Binomial Random Variables

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Binomial Random

Variables

Dr Tom Ilvento

Department of Food and Resource Economics

Overview

•

A special case of a Discrete Random Variable is the Binomial

•

This happens when the result of the experiment is a dichotomy

•

Success or Failure

•

Yes or No

•

Cured or not Cured

•

If the discrete random variable is a binomial, we have some easier ways to solve for probabilities

•

Formula

•

Probability Table

•

And the solution for the Mean and Variance is much easier to solve

2

Binomial Random Variable

•

In many cases the responses to an experiment are dichotomous

•

Sucess/Failure

•

Yes/No

•

Alive/Dead

•

Support/Don’t Support

•

When our focus is conducting an experiment n times independently and observing the number x of times that one of the two outcomes occurs (Success)

•

And the probability of success, p, remains the same from trial to trial

•

This X is a Binomial Random Variable

•

We can exploit this by using known formulas for a Binomial Probability Distribution

3

Conduct an experiment n times and observe the number x of times that Success occurs

Characteristics of a Binomial

Distribution

•

The experiment consists of n identical trials

•

There are only two outcomes on each trial. Outcomes can be denoted as

•

S for Success

•

F for Failure

•

The probability of S (success) remains the same from trial to trail

•

Denoted as p the proportion

•

The probability of F (failure)

•

Denoted as q q=(1-p)

•

The trials are independent of each other

•

The binomial random variable x is the number of

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Example of a Binomial Random

Variable: Marketing Survey

•

Marketing survey of 100 randomly chosen consumers

•

Record their preferences for a new and an old diet soda –

ask them to choose their preference

•

Let x be number of 100 who choose the new brand

•

This is a binomial random variable

5 Conduct an experiment 100 times and observe the

number x of times that the subject chooses the new brand

Example of a Binomial Random

Variable: Fitness Example

•

Heart Association says only 10% of adults over 30 can pass the fitness test

•

Suppose 4 people over 30 are selected at random

•

Let X be the number who pass the minimum requirements

•

Find the probability distribution for X

number x of times that pass occurs

How to solve this using the strategy

of a Discrete Random Variable

1. List the events

2. List the sample points that refer to that event 3. Calculate the probabilities

•

p = .1 and q = (1.0 - .1) = .9

•

I multiply through on the probabilities because each trial is independent of the others ₇

Event X Sample Points Probability

All Fail FFFF (.9)(.9)(.9)(.9) = .6561

Can you solve it – the probability that

exactly 1 person passes the test?

•

Count the ways we could have only one pass, and three failures

•

Assign probabilities to this event

8

•

SFFF FSFF FFSF FFFS

•

For each combination, the probabilities are:

•

.1*.9*.9*.9

•

And there are four ways to get one pass

•

4[.1*.9*.9*.9] = .2916

•

Another way to write it is

(3)

Let’s finish solving for the whole table

The number of times that an adult passes in a sample of four

9

Event X Sample Points Probability

0 All Fail FFFF (.9)(.9)(.9)(.9) = .6561 1 One passes SFFF FSFF FFSF FFFS 4[(.1)(.9)3] = .2916 2 Two Pass SSFF SFSF SFFS FSSF FSFS FFSS 6[(.1)2(.9)2] = .0486 3 Three Pass SSSF FSSS SFSS SSFS 4[(.1)3(.9)] = .0036 4 Four Pass SSSS (.1)(.1)(.1)(.1) = .0001

Probability Distribution

•

When x = 0 All Fail P = .6561

•

When x = 1 One Pass P = .2916

•

When x =2 Two pass P =.0486

•

When x=3 Three pass P = .0036

•

When x=4 Four pass P = .0001

10 0 0.175 0.350 0.525 0.700 0 1 2 3 4 P(X) X 0 1 2 3 4 P(X) 0.6561 0.2916 0.0486 0.0036 0.0001

Fitness Example

•

Find the probability that none of the adults pass the test

•

P(x=0) = .6561

•

Find the probability that 3 of 4 adults pass the test

•

P(x=3) = .0036

•

What is the probability that 2 or more adults pass the test?

•

P(x=2) + P(x=3) + P(x=4) = .0486 + .0036 + .0001 = .0523

11

X 0 1 2 3 4

P(X) 0.6561 0.2916 0.0486 0.0036 0.0001

Binomial Probability Distribution

Formula

•

Sometimes the number of trials gets large

•

We can also use the binomial probability distribution formula to generate the probabilities

•

It uses factorial notation

•

n! = n(n-1)(n-2)…(n-(n-1))

•

5! = 5x4x3x2x1 = 120

•

0! = 1, 1!=1, 2!=2x1=2, …

•

The formula for any x in n trials is:

12 x n x

_q

p

x

n

x

n

x

P

!

=

(

)

(

)

)!

(

!

)

(

(4)

What defines the Binomial

Distribution?

•

p = Probability of a success on a single trial

•

q = (1-p) probability of failure

•

n = number of trials

•

x = number of successes in n trials

13 x n x

_q

p

x

n

x

n

x

P

!

=

(

)

(

)

)!

