Binomial Dist Fnl

7/30/2019 Binomial Dist Fnl

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Modern filling machines are designed to work

efficiently and with high reliability. Machinecan fill toothpaste tubes to within 1 gram ofthe desired level 80 percent of the time. Avisitor to the plant, watching filled tubes beingplaced in to cartons, asked Whats thechance that exactly half the pumps in a cartonselected at random will be filled to within 1

gram of the desired level?We can not makean exact forecast, but probability distributionenable us to give a pretty good answer to thisquestion.


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Probability Distributions

G N Patel


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Toss of a Fair Coin

Possible

Outcomesfrom TwoTosses froma Ffir Coin

First Toss Second Toss Number ofTails on

Two Tosses

Probabilityof Possible

OutcomesT T 2 0.5x0.5=0.25

T H 1 0.5x0.5=0.25

H T 1 0.5x0.5=0.25

H H 0 0.5x0.5=0.25

Sum 1.00


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Probability Distribution

Values Probability

0 1/4 = .25

1 2/4 = .50

2 1/4 = .25

Probability Distribution of

Possible Number of Tails

Event: Toss 2 Coins. Count # Tails.

T

T

T T


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Random Variable

Random Variable

Outcomes of an experiment expressed numerically

e.g. Toss a die twice; count the number of timesthe number 4 appears (0, 1 or 2 times)

e.g. Toss a fair coin twice; count the number of

times the tail appears (0, 1 or 2 times)

Measure the time between customer arrival at a

retail outlet.


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Discrete Random Variable

Discrete Random Variable

Obtained by Counting (0, 1, 2, 3, etc.)

Usually a finite number of different values e.g. Toss a coin 5 times; count the number of tails

(0, 1, 2, 3, 4, or 5 times)


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Probability Distribution

A frequency distribution is a listing of theobserved frequencies of all the outcomes of

an experiment that actually occurred whenthe experiment was done, whereas aprobability distribution is a listing of theprobabilities of all the possible outcomes that

could result if the experiment were done.


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Discrete Probability Distribution

List of All Possible [Xj , P(Xj)]Pairs

Xj=Value of random variable

P(Xj) =Probability associated with value

Mutually Exclusive (Nothing in Common)

Collective Exhaustive (Nothing Left Out)

0 1 1j jP X P X


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Summary Measures

Expected value (The Mean)

Weighted average of the probability distribution

e.g. Toss 2 coins, count the number of tails,

compute expected value

j jj

E X X P X

0 .25 1 .5 2 .25 1

j j

j

X P X


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Summary Measures

Variance

Weighted average squared deviation about the mean

e.g. Toss 2 coins, count number of tails, computevariance

(continued)

222

j jE X X P X

22

2 2 20 1 .25 1 1 .5 2 1 .25 .5

j jX P X


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Bob Walter, who frequently invests in the stock market, carefullystudies any potential investment. He is currently examining thepossibility of investing in the Trinity Power Company. Through

studying past performance, Walter has broken the potential resultsof the investment into five possible outcomes with accompanyingprobabilities. The outcomes of annual rates of return on a singleshare of stock that currently costs $150. Find the expected valueof the return for investing in a single share of Trinity Power. If

Walter purchases stock whenever the expected rate of returnexceeds 10 percent, will he purchase the stock according to the

following.

ROI ($) 0.00 10.00 15.00 25.00 50.00Probability 0.20 0.25 0.30 0.15 0.10


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Consider a case of fruit and vegetable wholesaler who sellsstrawberries. This product has very limited useful life. If not soldon the day of delivery, it is worthless. One case of strawberries

costs $20, and the wholesaler receives $50 for it. The wholesalercan not specify the number of cases customers will cal for on anyday, but her analysis of past records has produced the followinginformation. Develop a conditional loss table consideringobsolescence lossesand opportunity losses. Find the optimal stock

action based on expected loss.

Sales During100 Days

Daily sales Number ofDays Sold

Probability ofeach Number

Being Sold

10 15 0.15

11 20 0.20

12 40 0.40

13 25 0.25


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Important Discrete ProbabilityDistributions

Discrete Probability

Distributions

Binomial


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Binomial Probability Distribution

n Identical Trials

e.g. 15 tosses of a coin

2 Mutually Exclusive Outcomes on Each Trial e.g. Head or tail in each toss of a coin; defective or

not defective light bulb

Trials are Independent

The outcome of one trial does not affect theoutcome of the other


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Binomial Probability Distribution

Constant Probability for Each Trial

e.g. Probability of getting a tail is the same eachtime we toss the coin

(continued)


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Bernoulli Process:1. Each trial has only two possible outcomes

(heads or tails, yes or no, success or failure, trueor false)2. The probability of the outcome of any trialremain fixed over time.3. The trials are statistically independent:

p= probability of successq= probability of failure = 1-pr= number of success desiredn= number of trials undertaken

Probability of r success in n trial =

rnr

r

n

qpc


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The binomial distribution describes discretedata, resulting from an experiment knownas Bernoulli Process.

