01.discrete prob. distribution

8/3/2019 01.Discrete Prob. Distribution

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Discrete Probability Distributions



Discrete vs Continuous Distributions

Random Variable -- a variable which containsthe outcomes of a chance experiment

Discrete Random Variable -- the set of all possible values is at most a finite or a countably

infinite number of possible values ± Number of new subscribers to a magazine

± Number of bad checks received by a restaurant

± Number of absent employees on a given day

Continuous Random Variable -- takes on values

at every point over a given interval ± Current Ratio of a motorcycle distributorship

± Elapsed time between arrivals of bank customers

± Percent of the labor force that is unemployed



Some Special Distributions

Discrete ± binomial ± Poisson ±

hyp

ergeometric Continuous ± normal ± uniform ± exponential

± t ± chi-square ± F



Discrete Distribution -- Example

0

1

2

34

5

0.37

0.31

0.18

0.090.04

0.01

Number of Crises

Probability

Distribution of DailyCrises

0

0.1

0.2

0.3

0.40.5

0 1 2 3 4 5

P

ro

b

a

b

i

l

it

yNumber of Crises



Requirements for a

Discrete Probability Function Probabilities are between 0 and 1,

inclusively

Total of all probabilities equals 1

0 1e e P X ( ) for all X

P X ( )over all x§ ! 1



Requirements for a Discrete

Probability Function -- ExamplesX P(X)

-10

1

2

3

.1.2

.4

.2

.11.0

X P(X)

-1

0

1

2

3

-.1

.3

.4

.3

.11.0

X P(X)

-10

1

2

3

.1.3

.4

.3

.11.2



Mean of a Discrete Distribution

Q ! ! § E X X P X ( )

X

-1

0

12

3

P(X)

.1

.2

.4

.2

.1

-.1

.0

.4

.4

.3

1.0

X P X ( )



Variance and Standard Deviation

of a Discrete Distribution

2.1)(22

!! § X P X QW W W! ! $2

12 110. .

X

-1

0

12

3

P(X)

.1

.2

.4

.2

.1

-2

-1

01

2

X Q

4

1

01

4

.4

.2

.0

.2

.4

1.2

)(2

Q

X 2

( ) ( ) X P X Q



Mean of the Crises Data Example

Q ! ! !§ E X X P X ( ) .115

X P(X) XyP(X)

0 .37 .00

1 .31 .31

2 .18 .36

3 .09 .27

4 .04 .16

5 .01 .05

1.15

0

0.1

0.20.3

0.4

0.5

0 1 2 3 4 5

P

r

o

b

a

b

i

l

i

t

y

Number of Crises

An executive is considering out of town business travel for a given

Friday. She recognizes that at least one crises could occur on the day

that she is gone and she is concerned about the possibilities. Table

shows a discrete distribution that contains the number of crisis that

could occur during the day she is gone and the probability that each

number will occur.



Variance and Standard Deviation

of Crises Data Example

22

141W Q! !§ X P X ( ) . W W! ! !2

141 119. .

X P(X) (X-Q) (X-Q)2 (X-Q)2yP(X)

0 .37 -1.15 1.32 .49

1 .31 -0.15 0.02 .01

2 .18 0.85 0.72 .13

3 .09 1.85 3.42 .31

4 .04 2.85 8.12 .32

5 .01 3.85 14.82 .15

1.41



Binomial Distribution

Experiment involves n identical trials Each trial has exactly two possible outcomes: successand failure

Each trial is independent of the previous trials

p is the probability of a success on any one trial

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

p and q are constant throughout the experiment

X is the number of successes in the n trials



Binomial Distribution

Probabilityfunction

Meanvalue

Variance

andstandarddeviation

P X

n

X n X X n

X n X

p q( )!

! !!

e e

for 0

Q ! n p

2

2

W

W W

!

! !

n p q

n p q



Binomial Distribution: Development

Experiment: randomly select, with replacement,two families from the residents of Tiny Town

Success is µChildren in Household:¶ p = 0.75 Failure is µNo Children in Household:¶ q = 1- p =

0.25 X is the number of families in the sample with

µChildren in Household¶

FamilyChildren in

Household

Number of

Automobiles

A

B

C

D

Yes

Yes

No

Yes

3

2

1

2

Listing of Sample Space

(A,B), (A,C), (A,D), (D,D),

(B,A), (B,B), (B,C), (B,D),

(C,A), (C,B), (C,C), (C,D),

(D,A), (D,B), (D,C), (D,D)




