 250 is three standard deviations (3 increments of 50 units) above the mean of 100.

3.050

100250σ

μxz =−=−=


Comparing x and z units

z100

3.00250 x

Note that the distribution is the same, only the scale has changed. We can express the problem in original units (x) or in standardized units (z)

μ = 100

σ = 50


The Standard Normal Table

The Standard Normal table in the textbook (Appendix D)

gives the probability from the mean (zero)up to a desired value for z

z0 2.00

.4772Example:P(0 < z < 2.00) = .4772


The Standard Normal Table

The value within the table gives the probability from z = 0up to the desired z value

z 0.00 0.01 0.02 …

0.1

0.2

.4772

2.0P(0 < z < 2.00) = .4772

The row shows the value of z to the first decimal point

The column gives the value of z to the second decimal point

2.0

...

(continued)

z

0.00

0.01

0.02

…

0.1

0.2


General Procedure for Finding Probabilities

Draw the normal curve for the problem interms of x

Translate x-values to z-values

Use the Standard Normal Table

To find P(a < x < b) when x is distributed normally:


Z Table example Suppose x is normal with mean 8.0 and

standard deviation 5.0. Find P(8 < x < 8.6)

P(8 < x < 8.6)

= P(0 < z < 0.12)

Z0.120x8.68

05

88σ

μxz =−=−=

0.125

88.6σ

μxz =−=−=

Calculate z-values:


Z Table example Suppose x is normal with mean 8.0 and

standard deviation 5.0. Find P(8 < x < 8.6)

P(0 < z < 0.12)

z0.120x8.68

P(8 < x < 8.6)

µ = 8σ = 5

µ = 0σ = 1

(continued)


Z

0.12

z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

Solution: Finding P(0 < z < 0.12)

.0478.02

0.1 .0478

Standard Normal Probability Table (Portion)

0.00

= P(0 < z < 0.12)P(8 < x < 8.6)



Suppose x is normal with mean 8.0 and standard deviation 5.0.

Now Find P(x < 8.6)

Z

8.68.0




Now Find P(x < 8.6)

(continued)

Z

0.12

.0478

0.00

.5000P(x < 8.6)

= P(z < 0.12)

= P(z < 0) + P(0 < z < 0.12)

= .5 + .0478 = .5478


Upper Tail Probabilities


Now Find P(x > 8.6)

Z

8.68.0


Now Find P(x > 8.6)…(continued)

Z

0.120

Z

0.12

.0478

0

.5000 .50 - .0478 = .4522

P(x > 8.6) = P(z > 0.12) = P(z > 0) - P(0 < z < 0.12)

= .5 - .0478 = .4522

Upper Tail Probabilities


Lower Tail Probabilities


Now Find P(7.4 < x < 8)

Z

7.48.0


Lower Tail Probabilities

Now Find P(7.4 < x < 8)…

Z

7.48.0

The Normal distribution is symmetric, so we use the same table even if z-values are negative:

P(7.4 < x < 8)

= P(-0.12 < z < 0)

= .0478

(continued)

.0478


Normal Probabilities in PHStat

We can use Excel and PHStat to quicklygenerate probabilities for any normaldistribution

We will find P(8 < x < 8.6) when x isnormally distributed with mean 8 andstandard deviation 5


The Uniform Distribution


Distributions


Normal

Uniform

Exponential


The Uniform Distribution

The uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable


The Continuous Uniform Distribution:

otherwise 0

bxaifab

1≤≤

−

wheref(x) = value of the density function at any x valuea = lower limit of the intervalb = upper limit of the interval

The Uniform Distribution(continued)

f(x) =


Uniform Distribution

Example: Uniform Probability DistributionOver the range 2 ≤ x ≤ 6:

2 6

.25

f(x) = = .25 for 2 ≤ x ≤ 66 - 21

x

f(x)


The Exponential Distribution


Distributions


Normal

Uniform

Exponential



Used to measure the time that elapses between two occurrences of an event (the time between arrivals)

Examples: Time between trucks arriving at an unloading

dock Time between transactions at an ATM Machine Time between phone calls to the main operator



aλe1a)xP(0 −−=≤≤

The probability that an arrival time is equal to or less than some specified time a is

where 1/λ is the mean time between events

Note that if the number of occurrences per time period is Poisson with mean λ, then the time between occurrences is exponential with mean time 1/ λ


Exponential Distribution

Shape of the exponential distribution(continued)

f(x)

x

λ = 1.0(mean = 1.0)

λ= 0.5 (mean = 2.0)

λ = 3.0(mean = .333)


Example

Example: Customers arrive at the claims counter at the rate of 15 per hour (Poisson distributed). What is the probability that the arrival time between consecutive customers is less than five minutes?

