The Binomial Probability
Distribution and Related Topics
Note:
 A quantitative variable x is a random variable if the value that x takes on in a
given experiment or observation is a chance or random outcome
Two types of random variable
 Discrete random variable: can take on only a finite number of values or a
countable number of values
 Continuous random variable: can take on any of the countless number of
values in a line interval
Example:
 Discrete random variable: number of students in my statistics class. I can
have 38, 34, 25, 20 students. I can NOT have 25.5 students.
 Continuous random variable: Pumping air in a tire. I can have 0 psi, 20.125
psi, 33.5 psi, etc.
Group Activity: Identity discrete or continue
random variable
 A) Measure the time when students, selected at random, submit college
applications.
 B) Count the number of bad checks on a day selected at random.
 C) Measure the amount of gasoline you put in a car when you fill up the
tank.
 D) Random sample of how many people voted in the Presidential Election
Answer
 A) Continuous random variable
 B) Discrete random variable
 C) Continuous Random Variable
 D) Discrete random variable
Probability Distribution
 Probability distribution is an assignment of probabilities to each distinct
value of a discrete random variable or to each interval of values of a
continuous random variable
Note:
 Features of the probability distribution of a discrete random variable
 1) The probability distribution has a probability assigned to each distinct
value of the random variable
 2) The sum of all the assigned probabilities must be 1
Example of discrete probability distribution
AP Calc BC Test Scores
Number of subjects
1
5000
2
15000
3
32000
4
23000
A) Convert this chart into the probability distribution
(relative frequency table) then graph it
B) P(3 or 4)=
Answer A
AP Score X
Probability P(X)
1
0.07
2
0.35
3
0.43
4
0.31
Probability P(X)
Probability distribution of AP Test
scores
0.5
0.4
0.3
0.2
0.1
0
1
2
3
AP Score
4
Answer B
 P(3 or 4) = P(3)+P(4)=.74
Group Activity: Frequencies of Alphabets in a
1000-Letter Sample
Letter
A
B
C
D
E
F
G
H
I
J
K
L
M
Freq
Prob
73
9
30
44
130
28
16
35
74
2
3
35
25
Letter
N
O
P
Q
R
S
T
U
V
W
X
Y
Z
Freq
Prob
78
74
27
3
77
63
93
27
13
16
5
19
1
Questions:
 A) Fill out the probability chart
 B)do the probabilities of all the letters add up to 1?
 C) If a letter is selected at random what is the probability that it will be a
vowel?
 A) Just do it
 B) yes of course
 C) P(a,e,i,o, or u) = 0.378
Mean and standard deviation again!?!
Yes! But it’s for discrete population probability distribution.
𝜇 = (𝑥 𝑃(𝑥)); 𝜇 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝜎=
𝑥 − 𝜇 2 𝑃(𝑥); 𝜎 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥
X is the value of a random variable,
P(x) is the probability of that variable, and sum is taken for all the values of the
random variable
Note:
 Why was that in population symbol? It is because sum is taken over all
values of the random variable (i.e. the entire sample space)
Example:  Sad fact
Number of
times
Student
study
before test
1
2
3
4
5
Percentage
of students
60%
26%
8%
5%
1%
Answer
X
P(x)
xP(x)
x-u
(x-u)^2
(x-u)^2*P(x)
1
.6
0.6
-.67
.4489
.26934
2
.26
0.52
0.33
.1089
.028314
3
.08
0.24
1.33
1.7689
.141512
4
.05
0.35
2.33
5.4289
.271445
5
.01
0.05
3.33
11.0889
.110889
u= 1.67
So the mean is 1.67
And the standard deviation
is .9064
(𝑥 −
Group Activity
 At a carnival, you pay $2.00 to play a coin flipping game with 3 fair coins.
One of each coin one side has the number 0 and another with number 1.
You flip the three coins at one time and you win $1.00 for every 1 that
appears on top. Are your expected earnings equal to the cost to play?
Fill this out and find 𝜇, so who is making money
here?
Number of 1s, x
0
1
2
3
Frequency
P(x)
xP(x)
Linear Functions of a Random Variable
 Let a and b be any constants, and let x be a random variable. Then the new
random variable L = a+bx is called a linear function of x
 𝐿𝑒𝑡 𝑥 𝑏𝑒 𝑎 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑤𝑖𝑡ℎ 𝑚𝑒𝑎𝑛 𝜇 𝑎𝑛𝑑 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝜎
Then the linear function L= a + bx has mean, variance, and standard deviation
as follows:
𝜇𝐿 = 𝑎 + 𝑏𝜇
𝜎𝐿2 = 𝑏 2 𝜎 2
𝜎𝐿 = 𝑏2 𝜎 2 = 𝑏 𝜎
Linear Combinations of Independent Random
variables