Continued Families A, B, and D have

children in the household;family C does not

Success is µChildren inHousehold:¶ p = 0.75

Failure is µNo Children inHousehold:¶ q = 1- p = 0.25

X is the number of families

in the sample withµChildren in Household¶

(A,B),

(A,C),

(A,D),

(D,D),

(B,A),

(B,B),

(B,C),

(B,D),

(C,A),

(C,B),

(C,C),

(C,D),

(D,A),

(D,B),

(D,C),

(D,D)

Listing of SampleSpace

2

1

2

2

2

2

1

2

1

1

0

1

2

2

1

2

X

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

P(outcome)




Continued

(A,B),

(A,C),(A,D),

(D,D),

(B,A),

(B,B),

(B,C),

(B,D),

(C,A),

(C,B),(C,C),

(C,D),

(D,A),

(D,B),

(D,C),

(D,D)

Listing of SampleSpace

2

12

2

2

2

1

2

1

10

1

2

2

1

2

X

1/16

1/161/16

1/16

1/16

1/16

1/16

1/16

1/16

1/161/16

1/16

1/16

1/16

1/16

1/16

P(outcome) X

0

1

2

1/16

6/16

9/16

1

P(X)

P X

n

X n X

x n x

p q( )!

! !!

P X ( )!

!

.. .! !

! !

02

0! 2 0

0 06251

16

0 2 0

75 25

P X ( )

!

! !.. .! !

! !

1

2

1 2 10375

3

16

1 2 1

75 25

P X ( )

!

! !.. .! !

! !

22

2 2 20 5625

9

16

2 2 2

75 25




Continued Families A, B, and Dhave children in thehousehold; family Cdoes not

Success is µChildren inHousehold:¶ p = 0.75

Failure is µNo Childrenin Household:¶ q = 1- p = 0.25

X is the number of families in the samplewith µChildren inHousehold¶

XPossible

Sequences

0

1

1

2

(F,F)

(S,F)

(F,S)

(S,S)

P(sequence)

(. )(. ) (. )25 25225!

(. )(. )25 75

(. )(. )75 25

(. )(. ) (. )75 75275!




ContinuedX

PossibleSequences

0

1

1

2

(F,F)

(S,F)

(F,S)

(S,S)

P(sequence)

(. )(. ) (. )25 25225!

(. )(. )25 75

(. )(. )75 25

(. )(. ) (. )75 75275!

X

0

1

2

P(X)

(. )(. )25 752 =0.375

(. )(. ) (. )75 75275! =0.5625

(. )(. ) (. )25 25225! =0.0625

P X

n

X n X

x n x

p q( )!

! !!

P X ( ) !

!.. .! !

!

02

0! 2 00 0625

0 2 0

75 25 P X ( ) !

! !.. .! !

!

1 2

1 2 10 375

1 2 1

75 25

P X ( )

!

! !.. .! !

!

2

2

2 2 20 5625

2 2 2

75 25



Binomial Distribution: Demonstration Problem

n

pq

P X P X P X P X

!

!

!

e ! ! ! !

! !

20

0694

2 0 1 2

2901 3703 2246 8850

.

.

( ) ( ) ( ) ( )

. . . .

P X ( )

)!

( )( )(. ) .. .! !

! !

020!

0!(20 0

1 1 2901 29010 20 0

06 94

P X ( )!( )!

( )(. )(. ) .. .! !

! !

120!

1 20 120 06 3086 3703

1 20 1

06 94

P X ( )

!( )!

( )(. )(. ) .. .! !

! !

220!

2 20 2

190 0036 3283 22462 20 2

06 94

According to the U.S. census Bureau, approximately 6% of all

workers in Jacson, Mississippi, are unemployed. In conducting a random telephone survey in Jacson, what is the probability

of getting two or fewer unemployed workers in a sample of 20?



BinomialTable

n = 20 PROBABILITY

X 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0 0.122 0.012 0.001 0.000 0.000 0.000 0.000 0.000 0.0001 0.270 0.058 0.007 0.000 0.000 0.000 0.000 0.000 0.000

