binomial probability distribution

BinomialProbability Distribution

Permutations

( 1)...( )( ) 1)...(1) !( )( 1)...(1) ( )!n r

n n n r r r nPr r n r

( 1( 2)...(1) !n nP n n n n

5 35 4 3 2 1 5! 5 4 3 60

2 1 2!P

Number of possible arrangements

Combinations

Number of sets or groups !

! !( )!n r

n r

n P nCr r r n r

##

Arrangements of n things taken r at a timeArrangements of r things

5 35! 5 4 3 60 10

3!(2)! 3 2 1 6C

Binomial Theorem

n nn k n k k n k

n k1 1

n(a b) a b C a b

k

3 3 0 2 1 1 2 0 33 3 3 3(x 2) x 2 x 2 x 2 x 2

3 2 1 0

3 3 2

3 2

(x 2) x 3x 2 3x 4 8x 6x 12x 8

Binomial Distribution The binomial distribution is the probability distribution that results from sampling a binomial population. Binomial samples have the following properties:

1. Fixed number of trials, represented as n.2. Each trial has two possible outcomes, a “success” and a

“failure”.3. P(success)=p (and thus: P(failure)=1–p), for all trials.4. The trials are independent, which means that the

outcome of one trial does not affect the outcomes of any other trials. May be population is large enough that sampling without replacement does not change values.

Success and Failure…

…are just labels for a binomial experiment, there is no value judgment implied.For example a coin flip will result in either heads or tails. If we define “heads” as success then necessarily “tails” is considered a failure (in as much as we attempting to have the coin lands heads up).Other binomial experiment notions:

An election candidate wins or loses An employee is male or female

ExamplePat Statsdud is a (not good) student taking a statistics course. Pat’s exam strategy is to rely on luck for the next quiz. The quiz consists of 10 multiple-choice questions. Each question has five possible answers, only one of which is correct. Pat plans to guess the answer to each question.• What is the probability that Pat gets no answers correct?

• What is the probability that Pat gets two answers correct?

Pat Statsdud…Algebraically then: n=10, and

P(success)= 1/5 = .20Pat plans to guess the answer to each question.

Is this a binomial experiment? Check the conditions:There is a fixed finite number of trials (n=10).

An answer can be either correct or incorrect.

The probability of a correct answer (P(success)=.20) does not change from question to question.

Each answer is independent of the others.

Pat Statsdud…n=10, and P(success) = .20

a. What is the probability that Pat gets no answers correct?i.e. # success, x, = 0; hence we want to know P(x=0)

Pat has about an 11% chance of getting no answers correctusing the guessing strategy.

Pat Statsdud…

n=10, and P(success) = .20b. What is the probability that Pat gets two answers correct?I.e. # success, x, = 2; hence we want to know P(x=2)

Pat has about a 30% chance of getting exactly two answerscorrect using the guessing strategy.

The binomial distribution with n trials and success probability p has:

Mean =

Variance =

Standard deviation =

2 1np p

2 1np p

np

Binomial Probability Distribution

Cumulative Probability…Thus far, we have been using the binomial probability distribution to find probabilities for individual values of x. To answer the question: (Example 10) “Find the probability that Pat fails the quiz”requires a cumulative probability, that is, P(X ≤ x)If a grade on the quiz is less than 50% (i.e. 5 questions out of 10), that’s considered a failed quiz.

Thus, we want to know what is: P(X ≤ 4) to answer

Pat Statsdud

P(X ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)We already know P(0) = .1074 and P(2) = .3020. Using the binomial formula to calculate the others:P(1) = .2684 , P(3) = .2013, and P(4) = .0881We have P(X ≤ 4) = .1074 + .2684 + … + .0881 = .9672

Thus, its about 97% probable that Pat will fail the test using the luck strategy and guessing at answers…

BinomialPDF & …CDF

Pat Statsdud Has Been Cloned

Suppose that a professor has a class full of students like Pat. What is the mean? What is the standard deviation?