Time between arrivals is exponentially distributed with mean time between arrivals of 4 minutes (15 per 60 minutes, on average)

1/λ = 4.0, so λ = .25

P(x < 5) = 1 - e-λa = 1 – e-(.25)(5) = .7135


Chapter Summary

Reviewed key discrete distributions binomial, poisson, hypergeometric

Reviewed key continuous distributions normal, uniform, exponential

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems

Comparing Binomial, Negative Binomial and Geometric

Distributions

63

Binomial Distribution

64

The Binomial Model:

Let X be the number of “successes” in n trials.

If1. The trials are identical and independent

2. Each trial results in one of two possible outcomes success or failure

3. The probability of success on a single trial is p,and is constant from trial to trial,

X ~ B (n,p)

65

The Binomial Model:

then X has the Binomial Distribution with Probability Mass Function given by:

where

( )!xn!x!n

xn

−=

n0,1,..., xq ),;()( x-n =

== xp

xn

pnxbxb

pq −=1and

66


Rule:

for x = 0, …, n - 1

Rule:

)p,n;x(bqp

1xxn )p,n;1x(b

+−

=+

( ) ∑=

−

=+=

n

0x

xnxn qpxn

qp1

67


• Mean or Expected Value

µ = np

• Standard Deviation

( )21

npqσ =

68

The Binomial Model in Excel

Excel provides an easier way of finding probabilities:

Click the Insert button on the menu bar (at the top of the Excel page) Go to the function option Choose Statistical from the Function Category

window (a list of all available statistical functions will appear in the Function Namewindow)

Choose the BINOMDIST function Type in parameters:

number_s => X Trials => N Probability_s => p Cumulative (logical) => TRUE for cumulative

function, FALSE for mass function.

69

Example - Binomial Distribution

When circuit boards used in the manufacture of compact disc players are tested, the long-run

percentage of defects is 5%. Let X=the number of defective boards in a random sample size n=25, so

X~B(25,0.05).

a) Determine P(X ≤ 2).b) Determine P(X ≥ 5).

c) Determine P(1 ≤ X ≤ 4).d) What is the probability that none of the 25

boards are defective?e) Calculate the expected value and standard

deviation of X.

70

Solution – Binomial Distribution

71

Negative Binomial Distribution

72

The Negative Binomial Model:

The negative binomial distribution is based on an experiment satisfying the following conditions:

1. The experiment consists of a sequence of independent and identical trials

2. Each trial can result in either a success, S, or aFailure, F.

3. The probability of success is constant from trialto trial, so P(S on trial i) = p for i = 1, 2, 3, …

73


4. The experiment continues (trials are performed)until a total of r successes have been observed,

where r is a specified positive integer.

The associated random variable is:

X = number of failures that precede the rth success

X is called the negative binomial random variablebecause, in contrast to the binomial random

variable, the number of successes is fixed and the number of trials is random.

Possible values of X are x = 0, 1, 2, ...

74


The probability mass function of the negativebinomial random variable X with parameters

r = number of successes and

p = probability of success on a single trial

( ) ( ) .0,1,2,.. x ,11

1,; =−

−−+

= xr ppr

rxprxNB

75


The negative binomial model can also beexpressed as:

X = total number of trials to get k successes

k = number of successes and

p = probability of success on a single trial

( ) ( ) .0,1,2,.. x ,111

,; =−

−−

= −kxk ppkx

pkxNB

76

Negative Binomial Probability Mass Function Derivation

Consider the probability of a success on the trial preceded by successes and failures in some specified order.

Since the trials are independent, we can multiply all the probabilities corresponding to each desired outcome.

Each success occurs with probability and each failure with probability

Therefore, the probability for the specified order, ending in a success, is

1−k kx −thx

ppq −=1

kxkkxk qppqp −−− =1

77

The total number of sample points in the experiment ending in a success, after the occurrence of successes and failures in any order, is equal to the number of partitions of

trials into two groups with successes corresponding to one group and failures corresponding to the other group.

The sample space consists of points, each

mutually exclusive and occurring with equal probability

We obtain the general formula by multiplying by

1−k kx −1−x 1−k kx −

−−

11

kx

kxk qp −kxk qp −

−−

11

kx

78

The Negative Binomial Model: Example

Find the probability that a person tossing three coinswill get either all heads or all tails for the second time

on the fifth toss.