2 0.285 0.137 0.028 0.003 0.000 0.000 0.000 0.000 0.000

3 0.190 0.205 0.072 0.012 0.001 0.000 0.000 0.000 0.000

4 0.090 0.218 0.130 0.035 0.005 0.000 0.000 0.000 0.000

5 0.032 0.175 0.179 0.075 0.015 0.001 0.000 0.000 0.000

6 0.009 0.109 0.192 0.124 0.037 0.005 0.000 0.000 0.000

7 0.002 0.055 0.164 0.166 0.074 0.015 0.001 0.000 0.000

8 0.000 0.022 0.114 0.180 0.120 0.035 0.004 0.000 0.000

9 0.000 0.007 0.065 0.160 0.160 0.071 0.012 0.000 0.000

10 0.000 0.002 0.031 0.117 0.176 0.117 0.031 0.002 0.000

11 0.000 0.000 0.012 0.071 0.160 0.160 0.065 0.007 0.000

12 0.000 0.000 0.004 0.035 0.120 0.180 0.114 0.022 0.000

13 0.000 0.000 0.001 0.015 0.074 0.166 0.164 0.055 0.002

14 0.000 0.000 0.000 0.005 0.037 0.124 0.192 0.109 0.009

15 0.000 0.000 0.000 0.001 0.015 0.075 0.179 0.175 0.032

16 0.000 0.000 0.000 0.000 0.005 0.035 0.130 0.218 0.090

17 0.000 0.000 0.000 0.000 0.001 0.012 0.072 0.205 0.19018 0.000 0.000 0.000 0.000 0.000 0.003 0.028 0.137 0.285

19 0.000 0.000 0.000 0.000 0.000 0.000 0.007 0.058 0.270

20 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.012 0.122



Using the

Binomial Table

Demonstration

Problem

n = 20 PROBABILITY

X 0.1 0.2 0.3 0.4

0 0.122 0.012 0.001 0.000

1 0.270 0.058 0.007 0.000

2 0.285 0.137 0.028 0.003

3 0.190 0.205 0.072 0.012

4 0.090 0.218 0.130 0.035

5 0.032 0.175 0.179 0.075

6 0.009 0.109 0.192 0.124

7 0.002 0.055 0.164 0.166

8 0.000 0.022 0.114 0.180

9 0.000 0.007 0.065 0.160

10 0.000 0.002 0.031 0.117

11 0.000 0.000 0.012 0.071

12 0.000 0.000 0.004 0.035

13 0.000 0.000 0.001 0.015

14 0.000 0.000 0.000 0.005

15 0.000 0.000 0.000 0.001

16 0.000 0.000 0.000 0.00017 0.000 0.000 0.000 0.000

18 0.000 0.000 0.000 0.000

19 0.000 0.000 0.000 0.000

20 0.000 0.000 0.000 0.000

1171.0)10(

40.

20

60.40.1010

1020 !!!

!

!

C X P

p

n



Binomial Distribution using Table:

Demonstration Problem

n

p

q

P X P X P X P X

!

!

!

e ! ! ! !

! !

20

06

94

2 0 1 2

2901 3703 2246 8850

.

.

( ) ( ) ( ) ( )

. . . .

P X P X ( ) ( ) . ." ! e ! !2 1 2 1 8850 1150

Q ! ! !n p ( )(. ) .20 06 1 202

2

20 06 94 1 128

1 128 1 062

W

W W

! ! !

! ! !

n p q ( )(. )(. ) .

. .

n = 20 PROBABILITY

X 0.05 0.06 0.07

0 0.3585 0.2901 0.2342

1 0.3774 0.3703 0.3526

2 0.1887 0.2246 0.2521

3 0.0596 0.0860 0.1139

4 0.0133 0.0233 0.0364

5 0.0022 0.0048 0.0088

6 0.0003 0.0008 0.0017

7 0.0000 0.0001 0.0002

8 0.0000 0.0000 0.0000

« « «

20 0.0000 0.0000 0.0000

«

According to a published data, Oreos control about 10% of themarket for cookie brands. Suppose 20 purchasers of cookies are

selected randomly from the population. What is the probability

that fewer than four purchasers choose Oreos?



Excel¶s Binomial Function

n = 20

p = 0.06

X P(X)

0 =BINOMDIST(A5,B$1,B$2,FALSE)





5 =BINOMDIST(A10,B$1,B$2,FALSE)6 =BINOMDIST(A11,B$1,B$2,FALSE)






Graphs of Selected Binomial Distributions

n = 4 PROBABILITY

X 0.1 0.5 0.9

0 0.656 0.063 0.000

1 0.292 0.250 0.004

2 0.049 0.375 0.049

3 0.004 0.250 0.292

4 0.000 0.063 0.656

P = 0.1

0.000

0.100

0.200

0.300

0.4000.500

0.600

0.700

0.800

0.900

1.000

0 1 2 3 4X

P ( X )

P = 0.5

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.9001.000

0 1 2 3 4X

P ( X )

P = 0.9

0.000

0.100

0.200

0.300

0.4000.500

0.600

0.700

0.800

0.900

1.000

0 1 2 3 4X

P ( X )



Poisson Distribution

A discrete distribution Describes discrete occurrences over a

continuum or interval

Describes rare events Each occurrence is independent any other

occurrences.

