Chapter 5
Discrete Probability Distributions
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Chapter Outline
• 5.1 Probability Distributions
• 5.2 Binomial Distributions
• 5.3 More Discrete Probability Distributions
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Section 5.1
Probability Distributions
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Section 5.1 Objectives
• Distinguish between discrete random variables and
continuous random variables
• Construct a discrete probability distribution and its
graph
• Determine if a distribution is a probability
distribution
• Find the mean, variance, and standard deviation of a
discrete probability distribution
• Find the expected value of a discrete probability
distribution
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Random Variables
Random Variable
• Represents a numerical value associated with each
outcome of a probability distribution.
• Denoted by x
• Examples
 x = Number of sales calls a salesperson makes in
one day.
 x = Hours spent on sales calls in one day.
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Random Variables
Discrete Random Variable
• Has a finite or countable number of possible
outcomes that can be listed.
• Example
 x = Number of sales calls a salesperson makes in
one day.
x
0
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2
3
4
5
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Random Variables
Continuous Random Variable
• Has an uncountable number of possible outcomes,
represented by an interval on the number line.
• Example
 x = Hours spent on sales calls in one day.
x
0
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2
3
…
24
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Example: Random Variables
Decide whether the random variable x is discrete or
continuous.
1. x = The number of stocks in the Dow Jones Industrial
Average that have share price increases
on a given day.
Solution:
Discrete random variable (The number of stocks
whose share price increases can be counted.)
x
0
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1
2
3
…
30
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Example: Random Variables
Decide whether the random variable x is discrete or
continuous.
2. x = The volume of water in a 32-ounce
container.
Solution:
Continuous random variable (The amount of water
can be any volume between 0 ounces and 32 ounces)
x
0
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2
3
…
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Discrete Probability Distributions
Discrete probability distribution
• Lists each possible value the random variable can
assume, together with its probability.
• Must satisfy the following conditions:
In Words
In Symbols
1. The probability of each value of
the discrete random variable is
between 0 and 1, inclusive.
0  P (x)  1
2. The sum of all the probabilities is
1.
ΣP (x) = 1
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Constructing a Discrete Probability
Distribution
Let x be a discrete random variable with possible
outcomes x1, x2, … , xn.
1. Make a frequency distribution for the possible
outcomes.
2. Find the sum of the frequencies.
3. Find the probability of each possible outcome by
dividing its frequency by the sum of the frequencies.
4. Check that each probability is between 0 and 1 and
that the sum is 1.
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Exercise 1: Constructing a Discrete
Probability Distribution
An industrial psychologist administered a personality
inventory test for passive-aggressive traits to 150
employees. Individuals were given a score from 1 to 5,
where 1 was extremely passive and 5 extremely
aggressive. A score of 3 indicated
Score, x Frequency, f
neither trait. Construct a
1
24
probability distribution for the
2
33
random variable x. Then graph the
3
42
distribution using a histogram.
4
30
5
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Solution: Constructing a Discrete
Probability Distribution
• Divide the frequency of each score by the total
number of individuals in the study to find the
probability for each value of the random variable.
P (1) 
24
 0.16
150
30
P (4) 
 0.20
150
P (2) 
33
 0.22
150
P (3) 
42
 0.28
150
21
P (5) 
 0.14
150
• Discrete probability distribution:
x
1
2
3
4
5
P(x)
0.16
0.22
0.28
0.20
0.14
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Solution: Constructing a Discrete
Probability Distribution
x
1
2
3
4
5
P(x)
0.16
0.22
0.28
0.20
0.14
This is a valid discrete probability distribution since
1. Each probability is between 0 and 1, inclusive,
0 ≤ P(x) ≤ 1.
2. The sum of the probabilities equals 1,
ΣP(x) = 0.16 + 0.22 + 0.28 + 0.20 + 0.14 = 1.
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Solution: Constructing a Discrete
Probability Distribution
• Histogram
Passive-Aggressive Traits
Probability, P(x)
0.3
0.25
0.2
0.15
0.1
0.05
0
1
2
3
4
5
Score, x
Because the width of each bar is one, the area of
each bar is equal to the probability of a particular
outcome.
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Mean
Mean of a discrete probability distribution
• μ = ΣxP(x)
• Each value of x is multiplied by its corresponding
probability and the products are added
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Exercise 2: Finding the Mean
The probability distribution for the personality
inventory test for passive-aggressive traits is given. Find
the mean.