The mean = μ = np = 10(0.2) = 2

The standard deviation isσ = √ np ( 1 – p ) = √ 10( .2)( 1 - .2)

= 1.26

Sampling Distribution of the Sample Proportion

Lock Section 6.1

Binomial PopulationTwo Choices:

SuccessFailure

Fixed ProbabilityIndependence

Sampling Distribution

P-Hat

p

P-Hat Definition

nXp /ˆ

Properties of p-hat When sample sizes are fairly large, the shape of

the p-hat distribution will be normal. The mean of the distribution is the value of the

population parameter p. The standard deviation of this distribution is

the square root of p(1-p)/n.

npppsd )1()ˆ(

Sampling Distribution of p-hat

For the sampling distribution to be normal, you must have:

np 10 and n(1- p) 10

Calculate Probabilities Because the shape of the distribution is

normal, we can standardize the variable p-hat to a Z standard normal distribution. Use Z-transform:

nppppZ

)1(ˆ

Confidence Intervals for

Population Proportions

AP Statistics Chap 10-25

Lock Section 6.2

Confidence Intervals


Population Mean

σ Unknown

ConfidenceIntervals

PopulationProportion

σ Known

Confidence Intervals for the Population Proportion, p Recall that the distribution of the sample

proportion is approximately normal if the sample size is large, with standard deviation

We will estimate this with sample data:


(1 )snp

p p

np)p(1σp

How Big???? Rule of Thumb 1: Formula for standard

deviation of p-hat only when the population, N, is at least 10 times the sample size: N ≥ 10 n

Rule of Thumb 2: The sampling distribution of p-hat is approximately Normal when

np ≥ 10 and n(1-p) ≥ 10

Confidence interval endpoints

Upper and lower confidence limits for the population proportion are calculated with the formula

where z is the standard normal value for the level of confidence desired p is the sample proportion n is the sample size


/2(1 )p pp zn

Example

A random sample of 100 people shows that 25 are left-handed.

Form a 95% confidence interval for the true proportion of left-handers


Example A random sample of 100 people shows

that 25 are left-handed. Form a 95% confidence interval for the true proportion of left-handers.


1.

2.

3.

p

25/100 .25

S (1 )/100 .25(.75)/100 .0433

p

p p

0.3349 . . . . . 0.1651(.0433) 1.96 .25

(continued)

1-PropZInterval

Interpretation

We are 95% confident that the true percentage of left-handers in the population is between

16.51% and 33.49%.

Although this range may or may not contain the true proportion, 95% of intervals formed from samples of size 100 in this manner will contain the true proportion.


Changing the sample size

Increases in the sample size reduce the width of the confidence interval.

Example: If the sample size in the above example is

doubled to 200, and if 50 are left-handed in the sample, then the interval is still centered at .25, but the width shrinks to

.19 …… .31


Example 2A random sample of 400 graduates showed 32 went to grad school. Set up a 95% confidence interval estimate for p.

ˆ ˆ ˆ ˆˆ ˆα/2 α/2p×( 1-p) p×( 1-p)p-Z × ≤p≤p+Z ×n n

.08×( 1-.08) .08×( 1-.08).08-1.96× ≤p≤.08+1.96×400 400 .053 p .107

Example 2 A random sample of 400 graduates showed 32 went to grad school. Set up a 95% confidence interval estimate for p.

Thinking ChallengeYou’re a production manager for a newspaper. You want to find the % defective. Of 200 newspapers, 35 had defects. What is the 90% confidence interval estimate of the population proportion defective?

Confidence Interval Solution

ˆ ˆ ˆ ˆˆ ˆa/2 a/2p×( 1-p) p×( 1-p)p-z × ≤p≤p+z ×n n

.175×( .825) .175×( .825).175-1.645× ≤p≤.175+1.645×200 200.1308≤p≤.2192

Required Sample Size


n)p(pze /2

1

Solve for n:

Define the margin of error:

2/2

e)p(pzn

12

p can be estimated with a pilot sample, if necessary (or conservatively use p = .50)

What sample size...?

How large a sample would be necessary to estimate the true proportion defective in a large population within 3%, with 95% confidence?

(Assume a pilot sample yields p-hat = .12)


What sample size...?


Solution:For 95% confidence, use Z = 1.96E = .03P-hat = .12, so use this to estimate p

So use n = 451

2 2

/22 2

(1 ) (1.96) (.12)(1 .12) 450.74(.03)

z p pne

(continued)

HYPOTHESIS TESTS FOR SINGLE

PROPORTIONS

Lock Section 6.3

Test of Proportions Old drug cures 80 percent of the time. New

drug for heartworm disease in dogs cures 90 percent of n=1000 dogs, so p-hat=.9.

What is the probability that this results occurred by random chance?

o op 1- p .8 .2SE = = = .0126n 1000

o

o o

p - p .9 - .8Z =p 1- p .8 .2

n 1000

7.9

p - value = 0

Proportion Test

binomial probability distribution

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