8.1 – The Binomial Distributions

8.1 – The Binomial Distributions

When there are two outcomes to a setting it is said to be a binomial setting. One of success and failure.

• The number of heads when flipping a coin

• The number of free throws made

• The number of girls born in a family

The Binomial Setting: (BINS)

B: Binomial, Success or Failure

I: The n observations are independent

N: There is a fixed number of observations, n

S: The same probability of success for each observation

If the conditions are met, then:

The random variable X = the number of successes in n observations.

It has a binomial distribution, and is denoted as X ~ B(n,p)

Example #1: Identify each as a binomial distribution.

According to the Textbook of Medical Physiology, 9% of the population has blood type B. Suppose we choose 8 people at random from the population and test the blood type of each. Assume that each measurement is independent from the next.

B: Binomial, Success = Blood type B or Failure: Not B

I: Says to assume

N: n = 8

S: Each person has a probability of 0.09 to have blood type B.

Yes, Binomial.X ~ B(8, 0.09)

Let X = # of people with blood type B.

Example #1: Identify each as a binomial distribution.

Privacy is a concern for many users of the Internet. One survey showed that 59% of all Internet users are somewhat concerned about the confidentiality of their email. Assume opinions of users are independent of each other and we select a random sample of 10 Internet users.

B: Binomial, Success = Concern for confidentiality or Failure: Not concerned

I: Says to assume

N: n = 10

S: Each person has a probability of 0.59 to be concerned

Yes, Binomial.

X ~ B(10, 0.59)

Let X = # of people with concern for confidentiality

Example #1: Identify each as a binomial distribution.

Deal 10 cards from a shuffled deck and count the number of X red cards. There are 10 observations, and each gives either a red or black card.

B: Binomial, Success = Red card or Failure: Not Red

I: Not independent! When you take one, it changes the probability of the next one

No, Not Binomial.

Binomial Coefficient:

The number of ways to arrange k success among n observations (n choose k):

n

k

!

! !

n

k n k

Math – Prob – nCr

nCk

Example #2

Find how many ways you can arrange 5 out of 7 objects.

n

k

42

2

!

! !

n

k n k

7!

5!(7 5)!

7!

5!2!

7 6 5!

5! 2 1

7 6

2 1

21

OR: 7 nCr 5 = 21

Example #3

Suppose I have a group of 4 students and I want to choose 1 of them as a volunteer. In how many ways can I choose 1 out of 4 students?

n

k

!

! !

n

k n k

4!

1!(4 1)!

4!

1!3!

4 3!

3!

4

OR: 4 nCr 1 = 4

Binomial Probability:

If X has the binomial distribution with n observations and probability p of success on each observation, the possible values of X are:

knk ppk

n

)1(P(X = k) =

2nd – Dist – binompdf(n, p, k)

Example #4According to the Textbook of Medical Physiology, 9% of the population has blood type B. Suppose we choose 8 people at random from the population and test the blood type of each. Assume that each measurement is independent of the next. The number of people we observe with blood type B has the following binomial distribution.

a. What is the probability that 3 of these people have blood type B?

(1 )k n knp p

k

P(X = 3) = 3 8 38

0.09 (1 0.09)3

5(56)(0.000729)(0.91) 0.02548

OR: binompdf( n, p, k)

binompdf( 8, 0.09, 3) = 0.02548

Example #4According to the Textbook of Medical Physiology, 9% of the population has blood type B. Suppose we choose 8 people at random from the population and test the blood type of each. Assume that each measurement is independent of the next. The number of people we observe with blood type B has the following binomial distribution.

b. What is the probability that no more than 1 of these people have blood type B?

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

(1 )k n knp p

k

0 8 080.09 (1 0.09)

0

8(1)(1)(0.91) 0.47025

P(X = 0) =

P(X = 1) = (1 )k n knp p

k

1 8 180.09 (1 0.09)

1

7(8)(0.09)(0.91) 0.372068

P(X = 0) + P(X = 1) = 0.47025 + 0.372068 = 0.84232

OR:binompdf( n, p, k)

binompdf( 8, 0.09, 0) + binompdf( 8, 0.09, 1) =

0.84232

Binomial Probability:

If you are looking for more than one value, you can find that number and less than using the cumulative density function.

2nd – Dist – binomcdf(n, p, k)

(Always finds the probability of less than or equal to)

Example #5A biologist is studying a new hybrid tomato. It is known that the seeds of this hybrid tomato have probability .30 of germinating. The biologist plants 7 seeds. What is the probability that less than 3 seeds will germinate?

B: Binomial, Success = Germinating Tomato or Failure: Not germinating

I: It is safe to assume independence

N: n = 7

S: Each plant has a probability of 0.30 of germinating

Yes, Binomial.

X ~ B(7, 0.30)

Let X = The number of germinating tomatoes

Example #5A biologist is studying a new hybrid tomato. It is known that the seeds of this hybrid tomato have probability .30 of germinating. The biologist plants 7 seeds. What is the probability that less than 3 seeds will germinate?