The number of occurrences in each interval

can vary from zero to infinity.

The expected number of occurrences musthold constant throughout the experiment.



Poisson Distribution: Applications

Arrivals at queuing systems ± airports -- people, airplanes, baggage ± banks -- people, automobiles, loan applications

Defects in manufactured goods ± number of defects per 1,000 feet of extruded

copper wire ± number of blemishes per square foot of painted

surface ± number of errors per typed page



Poisson Distribution

Probability function

P X

X

X

where

long run average

e

X

e( )!

, , , , ...

:

. . ..

! !

!

!

PP

P

for

(the base of natural logarithms)

0 1 2 3

2

7182

82

P

Mean valueMean value

P

Standard deviationStandard deviation VarianceVariance

P



Bank customers arrive randomly on weekday afternoons at an

average of 3.2 customers every 4 minutes. What is the

probability of getting exactly 10 customers during an 8

minutes interval on a weekday afternoon?

Poisson Distribution: Demonstration

Problem



Poisson Distribution: Demonstration

Problem

P

P

P

PP

!

!

3 2

6 4

1010

0 0528

6 4

.

!

!.

.

customers / 4 minutes

X = 10 customers / 8 minutes

Adjusted

= . customers / 8 minutes

P(X) =

( = ) =

X

10

6.4

e

e

X

P X

P

P

P

PP

!

!

3 2

6 4

66

0 1586

6 4

.

!

!.

.

customers / 4 minutes

X = 6 customers / 8 minutes

Adjusted

= . customers / 8 minutes

P(X) =

( = ) =

X

6

6.4

e

e

X

P X



Poisson Distribution: Probability Table

X 0.5 1.5 1.6 3.0 3.2 6.4 6.5 7.0 8.0

0 0.6065 0.2231 0.2019 0.0498 0.0408 0.0017 0.0015 0.0009 0.0003

1 0.3033 0.3347 0.3230 0.1494 0.1304 0.0106 0.0098 0.0064 0.0027

2 0.0758 0.2510 0.2584 0.2240 0.2087 0.0340 0.0318 0.0223 0.0107

3 0.0126 0.1255 0.1378 0.2240 0.2226 0.0726 0.0688 0.0521 0.0286

4 0.0016 0.0471 0.0551 0.1680 0.1781 0.1162 0.1118 0.0912 0.0573

5 0.0002 0.0141 0.0176 0.1008 0.1140 0.1487 0.1454 0.1277 0.09166 0.0000 0.0035 0.0047 0.0504 0.0608 0.1586 0.1575 0.1490 0.1221

7 0.0000 0.0008 0.0011 0.0216 0.0278 0.1450 0.1462 0.1490 0.1396

8 0.0000 0.0001 0.0002 0.0081 0.0111 0.1160 0.1188 0.1304 0.1396

9 0.0000 0.0000 0.0000 0.0027 0.0040 0.0825 0.0858 0.1014 0.1241

10 0.0000 0.0000 0.0000 0.0008 0.0013 0.0528 0.0558 0.0710 0.0993

11 0.0000 0.0000 0.0000 0.0002 0.0004 0.0307 0.0330 0.0452 0.0722

12 0.0000 0.0000 0.0000 0.0001 0.0001 0.0164 0.0179 0.0263 0.0481

13 0.0000 0.0000 0.0000 0.0000 0.0000 0.0081 0.0089 0.0142 0.0296

14 0.0000 0.0000 0.0000 0.0000 0.0000 0.0037 0.0041 0.0071 0.0169

15 0.0000 0.0000 0.0000 0.0000 0.0000 0.0016 0.0018 0.0033 0.0090

16 0.0000 0.0000 0.0000 0.0000 0.0000 0.0006 0.0007 0.0014 0.0045

17 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0003 0.0006 0.0021

18 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0002 0.0009

P



Poisson Distribution: Using the

Poisson Tables

X 0.5 1.5 1.6 3.0

0 0.6065 0.2231 0.2019 0.0498

1 0.3033 0.3347 0.3230 0.1494

2 0.0758 0.2510 0.2584 0.2240

3 0.0126 0.1255 0.1378 0.2240

4 0.0016 0.0471 0.0551 0.1680

5 0.0002 0.0141 0.0176 0.1008

6 0.0000 0.0035 0.0047 0.0504

7 0.0000 0.0008 0.0011 0.0216

8 0.0000 0.0001 0.0002 0.0081

9 0.0000 0.0000 0.0000 0.0027

10 0.0000 0.0000 0.0000 0.0008

11 0.0000 0.0000 0.0000 0.0002

12 0.0000 0.0000 0.0000 0.0001

P

P !

