11. Binomial Distribution

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Lecture 11

Binomial Probability

Distribution

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Reading Assignments:

Textbook: Read Chapter 6.2 (pg. 272 - 283)

Learning Objectives:

How to determine if a probability experiment is a binomial experiment.

How to find binomial probabilities using the binomial probability formula.

How to find binomial probabilities using calculator.

How to find the mean and standard deviation of a binomial probability distribution.

How to use binomial distribution to solve business problems.

BINOMIAL DISTRIBUTION

Dr. Raphael Djabatey

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There are many probability experiments for which the results of each trial can be

reduced to two outcomes: success or failure. Probability experiments such as these

are called binomial experiment.



a basket ball player attempts a free throw, he or she either makes the

basket or does not.

A product is classified as either acceptable or not acceptable by the

quality control department.

A company makes a profit or not. You win a lottery or not. You survive a

surgery or not.

A sales calls results in the customer either purchasing the product or

not purchasing the product.

A jury may return a guilty or not guilty verdict.

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The binomial distribution is a discrete probability distribution function with

many every day applications.

A binomial experimental process must satisfy the following five conditions:



1. The procedure has a fixed number of trials set forth in advance.

2. There are two possible outcomes in each trial result - success or failure.

3. The probability of a success and failures remains the same for each trial.

4. The trials are independent. The outcome of one trial has no effect on

subsequent trials.

5. The random variable (x) is the result of counts of the number of success trials.

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Probability of Success

P(S) = p

Probability of Failure

P(F) = 1 – p = q

p = probability of a success

q = probability of a failures

n = the fixed number of trials.

x = specific number of success in “n” trials.

p = probability of success in one of the “n” trials.

q = probability of failure in one of the “n” trials.

p(x) = probability of getting exactly x successes among the “n” trials.

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Probability of Success

P(S) = 0.20

Probability of Failure

P(F) = 1 – 0.20 = 0.80

p = probability of a success

q = probability of a failure

n = the fixed number of trials (20).

x = specific number of success in “n” trials (14).

p = probability of success in one of the “n” trials (0.20).

You took a test which consists of 20 multiple-choice questions. Each question

has five choice answers but only one correct answer. What is the probability that

you will select correct answer to 14 questions.

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Sampling without replacement involves dependent events, which violates the

second requirement in the preceding definition.

However, it is a rule of thumb that if the sample is very small relative to the

population size, the difference in results will be negligible if we treat the trials as

independent, even though they are actually dependent.

When sampling without replacement, the events can be considered to be

independent if the sample size is no more than 5% of the population size.

There are three methods for finding probabilities in a binomial experiment. The first

method involves calculations using the binomial probability formula. The second

method involves the use of binomial distribution table. The third method involves of

statistical software or calculator.



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FORMULA METHOD:

CALCULATOR METHOD:



P(X) = n! (px) (1 - p)n-x

(n - X)! X!

• Select STAT menu

• Press F5 to select DIST

• Press F5 to select BINM

• Press F1 to select Bpd

• For Data, press F2 to select VAR to input a variable

• For X, enter the value

• For Numtrial, enter the total sample size

• For p, enter the probability of success

• Press EXE

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The likelihood that an inventory card will contain an error is 10%. Suppose you

have 8 inventory cards, what is the probability that three of these cards will

contain an error.

Binomial Distribution:

P(X) = n! px (1 - p) n-x

X!(n - X)!

P(1) = 8! (0.10)3(1 - 0.10)8 - 3

3!(8 - 3)!

= 8! (0.10)3(0.90)5

3!5!

= (8)(7)(6)(5)(4)(3)(2)(1) (0.001)(0.5905)

(3)(2)(1)(5)(4)(3)(2)(1)

= 0.0331 or 3.31%

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• Variable (X) = 3

• Sample Size (n) = 8

• Probability (P) = 0.10



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Variable (X) = 3

Sample Size (N) = 8

Probability (P) = 0.10

Select F5

X

8

P 0.10

Numtrial

EXE

3

Answer = 0.0331

Calculator

Enter

Select F5

Select F1

Data F2 for VAR

Enter

Enter

Press

Press

DIST

BINM

BPD

3.31%OR

The likelihood that an inventory card will contain an error is 10%. Suppose you have 8

inventory cards, what is the probability that three of these cards will contain an error.

