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Lecture 11
Binomial Probability
Distribution
2
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.
BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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:
BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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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5
BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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.
4
7
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.
BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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FORMULA METHOD:
CALCULATOR METHOD:
BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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
BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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.
1
6
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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
Variable (X) = 1
Sample Size (N) = 8
Probability (P) = 0.10
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
The likelihood that an inventory card will contain an error is 10%. Suppose you have 8
inventory cards, what is the probability that only one card will contain an error.
2
12
BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
Variable (X) = 0
Sample Size (N) = 8
Probability (P) = 0.10
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
The likelihood that an inventory card will contain an error is 10%. Suppose you have 8
inventory cards, what is the probability that none of the cards will contain an error.
3
7
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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
Variable (X) = 2
Sample Size (N) = 5
Probability (P) = 0.20
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?
4
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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
Using LIST
Variable (X) = 0, 1, 2, 3, 4, 5
Sample Size (N) = 5
Probability (P) = 0.15
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
8
15
BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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
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 ………
5
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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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
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 ………
• Variable (X) = 0, 1, 2, 3, 4, 5
• Sample Size (n) = 5
• Probability (P) = 0.85
Using LISTLIST 1 LIST 2
0 0.0001
1 0.0022
2 0.0244
3 0.1382
4 0.3915
5 0.4437
6
9
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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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
Probability (P) = 0.60
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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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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 less than 48 drivers are
wearing seat belts?
Using CUMULATIVE function (BCP)
Variable (X) = 0, 1, 2 ……..47
Sample Size (N) = 80
Probability (P) = 0.60
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
8
10
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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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 at least 45 drivers are wearing
seat belts?
Using CUMULATIVE function (BCP)
Variable (X) = 45, 46, 47…….80
Sample Size (N) = 80
Probability (P) = 0.60
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
9
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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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 more than 45 drivers are
wearing seat belts?
Using CUMULATIVE function (BCP)
Variable (X) = 46, 47, 48…….80
Sample Size (N) = 80
Probability (P) = 0.60
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
11
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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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 at most 30 drivers are not
wearing seat belts?
Using CUMULATIVE function (BCP)
Variable (X) = 0, 1, 2, ……. 30
Sample Size (N) = 80
Probability (P) = 0.40
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
11
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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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 fewer than 30 drivers are not
wearing seat belts?
Using CUMULATIVE function (BCP)
Variable (X) = 0, 1, 2, ……. 29
Sample Size (N) = 80
Probability (P) = 0.40
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
12
12
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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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
• Variable (X) = 15
• Sample Size (N) = 50
• Probability (P) = 0.25
Using Cumulative Function
0.8369
• Variable (X) = 19
• Sample Size (N) = 50
• Probability (P) = 0.25
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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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
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
• Probability (P) = 0.25
• Sample Size (n) = 5
• Variable (X) = 0, 1, 2, 3, 4, 5
Var (X) BPD BCP
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:
14
13
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?
25
BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
• Mean (µ) = (n) (p)
= (5) (0.20) = 1
• Variable (X) = 2
Sample (N) = 5
Probability (P) = 0.20
• 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.
15
A statistics quiz consists of 10 multiple-choice questions, each with 5 possible
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?
26
BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
0.007• Probability (P) = 0.20
• Sample Size (n) = 10
• Variable (X) = 6, 7, 8, 9, 10
0.058• Probability (P) = 0.30
• Sample Size (n) = 8
• Variable (X) = 5, 6, 7, 8
16
14
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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BINOMIAL DISTRIBUTION
Dr. Raphael Djabatey
0. 179
• Probability (P) = 0.40
• Sample Size (n) = 5
• Variable (X) = 1
• Probability (P) = 0.60
• Sample Size (n) = 6
• Variable (X) = 0, 1, 2
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