1. The tossing of a fair coin a fixed number of timesis a Bernoulli process and the outcome of such tossescan be represented by binomial probability

distribution.

2. The success or failure in a test may also bedescribed by Bernoulli Process.


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Ex: What is the probability of getting 2 heads if a fair coinis tossed 3 times

n=3 p=0.5r=2 q=0.5Probability of 2 success in 3 trial

1. When p is small, the distribution is skewed to right2. When p=0.5, the distribution is symmetrical3. When p >0.5, the distribution is skewed to the left.

375.0

0.525.03

0.50.53c12

2


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Case let 1. A Survey found that 65% of allfinancial consumer were very satisfied with theirprimary financial institutions. If this figure stillholds true today, suppose 40 financial consumerssampled randomly. What is the probability thatexactly 23 of the 40 are very satisfied with their

primary financial institution ?p = 0.65 q= 1-p=0.35, n= 40,

r=23

0784.35.65.0 172323

40 c


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For Binomial Distribution

Mean =

npqnp

Case let 2. The latest nationwide political poll indicatedthat for Indian, who are randomly selected, theprobability that they are conservative is 070%, theprobability that they are liberal is 0.15% and 0.15%

are in middle of the road. Assuming that theseprobability are accurate, answer the followingpertaining to a randomly chosen group of 10 Indian.


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a). What is the probability that 4 are liberal?

b). What is the probability that none are

conservative?

c). What is the probability that two are middle ofthe road?

d). What is the probability that at least eight areliberal?


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Binomial Probability DistributionFunction

!1

! !

: probability of successes given and

: number of "successes" in sample 0,1, ,

: the probability of each "success"

: sample size

n XXnP X p p

X n X

P X X n p

X X n

p

n

Tai ls in 2 Tos ses o f Coin

X P(X)

0 1/4 = .25

1 2/4 = .50

2 1/4 = .25


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Binomial DistributionCharacteristics

Mean

E.g.

Variance andStandard Deviation

e.g.

E X np 5 .1 .5np

n = 5p = 0.1

0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

1 5 .1 1 .1 .6708np p

21

1

np p

np p


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Binomial Distribution:

Demonstration Problemn

p

q

P X P X P X P X

20

06

94

2 0 1 2

2901 3703 2246 8850

.

.

( ) ( ) ( ) ( )

. . . .

P X( ))!

( )( )(. ) .. .

020!

0!(20 01 1 2901 2901

0 20 0

06 94

P X( ) !( )! ( )(. )(. ) .. .

1

20!

1 20 1 20 06 3086 3703

1 20 1

06 94

P X( )!( )!

( )(. )(. ) .. .

220!

2 20 2190 0036 3283 2246

2 20 2

06 94

G N Patel


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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 0.000

1 0.270 0.058 0.007 0.000 0.000 0.000 0.000 0.000 0.0002 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 0.00010 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

G N Patel


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Demonstration

Use of theBinomial Table

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 0.000

17 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

n

p

P X C

20

40

10 0117120 1010 10

40 60

.

( ) .. .

G N Patel


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Binomial Distribution using Table:

Demonstration Problem

n

p

q

P X P X P X P X

20

06

94

2 0 1 2

2901 3703 2246 8850

.

.

( ) ( ) ( ) ( )

. . . .

P X P X( ) ( ) . . 2 1 2 1 8850 1150

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

2

20 06 94 1 128

1 128 1 062

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

. .

n = 20 PROBABILITY

X 0.05 0.06 0.07

0 0.35850.29010.2342

1 0.37740.37030.3526

2 0.18870.22460.2521

3 0.05960.08600.1139

4 0.01330.02330.0364

5 0.00220.00480.0088

6 0.00030.00080.0017

7 0.00000.00010.0002

8 0.00000.00000.0000

20 0.00000.00000.0000

G N Patel


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Excels Binomial Function

n = 20

p = 0.06

X P(X)

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










G N Patel


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Graphs of Selected Binomial Distributions

n = 4PROBABILITY

X 0.1 0.5 0.90 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

0.400

0.500

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

0.900

1.000

0 1 2 3 4

X

P(X)

P = 0.9

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.000

0 1 2 3 4X

P(X

)

G N Patel

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Binomial Dist Fnl