𝐿𝑒𝑡 𝑥1 𝑎𝑛𝑑 𝑥2 𝑏𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒 𝑚𝑒𝑎𝑛 𝜇1 𝑎𝑛𝑑 𝜇2
And 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝜎12 𝑎𝑛𝑑 𝜎22 . 𝐹𝑜𝑟 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒𝑎𝑟 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑊 = 𝑎𝑥1 +
𝑏𝑥2 , 𝑡ℎ𝑒 𝑚𝑒𝑎𝑛, 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒, 𝑎𝑛𝑑 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑒 𝑎𝑠 𝑓𝑜𝑙𝑙𝑜𝑤𝑠:
𝑢𝑤 = 𝑎𝜇1 + 𝑏𝜇2
𝜎𝑤2 = 𝑎2 𝜎12 + 𝑏 2 𝜎22
𝜎𝑤 =
𝑎2 𝜎12 + 𝑏 2 𝜎22
Uh….so what does that mean? Here are some
examples:
 Let
𝑥1 𝑎𝑛𝑑 𝑥2 𝑏𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒 𝑚𝑒𝑎𝑛 𝜇1 =
75 𝑎𝑛𝑑 𝜇2 = 50 𝑎𝑛𝑑 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛𝑠 𝜎1 = 16 and 𝜎2 = 9
 A) Let L = 3 + 2𝑥1 , compute the mean, variance, and SD of L
𝜇𝐿 = 3 + 2𝜇1 = 3 + 2(75)
𝜎𝐿2 = 𝑏 2 𝜎 2 = 22 162 = 1024
𝜎𝐿 = 𝑏 2 𝜎 2 = 𝑏 𝜎 = 2 ∗ 16 = 32
Continue
 Let
𝑥1 𝑎𝑛𝑑 𝑥2 𝑏𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒 𝑚𝑒𝑎𝑛 𝜇1 =
75 𝑎𝑛𝑑 𝜇2 = 50 𝑎𝑛𝑑 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛𝑠 𝜎1 = 16 and 𝜎2 = 9
 B) Let W=𝑥1 + 𝑥2 . Find the mean, variance, and standard deviation of W
𝑢𝑤 = 𝑎𝜇1 + 𝑏𝜇2 = (1)75 + (1)50 = 125
𝜎𝑤2 = 𝑎2 𝜎12 + 𝑏 2 𝜎22 = 12 162 + 12 92 = 337
𝜎𝑤 =
𝑎2 𝜎12 + 𝑏 2 𝜎22 = 18.36
Group Work
 Let
𝑥1 𝑎𝑛𝑑 𝑥2 𝑏𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒 𝑚𝑒𝑎𝑛 𝜇1 =
75 𝑎𝑛𝑑 𝜇2 = 50 𝑎𝑛𝑑 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛𝑠 𝜎1 = 16 and 𝜎2 = 9
 C) W= 𝑥1 − 𝑥2 . Find the mean, variance and standard deviation of W
 D) 𝑊 = 3𝑥1 − 2𝑥2 . Find the mean, variance and standard deviation of W
Answer
 C) 25, 337, 18.36
 D) 125, 2628, 51.26
Homework Practice
 Pg 178 #1-17 every other odd
Binomial Probabilities
Features of a binomial experiment
 1) There are a fixed number of trials. We denote this number by the letter n
 2) The n trials are independent and repeated under identical conditions
 3) Each trial has only two outcomes: success, denoted by S and failure,
denoted by F
 4) For each individual trial, the probability of success is the same. We
denote the probability of success by p and that of failure by q. Since each
trial results in either success or failure, p+q=1 and q=1-p
 5) The central problem of a binomial experiment is to find the probability of
r successes out of n trials.
Example: Read pg 183
 Read pg 183 example 4
Group Work:
 You are taking the final and you are running out of time. You have 3
multiple choice questions left and each question has 4 choices.
 1) What’s the probability of getting all 3 questions correct?
 2) What’s the probability of getting at least 1 correct?
Answers
 There are 8 outcomes. S is success, F is Failure.
 SSS SSF SFS FSS SFF FSF FFS FFF
 P(S)= .25
 P(F)= .75
 1) SSS = (.25)(.25)(.25)=.016
 2) Getting at least 1 correct means, all of the outcomes other than FFF. FFF =
(.75)(.75)(.75)= .422. So 1-.422=.578
Continue with the last one:
 1) What’s P(1)? (One correct)
 2) What’s P(2)? (two correct)
Answer
 1) .422
 2) .141
Are there any way to do this without doing all the
outcomes?
 Yes! It’s called Binomial probability distribution
What is Binomial Probability Distribution?
 You can use this when the outcomes is either success or fail.
 𝑃 𝑟 =
𝑛!
𝑝𝑟 𝑞 𝑛−𝑟
𝑟! 𝑛−𝑟 !
= (𝑛𝐶𝑟)𝑝𝑟 𝑞𝑛−𝑟
 N=number of trials
 p=probability of success on each trial
 q=1-p=probability of failure on each trial
 r=random variable representing the number of success out of n trials (0 ≤