(

!

)

(

Note: it uses the Combinatorial Rule as

the first part of the formula

This part reflects the probabilities with each

combination

For x=3 in the fitness example, n=4, p=.1

•

The four is how many combinations of 3 success in 4

•

The last part of the formula is the probability associated with each of these combinations

•

The probability, .0036, is the exact same one we

calculated earlier 14 x n x _q p x n x n x P ! ! = ( )( ) )! ( ! ! ) ( 3 4 3

_(.

₉

₎

)

1 (.

)!

3

4 (

!

3 !

4 )

3 (

!

=

P

P(3) =

4 ! 3! 2 !1

3! 2 !1

(

)

_{( )}

1 (.1)

3

_(.9)

4"3

P(3) =

24

6 (.1)

3

(.9)

4!3

P(3) = 4(.1)

3

(.9)

4!3

P(3) = 4(.001)(.9)

P(3) = 4(.0009) = .0036

Your Try it:

For x=2 in the fitness

example, n=4, p=.1

•

I will get you started

15 x n x _q p x n x n x P ! ! = ( )( ) )! ( ! ! ) (

!

P(2) =

4!

2!(4 " 2)!

(.1)

2

(.9)

4"2

P(2) = 6(.0081) = .0486

Mean and Variance for a

Binomial Random Variable

•

Since a binomial is only a dichotomy, the formulas for the mean and the standard deviation will simplify

•

From µ = !(x!P(x))

•

To µ = n!p

•

Our fitness example: µ = 4*.1 = .4

•

The Variance changes from

•

From "2_{= ![(x-µ)}2_!P(x))]

•

To !2_{= n*p*q}

•

Our fitness example: !2_{= 4*.1* .9 = .36}

(5)

I could have solved for the mean using the

formula for discrete random variables

•

To solve for the mean I would use this

formula from the discrete random variable lecture:

•

E(x) = (0)(.6561) + (1)(.2916) + (2) (.0486) + (3)(.0036) + (4)(.0001)

•

E(x) = .4

•

Binomial approach

•

E(x) = n·p = 4·(.1) = .4

•

The Binomial approach is much easier 17

µ

=

!

=

_"

= n i i i

P

x

E

1

)

(

)

(

If I know my Discrete Random Variable is distributed as a binomial random variable, it will make things much easier

I could have solved for the variance using

the formula for discrete random variables

•

To solve for the variance I would have:

•

E(x-µ)2_{= (0 -.4)}2_{(.6561) + (1-.4)}2 (.2916) + (2-.4)2_{(.0486) + (3-.4)}2 (.0036) + (4-.4)2_(.0001)

•

!2_{= .36}

•

Binomial approach

•

E(x) = n·p·q = 4·(.1)(.9) = .36

•

The Binomial approach is much easier

18

If I know my Discrete Random Variable is distributed as a binomial random variable, it will make things much easier

2 1 2 2

)

(

)

(

]

)

[(

"

µ

=

_#

"

µ

=

!

= n i i i

P

x

E

Return to the Nitrous Oxide

Example

•

Suppose we were recording the number of dentists that use nitrous oxide (laughing gas) in their practice

•

We know that 60% of dentists use the gas.

•

p = .6 and q = .4

•

Let X = number of dentists in a random sample of five dentists use use laughing gas.

•

n = 5

•

This is a Binomial Random Variable!

number x of times that use Nitrous Oxide

Nitrous Oxide Example

•

How to solve for these probabilities?

20 x n x

_q

p

x

n

x

n

x

P

!

=

(

)

(

)

)!

(

!

)

(

X 0 1 2 3 4 5 P(X) 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778

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Probability Distribution for the

Nitrous Oxide Example

• µ = 3.00

• !

2

_{= 1.20}

• ! = 1.01

• **µ = 5*.6 = 3.00**

• !

2

**_{= 5.6.4 = 1.20}**

• ! = SQRT(1.20) = 1.01

21 X 0 1 2 3 4 5 P(X) 0.0102 0.0768 0.2304 0.3456 0.2592 0.0778 0 0.1 0.2 0.3 0.4 0 1 2 3 4 5 Probability Distribution of X P(X)

Nitrous Oxide Example using

Excel

•

Open up the file, BINOM.xls

•

Click on the Worksheet Problem

•

This worksheet is designed to

solve problems up to n=50, for any value of p

•

You enter in:

•

p = .6

•

n = 5

•

The spreadsheet will do the rest!