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Variable (X) = 1

Sample Size (N) = 8


Select F5

X

8

P 0.10

Numtrial

EXE

1

Answer = 0.3826

Calculator

Enter

Select F5

Select F1

Data F2 for VAR

Enter

Enter

Press

Press

DIST

BINM

BPD

38.26%OR


inventory cards, what is the probability that only one card will contain an error.

2

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Variable (X) = 0

Sample Size (N) = 8


Select F5

X

8

P 0.10

Numtrial

EXE

0

Answer = 0.4305

Calculator

Enter

Select F5

Select F1

Data F2 for VAR

Enter

Enter

Press

Press

DIST

BINM

BPD

43.05%OR


inventory cards, what is the probability that none of the cards will contain an error.

3

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Variable (X) = 2

Sample Size (N) = 5


Select F5

X

5

P 0.20

Numtrial

EXE

2

Answer = 0.2048

Calculator

Enter

Select F5

Select F1

Data F2 for VAR

Enter

Enter

Press

Press

DIST

BINM

BPD

20.48%OR

There are five daily flights from Toronto to New York. Suppose the probability that any flight

arrives late is 0.20. What is the probability that two of the flights will arrive late today?

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Using LIST

Variable (X) = 0, 1, 2, 3, 4, 5

Sample Size (N) = 5


Select F5

LIST

5

P 0.15

Numtrial

EXE

LIST 1

Calculator

Enter

Select F5

Select F1

Data F1 for LIST

Enter

Enter

Press

Press

DIST

BINM

BPD

A warranty record shows the probability that a new car needs a warranty repair within

first year of purchase is 0.15. If a sample of 5 new cars is selected, what is the

probability that ………

LIST 1 LIST 2

0 0.4437

1 0.3915

2 0.1382

3 0.0244

4 0.0022

5 0.0001Save Res: LIST 2Enter

Var (X) Prob.

5

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none will need a warranty repair? = 0.44371

at least three will need a warranty repair? 0.0244 + 0.0022 + 0.0001 = 0.02672

at most two will need a warranty repair? 0.04437 + 0.3915 + 0.1382 = 0.97343

all will need a warranty repair? = 0.00014

between two and four will need a warranty repair? 0.1382 + 0.0244 + 0.0022 = 0.16475

LIST 1 LIST 2

0 0.4437

1 0.3915

2 0.1382

3 0.0244

4 0.0022

5 0.0001




5

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three will not need a warranty repair = 0.13821

three or more will not need a warranty repair? 0.1382 + 0.3915 + 0.4437 = 0.97342

less than two will not need a warranty repair? 0.0001 + 0.0022 = 0.00233

all will not need a warranty repair? = 0.44374




• Variable (X) = 0, 1, 2, 3, 4, 5



Using LISTLIST 1 LIST 2

0 0.0001

1 0.0022

2 0.0244

3 0.1382

4 0.3915

5 0.4437

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A survey revealed that three in five student drivers use their seat belts. A sample of 80

student drivers is selected. What is the probability that 50 or fewer drivers are wearing

seat belts?

Using CUMULATIVE function (BCP)

Variable (X) = 0, 1, 2………50

Sample Size (N) = 80


Select F5

X

80

P 0.60

Numtrial

EXE

50

Answer = 0.7139

Calculator

Enter

Select F5

Select F2

Data F2 for VAR

Enter

Enter

Press

Press

DIST

BINM

BCD

71.39%OR

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student drivers is selected. What is the probability that less than 48 drivers are

wearing seat belts?


Variable (X) = 0, 1, 2 ……..47



Select F5

X

80

P 0.60

Numtrial

EXE

47

Answer = 0.4516

Calculator

Enter

Select F5

Select F2

Data F2 for VAR

Enter

Enter

Press

Press

DIST

BINM

BCD

45.16%OR

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10

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student drivers is selected. What is the probability that at least 45 drivers are wearing

seat belts?