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

binomcdf(7, 0.30, 2) = 0.64707

Where n = 7, p = 0.30, k = 2,

Example #6The manager of The Burger Eatery notes that there is a 40% chance that the next person who comes in will order a hamburger and a 60% chance that they will order a cheeseburger. What is the probability that, out of the next five customers, at least two will order a hamburger?B: Binomial, Success = Next person orders a burger or Failure:

Not order a burgerI: It is safe to assume independence

N: n = 5

S: Each person has a probability of 0.40 of ordering a burger

Yes, Binomial.

X ~ B(5, 0.40)

Let X = The number hamburgers ordered

Example #6The manager of The Burger Eatery notes that there is a 40% chance that the next person who comes in will order a hamburger and a 60% chance that they will order a cheeseburger. What is the probability that, out of the next five customers, at least two will order a hamburger?

P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

1 – binomcdf(5, 0.40, 1) = 1 – 0.33696 = 0.66304

Where n = 5, p = 0.40, k = 1,

Mean and Standard Deviation of a Binomial RV:

If a count X has the binomial distribution with number of observations n and probability of success p, then:

(1 )X np p

X np

Example #7Jim enjoys playing basketball. He makes approximately 50% of his attempts during a game. If he attempts 6 shots in a game, what is the mean and standard deviation of baskets Jim will make?

(1 )X np p

X np 6 0.50 = 3

(6)(0.5)(1 0.5)

1.5 1.2247

Let X = # of baskets made in a game

Example #8The management for the mall claims that 20% of the parking space around the mall area is used by the local movie theater goers. A sample of 10 parking spaces was randomly selected. What is the expected number of spaces used by the movie theater goers in the sample? What is the standard deviation of spaces used by the movie theater goers in the sample?

(1 )X np p

X np 10 0.20 = 2

(10)(0.2)(1 0.2)

1.6 1.2649

Let X = # of parking spaces being used by movie goers

Normal Approximation to the Binomial Distribution:

You can use the normal distribution to estimate the probability of a binomial distribution if:

Rule of Thumb:

np ≥ 10 and n(1 – p) ≥ 10

Then: X ~ B(n, p) can be approximated by

( , ) ( , (1 ))X XN N np np p

Example #9A recent survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” Suppose that 60% of all adults would agree if asked this question.

a. Are we able to use the normal distribution?

Rule of Thumb:

np ≥ 10 and n(1 – p) ≥ 10

(2500)(0.60) ≥ 10

1500 ≥ 10

(2500)(1 – 0.60) ≥ 10

1000 ≥ 10

Yes

Example #9A recent survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” Suppose that 60% of all adults would agree if asked this question.

b. What is the mean and standard deviation?

(1 )X np p

X np 2500 0.60 = 1500

(2500)(0.6)(1 0.6)

600 24.494897

Let X = # of people who agree with the quote

Example #9A recent survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” Suppose that 60% of all adults would agree if asked this question.

c. What is the probability that less than 1450 adults would agree if asked this question?

=1500

= 24.49

1450

Z = x –

1450 – 1500 24.49=

– 50 .

24.49=

– 2.04=

Example #9A recent survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” Suppose that 60% of all adults would agree if asked this question.

c. What is the probability that less than 1450 adults would agree if asked this question?

P(Z < – 2.04) =

Example #9A recent survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” Suppose that 60% of all adults would agree if asked this question.

c. What is the probability that less than 1450 adults would agree if asked this question?

P(Z < – 2.04) = 0.0207

OR: normalcdf( -1000000000, 1450, 1500, 24.49) =

0.0206

Example #9A recent survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” Suppose that 60% of all adults would agree if asked this question.

d. Compare your answer to ‘c’ with the binomial probability setting.

binomcdf( n, p, k)

binomcdf( 2500, 0.60, 1449) = 0.0198

The binomial probability is slightly smaller, but it is close to the normal approximation.

Where n = 2500, p = 0.60, k = 1449,

Example #10The general ethnic profile of Denver is about 42% minority and 58% Caucasian. Suppose the city of Denver recently hired 60 new grounds and maintenance workers. It claimed that the hiring practice is completely impartial without regard to ethnic background. However, 12 of the new employees were minorities, and a complaint has been filed.

a. Are we able to use the normal approximation?

Rule of Thumb:

np ≥ 10 and n(1 – p) ≥ 10

(60)(0.42) ≥ 10

25.2 ≥ 10

(60)(1 – 0.42) ≥ 10

34.8 ≥ 10

Yes

Example #10The general ethnic profile of Denver is about 42% minority and 58% Caucasian. Suppose the city of Denver recently hired 60 new grounds and maintenance workers. It claimed that the hiring practice is completely impartial without regard to ethnic background. However, 12 of the new employees were minorities, and a complaint has been filed.

b. What is the mean and standard deviation?

(1 )X np p

X np 60 0.42 = 25.2

(60)(0.42)(1 0.42)

14.616 3.8231

Let X = # of grounds & maintenance workers that are minorities

Example #10The general ethnic profile of Denver is about 42% minority and 58% Caucasian. Suppose the city of Denver recently hired 60 new grounds and maintenance workers. It claimed that the hiring practice is completely impartial without regard to ethnic background. However, 12 of the new employees were minorities, and a complaint has been filed.

c. What is the probability that 12 or less minorities are hired as new grounds and maintenance workers?

