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Binomial Probability Distribution

Binomial Probability Distribution. Permutations Number of possible arrangements

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Page 1: Binomial Probability Distribution. Permutations Number of possible arrangements

Binomial

Probability

Distribution

Page 2: Binomial Probability Distribution. Permutations Number of possible arrangements

Permutations

( 1)...( )( ) 1)...(1) !

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

n n n r r r nP

r r n r

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

5 3

5 4 3 2 1 5!5 4 3 60

2 1 2!P

Number of possible arrangements

Page 3: Binomial Probability Distribution. Permutations Number of possible arrangements

Combinations

Number of sets or groups

!

! !( )!n r

n r

n P nC

r r r n r

#

#

Arrangements of n things taken r at a time

Arrangements of r things

5 3

5! 5 4 3 6010

3!(2)! 3 2 1 6C

Page 4: Binomial Probability Distribution. Permutations Number of possible arrangements

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 23 2 1 0

3 3 2

3 2

(x 2) x 3x 2 3x 4 8

x 6x 12x 8

Page 5: Binomial Probability Distribution. Permutations Number of possible arrangements

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.

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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

Page 7: Binomial Probability Distribution. Permutations Number of possible arrangements

Example

Pat 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?

Page 8: Binomial Probability Distribution. Permutations Number of possible arrangements

Pat Statsdud…

Algebraically then: n=10, and P(success)= 1/5 = .20

Pat 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.

Page 9: Binomial Probability Distribution. Permutations Number of possible arrangements

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.

Page 10: Binomial Probability Distribution. Permutations Number of possible arrangements

Pat Statsdud…

n=10, and P(success) = .20

b. 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.

Page 11: Binomial Probability Distribution. Permutations Number of possible arrangements

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

Page 12: Binomial Probability Distribution. Permutations Number of possible arrangements

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

Page 13: Binomial Probability Distribution. Permutations Number of possible arrangements

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) = .0881

We 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…

Page 14: Binomial Probability Distribution. Permutations Number of possible arrangements

BinomialPDF & …CDF

Page 15: Binomial Probability Distribution. Permutations Number of possible arrangements

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

Page 16: Binomial Probability Distribution. Permutations Number of possible arrangements

Sampling Distribution of the Sample Proportion

Lock Section 6.1

Page 17: Binomial Probability Distribution. Permutations Number of possible arrangements

Binomial Population

Two Choices:SuccessFailure

Fixed ProbabilityIndependence

Page 18: Binomial Probability Distribution. Permutations Number of possible arrangements

Sampling Distribution

Page 19: Binomial Probability Distribution. Permutations Number of possible arrangements

P-Hat

p

Page 20: Binomial Probability Distribution. Permutations Number of possible arrangements

P-Hat Definition

nXp /ˆ

Page 21: Binomial Probability Distribution. Permutations Number of possible arrangements

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.

n

pppsd

)1()ˆ(

Page 22: Binomial Probability Distribution. Permutations Number of possible arrangements

Sampling Distribution of p-hat

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

np 10 and n(1- p) 10

Page 23: Binomial Probability Distribution. Permutations Number of possible arrangements

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:

npp

ppZ

)1(

ˆ

Page 24: Binomial Probability Distribution. Permutations Number of possible arrangements
Page 25: Binomial Probability Distribution. Permutations Number of possible arrangements

Confidence Intervals

for

Population Proportions

AP Statistics Chap 10-25

Lock Section 6.2

Page 26: Binomial Probability Distribution. Permutations Number of possible arrangements

Confidence Intervals

AP Statistics Chap 10-26

Population Mean

σ Unknown

Confidence

Intervals

PopulationProportion

σ Known

Page 27: Binomial Probability Distribution. Permutations Number of possible arrangements

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:

AP Statistics Chap 10-27

(1 )s

np

p p

n

p)p(1σp

Page 28: Binomial Probability Distribution. Permutations Number of possible arrangements

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

Page 29: Binomial Probability Distribution. Permutations Number of possible arrangements

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

AP Statistics Chap 10-29

/2

(1 )p pp z

n

Page 30: Binomial Probability Distribution. Permutations Number of possible arrangements

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

AP Statistics Chap 10-30

Page 31: Binomial Probability Distribution. Permutations Number of possible arrangements

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.

AP Statistics Chap 10-31

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)

Page 32: Binomial Probability Distribution. Permutations Number of possible arrangements

1-PropZInterval

Page 33: Binomial Probability Distribution. Permutations Number of possible arrangements

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.

AP Statistics Chap 10-33

Page 34: Binomial Probability Distribution. Permutations Number of possible arrangements

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

AP Statistics Chap 10-34

Page 35: Binomial Probability Distribution. Permutations Number of possible arrangements

Example 2

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

ˆ ˆ ˆ ˆˆ ˆ

α/2 α/2

p×( 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

Page 36: Binomial Probability Distribution. Permutations Number of possible arrangements

Example 2

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

Page 37: Binomial Probability Distribution. Permutations Number of possible arrangements

Thinking Challenge

You’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?

Page 38: Binomial Probability Distribution. Permutations Number of possible arrangements

Confidence Interval Solution

ˆ ˆ ˆ ˆˆ ˆ

a/2 a/2

p×( 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

Page 39: Binomial Probability Distribution. Permutations Number of possible arrangements

Required Sample Size

AP Statistics Chap 10-39

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)

Page 40: Binomial Probability Distribution. Permutations Number of possible arrangements

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)

AP Statistics Chap 10-40

Page 41: Binomial Probability Distribution. Permutations Number of possible arrangements

What sample size...?

AP Statistics Chap 10-41

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 pn

e

(continued)

Page 42: Binomial Probability Distribution. Permutations Number of possible arrangements
Page 43: Binomial Probability Distribution. Permutations Number of possible arrangements

HYPOTHESIS TESTS FOR SINGLE

PROPORTIONS

Lock Section 6.3

Page 44: Binomial Probability Distribution. Permutations Number of possible arrangements

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 = = = .0126

n 1000

o

o o

p - p .9 - .8Z =

p 1- p .8 .2

n 1000

7.9

p - value = 0

Page 45: Binomial Probability Distribution. Permutations Number of possible arrangements

Proportion Test