Variable (X) = 45, 46, 47…….80



Select F5

X

80

P 0.60

Numtrial

EXE

44

0.2115

Answer = 1 – 0.2115 = 0.7885

Calculator

Enter

Select F5

Select F2

Data F2 for VAR

Enter

Enter

Press

Press

DIST

BINM

BCD

78.85%OR

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student drivers is selected. What is the probability that more than 45 drivers are

wearing seat belts?


Variable (X) = 46, 47, 48…….80



Select F5

X

80

P 0.60

Numtrial

EXE

45

0.2825

Answer = 1 – 0.2825 = 0.7175

Calculator

Enter

Select F5

Select F2

Data F2 for VAR

Enter

Enter

Press

Press

DIST

BINM

BCD

71.75%OR

10

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student drivers is selected. What is the probability that at most 30 drivers are not

wearing seat belts?


Variable (X) = 0, 1, 2, ……. 30



Select F5

X

80

P 0.40

Numtrial

EXE

30

Answer = 0.3687

Calculator

Enter

Select F5

Select F2

Data F2 for VAR

Enter

Enter

Press

Press

DIST

BINM

BCD

36.87%OR

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student drivers is selected. What is the probability that fewer than 30 drivers are not

wearing seat belts?


Variable (X) = 0, 1, 2, ……. 29



Select F5

X

80

P 0.40

Numtrial

EXE

29

Answer = 0.2861

Calculator

Enter

Select F5

Select F2

Data F2 for VAR

Enter

Enter

Press

Press

DIST

BINM

BCD

28.61%OR

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Find the probability that the student will get less than 16 correct answers.

Find the probability that the student will get at least 20 correct answers.

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• Sample Size (N) = 50


Using Cumulative Function

0.8369


• Sample Size (N) = 50


Using Cumulative Function

1 – 0. 9861 = 0.0139

A statistics quiz consists of 50 multiple-choice questions, each with 4 possible

answers. For a student who makes random guesses for all of the answers:

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less than three of the answers are correct = 0.89641

three or less of the answers are correct ? = 0.98432

at most one of the answers are correct? = 0.63283

at least four of the answers are correct? = 1 – 0.9843 = 0.01574

LIST 1 LIST 2 LIST 3

0 0.2373 0.2373

1 0.3955 0.6328

2 0.2636 0.8964

3 0.0878 0.9843

4 0.0146 0.9990

5 0.0010 1.0000



• Variable (X) = 0, 1, 2, 3, 4, 5

Var (X) BPD BCP


answers. For a student who makes random guesses for all of the answers:

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What is the probability that two flights will arrive late today?

What is the expected (mean) number of late flights?

What is the standard deviation?

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• Mean (µ) = (n) (p)

= (5) (0.20) = 1


Sample (N) = 5


• Standard Deviation() = √(n)(p)(1 – p)

= √(5)(0.20)(1 – 0.20)

= √0.8 = 0.8944

20.48%

There are five daily flights from Toronto to New York. Suppose the probability that any

flight arrives late is 0.20.

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answers. For someone who makes random guesses for all of the answers, find the

probability of passing if the minimum passing grade is 60%.

A quality control manager of a meat packaging plant has determined that 15 of every

50 meat packages have defective labels. If the meat packaging equipment processes

8 meat packages at any given time. What are the chances that more than four meat

packages will have a defective label?

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0.007• Probability (P) = 0.20


• Variable (X) = 6, 7, 8, 9, 10

0.058• Probability (P) = 0.30


• Variable (X) = 5, 6, 7, 8

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A study of weight loss between male and female shows that 40% of females lost 5

pounds in the first month of a weight loss program as compared with 60% of males

who lost the same amount of weight in the first month. Suppose five females are

selected, what is the probability that only one female will lose 5 pounds in the first

month?

Suppose six males are selected, what are the chances that less than three males will

lose 5 pounds in the first month?

0.259

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0. 179






• Variable (X) = 0, 1, 2

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11. Binomial Distribution