Z = x –

12 – 25.2 3.823=

–13.2.

3.823=

– 3.45==25.2

= 3.823

12

P(Z < – 3.45) =

Example #10The general ethnic profile of Denver is about 42% minority and 58% Caucasian. Suppose the city of Denver recently hired 60 new grounds and maintenance workers. It claimed that the hiring practice is completely impartial without regard to ethnic background. However, 12 of the new employees were minorities, and a complaint has been filed.

c. What is the probability that 12 or less minorities are hired as new grounds and maintenance workers?

P(Z < – 3.45) = 0.0003

OR: normalcdf( -1000000000, 12, 25.2, 3.823) =

0.0002775

Example #10The general ethnic profile of Denver is about 42% minority and 58% Caucasian. Suppose the city of Denver recently hired 60 new grounds and maintenance workers. It claimed that the hiring practice is completely impartial without regard to ethnic background. However, 12 of the new employees were minorities, and a complaint has been filed.

c. What is the probability that 12 or less minorities are hired as new grounds and maintenance workers?

8.2 – The Geometric Distributions

The Geometric Setting: (BINS)

B: Binomial: Success and Failure

I: Observations are independent

N: The number of trials needed to get a success

S: The probability of success is the same for each observation

If the conditions for a geometric distribution are met, then:

X = # of trials needed for the first success

Geometric Probability:

The probability of the first success on the nth trial is:

P(X n) (1 p)n1p

2nd – Dist – geometpdf(p, k)

For first success:

2nd – Dist – geometcdf(p, k)

For success at or before a number:

Geometric Probability:

The probability that it takes more than n trials to see the first success:

P(X n) (1 p)n

Geometric Probability:

X 1

p

X 1 pp2

Example #1Suppose the proportion of people who own a pet in Pleasant City is 0.87. a. Identify the distribution and define X.

B: Binomial: Success= own pet and Failure = don’t

I: Safe to assume independence.

N: The number of trials needed to find a pet owner

S: The probability of a pet owner is the same, 0.87

Lex X = # of people needed to find a pet owner

Example #1Suppose the proportion of people who own a pet in Pleasant City is 0.87.

b. If a person is chosen at random, what is the probability of choosing a pet owner on the 3rd try?

P(X 3) (1 p)n 1 p

(1 0.87)3 1(0.87)

0.132 • 0.87 = 0.01473

OR: geometpdf(0.87, 3) = 0.014703

Example #1Suppose the proportion of people who own a pet in Pleasant City is 0.87.

c. Before the 3rd try?

geometcdf(0.87, 2) = 0.9831

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

Let p = 0.87 and k = 0, 1, and 2

Example #1Suppose the proportion of people who own a pet in Pleasant City is 0.87.

d. What is the probability that you find such a pet owner after the 3rd try?

P(X 3) (1 p)n

(1 0.87)3 0.002197

Example #1Suppose the proportion of people who own a pet in Pleasant City is 0.87.

e. How many people would you expect to ask before finding a pet owner?

X 1

p

1

0.87 1.149

Example #1Suppose the proportion of people who own a pet in Pleasant City is 0.87.

f. If 30 people are chosen at random, how many pet owners should be expected in the sample?

X np

Note: Now n is fixed, and it is binomial

30 • 0.87 = 26.1

Example #2Suppose that it is known that only 14% of all US citizens who take trips go outside the United States. You choose a traveler randomly until you find one who has gone outside the US for a vacation.

a. Identify the distribution and define X.

B: Binomial: Success= take trips outside the US and Failure = don’t

I: Safe to assume independence.

N: The number of trials needed to find a person who has gone outside the US

S: The probability of a traveler is the same, 0.14

Let X = # of people needed to find a person who travels outside the US

Example #2Suppose that it is known that only 14% of all US citizens who take trips go outside the United States. You choose a traveler randomly until you find one who has gone outside the US for a vacation.

b. What is the probability that you find such a traveler on the 5th try?

P(X 5) (1 p)n 1 p

(1 0.14)5 1(0.14)

0.864 • 0.14 = 0.0766

OR: geometpdf(0.14, 5) = 0.0766

Example #2Suppose that it is known that only 14% of all US citizens who take trips go outside the United States. You choose a traveler randomly until you find one who has gone outside the US for a vacation.

c. What is the probability that you find such a traveler on or before the 5th try?

geometcdf(0.14, 5) = 0.5296

P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

Let p = 0.14 and k = 0, 1, 2, 3, 4, and 5

Example #2Suppose that it is known that only 14% of all US citizens who take trips go outside the United States. You choose a traveler randomly until you find one who has gone outside the US for a vacation.

d. What is the probability that you find such a traveler after the 5th try?

P(X 5) (1 p)n

(1 0.14)5 0.4704

Example #2Suppose that it is known that only 14% of all US citizens who take trips go outside the United States. You choose a traveler randomly until you find one who has gone outside the US for a vacation.

e. How many would you expect to choose before finding a traveler?

X 1

p

1

0.14 7.1429