22 Reverse p = 0.6000 X p(X) Cum p(X) Cum p(>=X) q = 0.4000 0 0.0102 0.0102 1.0000 n = 5 1 0.0768 0.0870 0.9898 2 0.2304 0.3174 0.9130 Mean 3.0000 3 0.3456 0.6630 0.6826 Variance 1.2000 4 0.2592 0.9222 0.3370 Std Dev 1.0954 5 0.0778 1.0000 0.0778

Binomial Formula using Excel

•

In Excel, the formula for the Binomial Distribution function is:

•

BINOMDIST(X,N,P,cumulative)

•

X is the number of successes

•

N is the number of independent trials

•

P is the probability of success on each trial

•

Cumulative is an argument - you enter TRUE or FALSE

•

Entering TRUE gives a cumulative probability up to and including X successes (or 1)

•

Entering FALSE gives the exact probability of X successes in N trials (or 0)

23 BINOMDIST(3,5,.6,TRUE)

Binomial Formula using Excel

•

For our example of dentists

•

BINOMDIST(2,5,.6,TRUE)

•

cumulative probability up to and including 2 successes

•

= .3174

•

BINOMDIST(2,5,.6,FALSE)

•

the exact probability of X successes in N trials

•

= .2304

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Binomial Table

•

Another way to get probabilities form Binomial Random Variables is via a table

•

In exams, I will give you a table which contains cumulative probabilities for n= 5, 6, 7, 8, 9, 10, 15, 20, and 25

•

Each table lists values of P across the top

•

P = .01, .05, .1, .2, .3, …, .95, .99

•

x = # of successes as the rows

•

It is a Cumulative Table

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Binomial Table

for n = 5

•

The probability associated with p=.3 and x = 4 is .998

•

This means that the cumulative probability, or P(x " 4) = .998

•

The actual probability of P(x = 4) = .998 - .969 = .029

•

You must subtract two values to get the actual probability of x

26 The values shown are cumulative probabilities for the probability of x (denoted as k in the table)

PROBABILITIES (p) x 0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.99 0 0.951 0.774 0.590 0.328 0.168 0.078 0.031 0.010 0.002 0.000 0.000 0.000 0.000 1 0.999 0.977 0.919 0.737 0.528 0.337 0.188 0.087 0.031 0.007 0.000 0.000 0.000 2 1.000 0.999 0.991 0.942 0.837 0.683 0.500 0.317 0.163 0.058 0.009 0.001 0.000 3 1.000 1.000 1.000 0.993 0.969 0.913 0.813 0.663 0.472 0.263 0.081 0.023 0.001 4 1.000 1.000 1.000 1.000 0.998 0.990 0.969 0.922 0.832 0.672 0.410 0.226 0.049 5 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 •The table is arranged cumulatively

•For each probability, the value in the cell is the cumulative probability up to and including X

•The last row (in this case for x = 5), the cumulative probability is 1.000

Nitrous Oxide Example

•

Use the n = 5 Table for p = .6

•

Solve the probability for x = 3

•

P(x " 3) = .663

•

P(x " 2) = .317

•

P(x=3) = .663 - .317 = .346

•

This is the same value (with some rounding error) that we calculated

using the formula (.3456) ₂₇

The values shown are cumulative probabilities for the probability of x (denoted as k in the table)

PROBABILITIES (p) x 0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.99 0 0.951 0.774 0.590 0.328 0.168 0.078 0.031 0.010 0.002 0.000 0.000 0.000 0.000 1 0.999 0.977 0.919 0.737 0.528 0.337 0.188 0.087 0.031 0.007 0.000 0.000 0.000 2 1.000 0.999 0.991 0.942 0.837 0.683 0.500 0.317 0.163 0.058 0.009 0.001 0.000 3 1.000 1.000 1.000 0.993 0.969 0.913 0.813 0.663 0.472 0.263 0.081 0.023 0.001 4 1.000 1.000 1.000 1.000 0.998 0.990 0.969 0.922 0.832 0.672 0.410 0.226 0.049 5 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

Nitrous Oxide Example

•

Use the n = 5 Table for p = .6

•

Solve the probability for x > 3

•

P(x " 3) = .663

•

P(x>3) = 1 - .663 = .337

•

Solve the probability for x " 2

•

P(x " 2) = .317

28 The values shown are cumulative probabilities for the probability of x (denoted as k in the table)

PROBABILITIES (p) x 0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.99 0 0.951 0.774 0.590 0.328 0.168 0.078 0.031 0.010 0.002 0.000 0.000 0.000 0.000 1 0.999 0.977 0.919 0.737 0.528 0.337 0.188 0.087 0.031 0.007 0.000 0.000 0.000 2 1.000 0.999 0.991 0.942 0.837 0.683 0.500 0.317 0.163 0.058 0.009 0.001 0.000 3 1.000 1.000 1.000 0.993 0.969 0.913 0.813 0.663 0.472 0.263 0.081 0.023 0.001 4 1.000 1.000 1.000 1.000 0.998 0.990 0.969 0.922 0.832 0.672 0.410 0.226 0.049 5 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

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The Rare Event Approach

•

What if we had 5 dentists selected randomly and none of them used nitrous oxide?

•

Given p=.6, this would be a very rare event

•

P(x=0) = .010

•

This is possible, but not probable

•

Was this just by chance????

•

Or was the assumption wrong – that p =.6?

29

Summary

•

The Binomial is a special form of the discrete random variable

•

There are other discrete random variables - poisson

•

If you know it is a Binomial Random Variable it makes it

easy to solve for probabilities, the mean and the variance

•

For probabilities you can use:

•

The Binomial Formula

•

The Binomial Tables

•

Excel also has functions to solve for binomials