𝑟 ≤ 𝑛)
Example: Basketball shooting example
Mr. Liu makes 30% of his shots. What’s the probability that he makes at least 2
fg out of 4 shots?
n=4
p=.3
q=.7
r=2,3,4
4𝐶2 (. 3)2 (. 7)2 + 4𝐶3 (. 3)3 (. 7)1 + 4𝐶4 (. 3)4 (. 7)0
= .2646 + .0756 + .0081 = .3483
Group Work:
 Privacy is a concern for many users of the Internet. One survey showed that
62% of Internet users are somewhat concerned about the confidentiality of
their e-mail. Based on this information, what is the probability that for a
random sample of 10 internet users, 6 are concerned about the privacy of
their e-mail?
Answer
 10𝐶6 .62
6
.38
4
= .2487
Group Work
 A biologist is studying a new hybrid tomato. It is known that the seeds of
this hybrid tomato have probability of .70 of germinating. The biologist
plants 10 seeds.
 A) What’s the probability that exactly 8 seeds germinate?
 B) What’s the probability that 4 or less will germinate?
Homework Practice
 Pg 191 #2-24 every other even
Additional Properties of the
Binomial Distribution
How to Graph a Binomial Distribution
 1) Place r values on the horizontal axis
 2) Place P(r) values on the vertical axis
 3) Construct a bar over each r value extending from r-0.5 to r+0.5. The
height of the corresponding bar is P(r).
Example of a graph of binomial distribution
Group Work:
 A waiter at the restaurant has learned from experience that the probability
that a lone diner will leave a tip is only 0.7. During one lunch hour, the
waiter serves six people who are dining by themselves. Calculate the
probabilities and then make a graph of the binomial probability distribution
that shows the probabilities that 0,1,2,3,4,5, or all 6 lone diners leave tips
 Hint: Use binomial probability formula
Answer
r (people leaving tip)
P(r)
0
.001
1
.010
2
.060
3
.185
4
.324
5
.303
6
.118
Group Work:
 Jamal enjoys playing basketball. He figures that he makes about 45% of his
shot attempts during the game. Calculate the probabilities and make a
histogram showing the probability that Jamal will make 0,1,2,3,4,5 shots out
of 5 attempted field goals
 Based on the probabilities, make a prediction how many shots Jamal will
make out of 5?
How to calculate the mean and standard
deviation.
 𝜇 = 𝑛𝑝 expected number of successes for random variable r
 𝜎 = 𝑛𝑝𝑞 is the standard deviation for the random variable r
 r is a random variable representing the number of successes in a binomial
distribution
 n is the number of trials
 p is the probability of success on a single trial
 q=1-p is the probability of failure on a single trial
Practice!
 𝑛 = 6, 𝑝 = 0.7, 𝑞 = 0.3
 𝜇=
 𝜎=
Answer
 𝜇 = 4.2
 𝜎 = 1.12
Find out the expected value and standard
deviation for Jamal’s basketball shooting.
 A) What is the expected value for Jamal? Is it the same as you predicted
before?
 B) What is the standard deviation of the binomial distribution?
Unusual Value
 Note:
 Chebysev’s Theorem tells us that no matter what the data distribution looks like, at least
75% will fall within 2 standard deviations of the mean.
 When distribution is mound-shaped and symmetrical, about 95% of the data are within 2
standard deviations of the mean.
 Data value might be an outlier when it is more than 2.5 standard deviations away from the
mean.
 For binomial distribution, it is unusual for the number of successes r to be
higher than 𝜇 + 2.5𝜎 𝑜𝑟 𝑙𝑜𝑤𝑒𝑟 𝑡ℎ𝑎𝑛 𝑢 − 2.5𝜎
What does it mean?
 If
 𝜇 = 4.2
 𝜎 = 1.12

