M15- Binomial Distribution 1 Department of ISM, University of Alabama, 1992-2003 Lesson Objectives ...

M15- Binomial Distribution 1 Department of ISM, University of Alabama, 1992-2003

Lesson Objectives Learn when to use the

Binomial distribution.

Learn how to calculate probabilitiesfor the Binomial using the formulaand the two tables in the book, andin Excel.

Bernoulli and BinomialDistributions

Chapter 6.2, 7.4


Two people in different rooms. “A” is shown one of the five cards,

selected randomly. “A” transmits his thoughts. “B” selects the card she thinks is being

sent to her, and records it. If the two cards match, a success occurs.

Does a person have ESP?

Experiment:

X = 1 success= 0 failure

Bernoulli (= ____ )

P(P(XX = 1) = ___ = 1) = ___ = = P(P(XX = 0) = ___ = 0) = ___ = 1- = 1-

X ~


A discrete data distribution used to describe a population

of binary variable values.

Binary one of only two outcomescan occur, coded as “0” or “1”

Bernoulli Distribution


Bernoulli distribution, “one” trial.

k 0 1k 0 1P(P(XX=k)=k)

= =

==The mean of of the Bernoulli is

The standard deviation is

Bernoulli has one parameter: = the probability of success.

Bernoulli ()X ~

1 1


Bernoulli distribution, “one” trial.

k 0 1k 0 1P(P(XX=k) =k) 1 1

22 = (0 - = (0 - ))22 • • (1 - (1 - )) + (1 - + (1 - ))22 • • = = 22 • (1 - • (1 - ) + (1 - ) + (1 - ))22 • • = = (1 - (1 - ) • [ ) • [ + (1 - + (1 - ) ] ) ]

= = 1 -1 - = = (1 - (1 - ))

0•(0•(1-1-) + 1•) + 1• = = ==

How are the “mean” and “standard deviation” determined?

How are the “mean” and “standard deviation” determined?

Once we know these results, we don’t need to derive it again.

Once we know these results, we don’t need to derive it again.

Use the same equations as the previous section.

a little algebra . . .


Examples of Bernoulli Variables

• Sex (male or female)

• Major (business or not business)

• Defective? (defective or non-defective)

• Response to a T-F question (true or false)

• Where student lives(on-campus or off-campus)

• Credit application result (accept or deny)

• Own home? (own or rent)

• Course result (Pass or Fail)


Bar Chart of Population for ESP

X0 1

1–

0

1.0


A discrete data distribution used to model a population of counts for

“n” independent repetitions of a Bernoulli experiment.

Binomial Distribution

Conditions:

1. a fixed number of trials, “n”.

2. all n trials must be independent of each other.

3. the same probability of success on each trial.

4. X = count of the number of successes.


Binomial distribution, “n trials”

k 0 1 2 3 k 0 1 2 3 n nP(P(XX = k) = k)

==

==The mean of of the Binomial is

The standard deviation is

Binomial has two parameters: n = the fixed number of trials, = the probability of success for each trial.

to be determined

X ~ Bino( n, )


Computing Binomial probabilities

Formula gives P(X = k), the probability for exactly one value.

Tables Table A.1 gives P(X = k),the probability for exactly one value.

Excel (BINOMDIST function),gives either individual or cumulative.

or Table A.2 gives P(X < k),the cumulative probability forX = 0 through X = k.

For selected values of n and .


Examples of Binomial Variables

A count of the number of females in a sampleof 80 fans at a Rolling Stones concert.

A count of the number of defectives in sampleof 50 tires coming off a production line.

A count of the number of the number of corrects answers on 10 true-false questionsfor which everybody guessed.

A count of the number of credit applications that are denied from a sample of 200.


Are these situation Binomial?

A count of the number of defectives tires coming off a production line in one year.

Count of people choosing Dr Pepper over Pepsi in a blind taste-test with 20 people.

A count of the number of playing cards that are diamonds in a sample of 13 cards.

A count of the number of credit applications that are accepted from a sample of 200.

A count of the number of fans at a Stones concert needed until we find 50 females.


If the population of X-values has a binomial data distribution, then the proportion of the population having the value k is given by:

This is also the probability that a single value of X will be exactly equal to x.

P( X = k ) = ( )nk

pk (1 – p)n–k

for k = 0, 1, 2, ..., n


P( X = k ) = ( )nk

pk (1 – p)n–k

Big “X” is the Big “X” is the “random variable.”“random variable.”