! !

1 6

4 0 0551

.

( ) . P X



Poisson

Distribution:

Using the

Poisson

Tables

P

X 0.5 1.5 1.6 3.0

0 0.6065 0.2231 0.2019 0.0498

1 0.3033 0.3347 0.3230 0.14942 0.0758 0.2510 0.2584 0.2240

3 0.0126 0.1255 0.1378 0.2240

4 0.0016 0.0471 0.0551 0.1680

5 0.0002 0.0141 0.0176 0.1008

6 0.0000 0.0035 0.0047 0.0504

7 0.0000 0.0008 0.0011 0.0216

8 0.0000 0.0001 0.0002 0.0081

9 0.0000 0.0000 0.0000 0.0027

10 0.0000 0.0000 0.0000 0.0008

11 0.0000 0.0000 0.0000 0.0002

12 0.0000 0.0000 0.0000 0.0001

P !

" ! ! ! ! !

! !

1 6

5 6 7 8 9

0047 0011 0002 0000 0060

.

( ) ( ) ( ) ( ) ( )

. . . . .

P X P X P X P X P X



PoissonDistribution:

Using the

PoissonTables

P !

u ! ! ! !

! !

1 6

2 1 2 1 0 1

1 2019 3230 4751

.

( ) ( ) ( ) ( )

. . .

P X P X P X P X

P

X 0.5 1.5 1.6 3.0

0 0.6065 0.2231 0.2019 0.04981 0.3033 0.3347 0.3230 0.1494

2 0.0758 0.2510 0.2584 0.2240

3 0.0126 0.1255 0.1378 0.2240

4 0.0016 0.0471 0.0551 0.1680

5 0.0002 0.0141 0.0176 0.1008

6 0.0000 0.0035 0.0047 0.0504

7 0.0000 0.0008 0.0011 0.0216

8 0.0000 0.0001 0.0002 0.00819 0.0000 0.0000 0.0000 0.0027

10 0.0000 0.0000 0.0000 0.0008

11 0.0000 0.0000 0.0000 0.0002

12 0.0000 0.0000 0.0000 0.0001



Poisson Distribution: Graphs

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0 1 2 3 4 5 6 7 8

P ! 1 6.

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0 2 4 6 8 10 12 14 16

P ! 6 5.



Excel¶s Poisson Function

P = 1.6

X P(X)

0 =POISSON(D5,E$1,FALSE)












Poisson Approximation

of the Binomial Distribution

Binomial probabilities are difficult tocalculate when n is large.

Under certain conditions binomial

probabilities may be approximated byPoisson probabilities.

Poisson approximation

If and the approximation is acceptable.n n p" e20 7,

Use P ! n p.



Poisson Approximation

of the Binomial Distribution

X Error

0 0.2231 0.2181 -0.0051

1 0.3347 0.3372 0.0025

2 0.2510 0.2555 0.0045

3 0.1255 0.1264 0.0009

4 0.0471 0.0459 -0.0011

5 0.0141 0.0131 -0.0010

6 0.0035 0.0030 -0.0005

7 0.0008 0.0006 -0.0002

8 0.0001 0.0001 0.0000

9 0.0000 0.0000 0.0000

P oisson

P ! 1 5.

Binomial

n

p

!

!

50

03.X Error

0 0.0498 0.0498 0.0000

1 0.1494 0.1493 0.0000

2 0.2240 0.2241 0.0000

3 0.2240 0.2241 0.0000

4 0.1680 0.1681 0.0000

5 0.1008 0.1008 0.0000

6 0.0504 0.0504 0.0000

7 0.0216 0.0216 0.0000

8 0.0081 0.0081 0.0000

9 0.0027 0.0027 0.0000

10 0.0008 0.0008 0.0000

11 0.0002 0.0002 0.0000

12 0.0001 0.0001 0.0000

13 0.0000 0.0000 0.0000

P oisson

P ! 3 0.

Binomial

n

p

!

!

10 000

0003

,

.

01.discrete prob. distribution

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