then 𝜇 + 2.5𝜎 = 4.2 + 2.5(1.12)=7 and 𝜇 − 2.5𝜎 = 4.2 − 2.5(1.12)=1.4
 That means anything higher than 7 and lower than 1.4 are outliers.
How to express binomial probabilities using
equivalent formulas
 𝑃 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑠𝑢𝑐𝑐𝑒𝑠𝑠 = 𝑃 𝑟 ≥ 1 = 1 − 𝑃 0
 P at least two success = P r ≥ 2 = 1 − 𝑃 0 − 𝑃 1
 𝑃 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑡ℎ𝑟𝑒𝑒 𝑠𝑢𝑐𝑐𝑒𝑠𝑠 = 𝑃 𝑟 ≥ 3 = 1 − 𝑃 0 − 𝑃 1 − 𝑃 2
 𝑃 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑚 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠 = 𝑃 𝑟 ≥ 𝑚 = 1 − 𝑃 0 − 𝑃 1 − ⋯ −
𝑃 𝑚 − 1 , 𝑤ℎ𝑒𝑟𝑒 1 ≤ 𝑚 ≤ 𝑛𝑢𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠
Example:
 Investments can be profitable as well as risky. Suppose you consider
companies with a 35% estimated risk of default and suppose you want to be
95% certain of meeting your goal of at least 4 good stock. How many stocks
should you buy to meet this goal?
 Please look at the table Appendix II starting in A11. It is a binomial table.
 We want 𝑃 𝑟 ≥ 4 ≥ .95
 Since the probability of success is .65 , we look at the binomial table under
p=0.65 and different values of n to find the smallest value of n that will
satisfy the relation.
Answer
 N=10 because 1-0-0-0.004-0.021=.975
Homework Practice
 Pg 203 #1-23 odd
The Geometric and Poisson
Probability Distributions
Why do we use Geometric Distribution?
 Suppose we have an experiment in which we repeat binomial trials until we
get our first success, and then we stop. Let n be the number of the trial on
which we get our first success. In this context, n is not a fixed number. In
fact, n could be any of the numbers 1,2,3… so on. What is the probability
that our first success comes on the nth trials? Geometric probability
distribution tells us that.
Geometric Probability distribution
 𝑃 𝑛 =𝑝 1−𝑝
𝑛−1
 Where n is the number of the trial on which the first success occurs
(n=1,2,3…) and p is the probability of success on each trial. Note: p must be
the same for each trial.
 𝜇=
1
𝑝
𝑎𝑛𝑑 𝜎 =
1−𝑝
𝑝
Note:
 In many real-life situations, we keep on trying until we achieve success. This
is true in areas as diverse as diplomacy, military science, real estate sales,
general marketing strategies, medical science, engineering, and technology
Example:
 Robotics class made a robot designed to look for ultrasonic sensor within a
certain time frame. If it does not locate it, it will try again and find it. From
experience, it is only 80% successful. The robot will keep trying until it finds
the sensor or the time is up and it ends.
 A) what is the probability that the robot’s first success will be on attempts
n=1, 2, or 3?
 B) If the robot has a max of 3 tries, what is the probability that the robot will
find the sensor?
Answer
 A)
n
𝑃(𝑛) = 𝑝 𝟏 − 𝒑
1
(.80)(.20)^0=.80
2
(.80)(.20)^1=.16
3
(.80)(.20)^2=.032
𝒏−𝟏
 B) P(n=1or2or3)=P(1)+P(2)+P(3)=.80+.16+.032=.992 or 99.2% of the time it
will find the sensor
Poisson Probability Distribution
 Poisson distribution applies to accident rates, arrival times, defect rates, the
occurrence of bacteria in the air, and many other areas of everyday life.
Poisson Probability Distribution
 Let 𝜆 (lambda) be the mean number of successes over time, volume, area,
and so forth. Let r be the number of successes (r=0,1,2,3,…) in a
corresponding interval of time, volume, area and so forth. Then the
probability of r successes in the interval is
 𝑃 𝑟 =
𝑒 −𝜆 𝜆𝑟
𝑟!
 e is approximately equal to 2.7183
 Using some mathematics involving infinite series, it can be shown that the
population mean and standard deviation of the Poisson distribution are
 𝜇=𝜆
𝑎𝑛𝑑
𝜎= 𝜆
Example:
 You are Indiana Jones. You went to an excavation site to try to dig up
different things from the past. You have a record of getting .68 things per
hour. Suppose you decide to find out how many different things you can dig
up in 7 hours.
 A) Use the information provided to find a probability distribution for r, in a
period of 7 hours
 B) What is the probability that in 7 hours, you will dig 0,1,2 or 3 things?
 C) What’s the probability of digging 4 or more things?
Answer