Solution( )successp P=

)]()[( TTTorHHHP=

81

81+=

41

=

79

Since a success must occur on the fifth toss of the 3 coins, the following outcomes are possible:

1 2 3 4 5S F F F SF S F F SF F S F SF F F S S

Toss probability

102427

102427

102427

102427

80

Therefore, the probability of obtaining all heads or all tails for the second time on the fifth toss is

25627

102427.4 =

81

The Negative Binomial Model: Example Solution continued

Or, using the Negative Binomial Distribution with r = 2, p = 0.25, and x = 3 gives

( ) ( ) 105.02562775.025.0

14

25.0,2;3 32 ==

=NB

82

The Negative Binomial Model

If X is a negative binomial random variable with probability mass function nb(x;r,p) then

and

( ) ( )p

prXE −= 1

( ) ( )21p

prXVar −=

83

The Negative Binomial Model

Note: By expanding the binomial coefficient in front of pr(1 - p)x and doing some cancellation, it can be seen that NB(x;r,p) is well defined even

when r is not an integer. This generalized negative binomial distribution has been found to fit the observed data quite well in a wide variety of

applications.

84

The Negative Binomial Model in Excel

Similar to the Binomial Dist in Excel: Click the Insert button on the menu bar

(at the top of the Excel page) Go to the function option Choose Statistical from the Function Category

window (a list of all available statistical functions will appear in the Function Namewindow)

Choose the NEGBINOMDIST function Type in parameters:

number_f => X Number_s => r Probability_s => p

85

The Geometric Distribution

If repeated independent and identical trials can result in a success with probability p and a failure

with probability q = 1 - p, then the probability mass function of the random variable x, the number of the trial on which the first success

occurs, is:

g(x;p) = pqx-1, x = 1,2,3…

86

The Geometric Distribution

The mean and variance of a random variable following the geometric distribution are

22

pp1σ ,

p1μ −==

87

The Geometric Distribution - Example

At “busy time” a telephone exchange is very near capacity, so callers have difficulty placing their

calls. It may be on interest to know the number of attempts necessary in order to gain a connection. Suppose that we let p = 0.05 be the probability of

a connection during a busy time. We are interested in knowing the probability that 5 attempts are necessary for a successful call.

88

The Geometric Distribution - Example Solution

The random variable X is the number of attempts for a successful call. Then

X~G(0.05),So that for with x = 5 and p = 0.05 yields:

P(X=x) = g(5;0.05)= (0.05)(0.95)4

= 0.041And the expected number of attempts is

200.05

1μ ==

Chapter 5�Discrete and Continuous �Probability DistributionsChapter GoalsSlide Number 3Discrete Probability DistributionsContinuous Probability DistributionsSlide Number 6The Binomial DistributionBinomial Distribution SettingsCounting Rule for CombinationsBinomial Distribution FormulaBinomial DistributionBinomial Distribution CharacteristicsBinomial CharacteristicsUsing Binomial TablesSlide Number 15The Poisson DistributionPoisson Distribution FormulaPoisson Distribution CharacteristicsUsing Poisson TablesGraph of Poisson ProbabilitiesPoisson Distribution ShapeSlide Number 22The Hypergeometric DistributionHypergeometric Distribution FormulaHypergeometric Distribution FormulaSlide Number 26The Normal DistributionSlide Number 28The Normal Distribution ShapeFinding Normal Probabilities Probability as �Area Under the CurveEmpirical RulesThe Empirical RuleImportance of the RuleThe Standard Normal DistributionThe Standard NormalTranslation to the Standard Normal DistributionExampleComparing x and z unitsThe Standard Normal TableThe Standard Normal TableGeneral Procedure for Finding ProbabilitiesZ Table exampleZ Table exampleSolution: Finding P(0 < z < 0.12)Finding Normal ProbabilitiesFinding Normal Probabilities�Upper Tail Probabilities�Upper Tail ProbabilitiesLower Tail ProbabilitiesLower Tail ProbabilitiesNormal Probabilities in PHStatSlide Number 53The Uniform DistributionThe Uniform DistributionUniform DistributionSlide Number 57The Exponential DistributionThe Exponential DistributionExponential DistributionExampleChapter SummaryComparing Binomial, Negative Binomial and Geometric DistributionsSlide Number 64Slide Number 65Slide Number 66Slide Number 67 The Binomial Model in ExcelSlide Number 69Slide Number 70Slide Number 71Slide Number 72Slide Number 73Slide Number 74Slide Number 75Negative Binomial Probability Mass Function DerivationSlide Number 77Slide Number 78Slide Number 79Slide Number 80Slide Number 81Slide Number 82Slide Number 83The Negative Binomial Model in ExcelSlide Number 85Slide Number 86Slide Number 87Slide Number 88

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Business Statistics: A Decision-Making Approach · 2019. 8. 23. · Business Statistics: A Decision-Making Approach, 6e © 2005 Prentice-Hall, Inc. Chap 5-1 Business Statistics: A