Little “k” is a Little “k” is a “specific value”“specific value”

of Big X.of Big X. Example:Example:

P( P( XX = k) = k) P( Count = 4)P( Count = 4)


P( X = k ) = ( )nk

pk (1 – p)n–k

( )( )nnkk

= =

n! =n! =

0! =0! =


P( X = x ) = ( )nx

px (1 – p)n–x

These are the These are the possible values of X;possible values of X;

each value has each value has its own probability.its own probability.

x = 0, 1, 2, ..., n


X = a count of the number of successes.X ~Bino(n=5, =.20)


Experiment:

Repeat the experiment 5 times.

Find the probability of exactly oneone success.

P(X=1) = 51

.20 .801 4

= 5 • .20 • .4096= .4096

P(X=3) =


Bino(n=5, =.20)

ESP Experiment

Find the probability of two or fewertwo or fewer successes.

+ 5•.20•.4096

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)

50

.20 .800 5= +

51

.20 .801 4 +

52

.20 .802 3

= 1• 1•.32768 + 10•.04•.5120

= .32768 + .4096 + .2048

= .94208.94208

Table A.1 gives P( X = k )

P( X = 2 ) =

Bino(n=5, =.20)ESP Experiment

.2048 .9421P( X < 2 ) =

Table A.2 gives P( X < k )

CumulativeIndividual

k:

0.2

.3277 .4096 .2048 .0512 .0064 .0003

012345

n = 5

k:

0.2

012345

n = 5

1.0000

.3277

.7373

.9421 .9933 .9997


k 0.2

.1678 .5033 .7969 .9437 .9896 .9988 .99991.000-1.0000

012345678

For the multiple choice test with 8 questions, 5 choices for each, find the probability of getting two or fewer correct.

About 80% of the classshould have two orfewer correct;about 20% should havethree or more correct.

About 80% of the classshould have two orfewer correct;about 20% should havethree or more correct.

Table A.2 gives P( X k )Bino(n=8, =.20)



012345678

.1678 .5033 .7969 .9437 .9896 .9988 .99991.000-1.0000

k 0.2 For the multiple choice test, find the probability of more the one but no more than three correct.

Same question as . . . “two or more but less than four;”or “exactly two or three.”

Same question as . . . “two or more but less than four;”or “exactly two or three.”


k

P( X 3)

= 1.000 – P(X 2)= 1.000 – .7969 = .2031


012345678

.1678 .5033 .7969 .9437 .9896 .9988 .99991.000-1.0000

k 0.2 For the multiple choice test find the probability of three or more.

Same question as . . . “more than two”Same question as . . . “more than two”


P( 3 or more correct)=

P(more than 3 correct)

=

012345678

.1678 .5033 .7969 .9437 .9896 .9988 .99991.000-1.0000

k 0.2

Watch the wording!Watch the wording!



Table A.2 gives P( X c )Bino(n=8, =.20)

012345678

.1678 .5033 .7969 .9437 .9896 .9988 .99991.000-1.0000

k 0.2 Find the probability ofat least one correct.


X = a count of the number of successes.X ~Bino(n=20, =.20)


Experiment:

Repeat the experiment 20 times.

Beth and Mike matched 10 times? Is this unusual?

P(X 10) =

=

=

This is a rare event! The evidence indicatesthat they do better thanguessing.

This is a rare event! The evidence indicatesthat they do better thanguessing.

The probability of observing a result this extreme is 0.0026.

The probability of observing a result this extreme is 0.0026.


Binomial Distribution

0.000

0.050

0.100

0.150

0.200

0.250

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

X

P(X

= x

)

X = a count of the number of successes.X ~


Bino(n=20, =.20)Expected value = _______

Standard deviation =


Free throws

Are shots independent? Evidence says yes. Suppose probability

of Mo making one free throw is .70.

Mo will shoot 9 shots.


Let X = count of shots made,

Find probability he misses fewer than three.

So, let Y = count of misses.

Y ~ B ( )

n = 9, = .70

n = 9, = _____

X ~ B ( )


Y 0 1 2 3 4 5 6 7 8 9 Misses

X 9 8 7 6 5 4 3 2 1 0 Hits

P( Y < 3 ) =

?

.70

Count

P( Y 2 ) =

P( X > 6 ) = 1 - P( X 6 )

= =Lookup in Tables.


Moral:

Pay attention to what you are counting!

M15- Binomial Distribution 1 Department of ISM, University of Alabama, 1992-2003 Lesson Objectives ...

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Transcript of M15- Binomial Distribution 1 Department of ISM, University of Alabama, 1992-2003 Lesson Objectives ...