0.68


0.68
7 ℎ𝑜𝑢𝑟𝑠
A) 𝜆 = 1ℎ𝑜𝑢𝑟 , 𝑖𝑛 𝑎 7 ℎ𝑜𝑢𝑟 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝜆 = 1 ℎ𝑜𝑢𝑟 ∗ 7 ℎ𝑜𝑢𝑟𝑠 = 4.76
𝑃 𝑟 =
B) 𝑃 0 =
𝑒 −𝜆 𝜆𝑟
𝑟!
=
𝑒 −4.76 4.76 𝑟
𝑟!
𝑒 −4.76 4.76 0
0!
≈
𝑃 1 =
𝑒 −4.76 4.76 1
1!
≈
𝑃 2 =
𝑒 −4.76 4.76 2
2!
≈

𝑃 3 =
𝑒 −4.76 4.76 3
3!
≈

C) 𝑃 𝑟 ≥ 4 = 1 − 𝑃 0 − 𝑃 1 − 𝑃 2 − 𝑃(3) =


Group Work
 You went to a frozen lake called I-Get-To-Catch-Fish Lake. It is told that on
average, fishermen can fish up to .76 fishes per hour. Suppose you and your
buddy went to go fishing for 6 hours
 A) Find a probability distribution for r, the number of fish you catch in a
period of 6 hours
 B) What the probability that in 6 hours you will get 0,1,2,3, or 4 fish?
 C) What’s the probability you catch 5 of more fish?
Answer

0.76


0.68
6 ℎ𝑜𝑢𝑟𝑠
A) 𝜆 = 1ℎ𝑜𝑢𝑟 , 𝑖𝑛 𝑎 6 ℎ𝑜𝑢𝑟 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝜆 = 1 ℎ𝑜𝑢𝑟 ∗ 6 ℎ𝑜𝑢𝑟𝑠 = 0.045
𝑃 𝑟 =
B) 𝑃 0 =
𝑒 −𝜆 𝜆𝑟
𝑟!
=
𝑒 −0.045 0.045 𝑟
𝑟!
𝑒 −0.045 0.045 0
0!
≈

𝑃 1 =
𝑒 −0.045 0.045 1
1!

𝑃 2 =
𝑒 −0.045 0.045 2
2!
≈

𝑃 3 =
𝑒 −0.045 0.045 3
3!
≈

C) 𝑃 𝑟 ≥ 5 = 1 − 𝑃 0 − 𝑃 1 − 𝑃 2 − 𝑃 3 − 𝑃(4) =
≈
How to Approximate Binomial Probabilities Using
Poisson Probabilities
 Poisson distribution can be used as a probability distribution for “rare” events.
 Suppose you have a binomial distribution with
 n= number of trials
 r= number of successes
 p=probability of success on each trial
 If 𝑛 ≥ 100 𝑎𝑛𝑑 𝑛𝑝 < 10 then r has a binomial distribution that is approximated
by a Poisson distribution with 𝜆 = 𝑛𝑝(expected value)
 𝑃(𝑟) ≈
𝑒 −𝜆 𝜆𝑟
𝑟!
Example:
 INFJ is a rare personality type. It occurs in only about 2.1% of the
population. Suppose a high school graduating class has 170 students, and
suppose success of the event is the personality type INFJ
 A) Let r be the number of success in the n=170 trials. p=P(S)=.021, will the
Poisson distribution be a good approximation to the binomial?
 B) Estimate the probability that this graduating class has 0,1,2,3 or 4 people
who have the INFJ personality type
 C) Estimate the probability that this class has five or more INFJ personality
types.
Answer

A) Since n=170 is greater than 100 and 𝜆 = 170 0.021 = 0.045 < 10. Poisson distribution
should be a good approximation to the binomial.
𝑒 −0.045 0.045 0
0!

B) 𝑃 0 =

𝑃 1 =
𝑒 −0.045 0.045 1
1!
≈

𝑃 2 =
𝑒 −0.045 0.045 2
2!
≈

𝑃 3 =
𝑒 −0.045 0.045 3
3!
≈

𝑃 4 =
𝑒 −0.045 0.045 4
4!
≈

C) 1 − 𝑃 𝑟 ≤ 4 =
≈
Select appropriate distribution
 Identify the type of probability distribution needed to solve the problem:
binomial, geometric, Poisson, or Poisson approximation to the binomial.
Then solve the problem.
 Denver, Colorado, is prone to severe hailstorms. Insurance agents claim that
a homeowner in Denver can expect to replace his or her rough once every
10 years. What is the probability that in 12 years a homeowner in Denver
will need to replace the roof twice?
Answer
 Poisson Distribution
 Since 1 per 10 years = 0.1 per year
 𝜆=
0.1
1𝑦𝑒𝑎𝑟
 𝑝 2 =
∗
12
12
=
𝑒 −1.2 1.2 2
2!
1.2
12𝑦𝑒𝑎𝑟𝑠
≈ .2169 𝑜𝑟 21.7%
Select appropriate distribution
 Identify the type of probability distribution needed to solve the problem:
binomial, geometric, Poisson, or Poisson approximation to the binomial.
Then solve the problem.
 A telephone network substation will keep trying to connect a long-distance
call to a trunk line until the fourth attempt has been made. After the fourth
unsuccessful attempt, the call number goes into a buffer memory bank, and
the caller gets a recorded message to be patient. During peak calling
periods, the probability of a call connecting to trunk line is 65% on each try.
What percentage of all calls made during peak time will wind up in the
buffer memory bank?
Answer
 Geometric probability
 P(S)=0.65=p
 𝑃(𝑛) = 𝑝 𝟏 − 𝒑
𝒏−𝟏
 𝑃 1 = .65 .35
0
= .65
 𝑃 2 = .65 .35
1
= .2275
 𝑃 3 = .65 .35
2
= .0796
 𝑃 4 = .65 .35
3
= .0279
 P(5 or more)=1-p(4 or less)=1-.65-.2275-.0796-.0276=.015 or 1.5%
Select appropriate distribution
 Identify the type of probability distribution needed to solve the problem:
binomial, geometric, Poisson, or Poisson approximation to the binomial.
Then solve the problem.
 The murder rate is 3.6 murders per 100,000 inhabitants. In a community of
1254 people, what is the probability that at least one person will be
murdered?
Answer

Poisson approximation to the binomial

n=1254

p=3.6/100000 = .000036

np=.045

Since n=1254 >100 and np=.045 <10 we can use the poisson approximation

𝑃 0 =

P(1 or more)=1-p(0)=1-.956=.044 or 4.4%
𝑒 −0.045 0.045 0
0!
≈.956
Homework Practice
 P217 #1-26 every other even