Solution:
x
P(x)
xP(x)
1
2
3
4
0.16
0.22
0.28
0.20
1(0.16) = 0.16
2(0.22) = 0.44
3(0.28) = 0.84
4(0.20) = 0.80
5
0.14
5(0.14) = 0.70
μ = ΣxP(x) = 2.94
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Variance and Standard Deviation
Defining Formulas
Variance of a discrete probability distribution
σ2 = Σ(x – μ)2P(x)
Standard deviation of a discrete probability
distribution
   2  ( x   ) 2 P ( x )
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Exercise 3: Finding the Variance and
Standard Deviation – Defining Formula
The probability distribution for the personality
inventory test for passive-aggressive traits is given. Find
the variance and standard deviation. ( μ = 2.94)
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x
P(x)
1
2
3
4
0.16
0.22
0.28
0.20
5
0.14
𝜎2 =
𝑥− 𝜇
2
∙ 𝑃(𝑥
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Solution: Finding the Variance and
Standard Deviation – Defining Formula
• Recall μ = 2.94
Score, x
1
f
P(x)
24
2
33
3
42
4
30
5
21
TOTAL
150
(x -µ)
2
(x -µ)
2*
P(x)
0.16
3.76
0.602
0.22
0.88
0.194
0.28
0.00
0.001
0.20
1.12
0.225
0.14
4.24
0.594
1.00
10.02
1.616
Standard Deviation:     1.616  1.3
2
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Exercise 4: Variance and Standard Deviation
using Computational Formula
The probability distribution for the personality
inventory test for passive-aggressive traits is given. Find
the variance and standard deviation. ( μ = 2.94)
Larson/Farber 4th ed
x
P(x)
1
2
3
4
0.16
0.22
0.28
0.20
5
0.14
𝜎2 =
𝑥2 ∙ 𝑃 𝑥
− 𝜇2
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Finding the Variance and Standard
Deviation – Computational Formula
Score, x
f
P(x)
1
2
3
4
5
24
33
42
30
21
TOTAL
150
Variance: 𝜎 2 =
x * P(x)
0.16
0.22
0.28
0.20
0.14
1.00
𝑥2 ∙ 𝑃 𝑥
0.16
0.44
0.84
0.80
0.70
2.94
2
x * P(x)
0.16
0.88
2.52
3.20
3.50
10.26
− 𝜇2 = 10.26 – 𝟐. 𝟗𝟒𝟐 = 1.616
Standard Deviation:     1.616  1.3
2
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Exercise 5: Finding the Variance and
Standard Deviation via TI-84
L1
Score, x
1
2
3
4
5
f
24
33
42
30
21
L2
P(x)
0.16
0.22
0.28
0.20
0.14
1-Var Stats L1, L2
𝑥 = 2.9
𝜎 = 1.271377206
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Expected Value
Expected value of a discrete random variable
• Equal to the mean of the random variable.
• E(x) = μ = Σ xP(x)
• E(x) represents the “average” value of all the
outcomes
• E(x) represents the value that we would expect to get
if the trials could continue forever!
• Mathematical expectation plays an important role in
an area called decision theory
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Example: Finding an Expected Value
At a raffle, 1500 tickets are sold at $2 each for four
prizes of $500, $250, $150, and $75. You buy one
ticket. What is the expected value of your gain?
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Solution: Finding an Expected Value
• To find the gain for each prize, subtract
the price of the ticket from the prize:
 Your gain for the $500 prize is $500 – $2 = $498
 Your gain for the $250 prize is $250 – $2 = $248
 Your gain for the $150 prize is $150 – $2 = $148
 Your gain for the $75 prize is $75 – $2 = $73
• If you do not win a prize, your gain is $0 – $2 = –$2
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Solution: Finding an Expected Value
• Probability distribution for the possible gains
(outcomes)
Gain, x
P(x)
$498
1
1500
$248
1
1500
$148
1
1500
$73
1
1500
–$2
1496
1500
E (x )  xP (x )
1
1
1
1
1496
 $248 
 $148 
 $73 
 ($2) 
1500
1500
1500
1500
1500
 $1.35
 $498 
You can expect to lose an average of $1.35 for each ticket
you buy.
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Exercise 6: Finding an Expected Value
• If you bet $1 in Kentucky’s Pick 4 Lottery, you either
lose $1, or gain $4999.
• The winning prize is $5000, but your $1 bet is not
returned, so the net gain is $4999.
• The game is played by selecting a four-digit number
between 0000 and 9999.
If you bet $1 on 1234, what is your expected gain (or
loss)?
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Solution: Finding an Expected Value
To find the gain for each prize, subtract the price of the ticket from the value of
the prize.
Expected
Description
Winner
Losing Ticket
Payout
$5,000
$0
Cost
$1
$1
TOTAL
Gain (x)
Fraction
Decimal
P(x)
P(x)
$4,999 1/10000
($1)9999/10000
Value
x*P(x)
0.0001
$0.50
0.9999
($1.00)
1
($0.50)
Interpretation: In the long run, you can expect to lose an average of $ 0.50 for
each $1 bet.
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Section 5.1 Summary
• Distinguished between discrete random variables and
continuous random variables
• Constructed a discrete probability distribution and its
graph
• Determined if a distribution is a probability
distribution
• Found the mean, variance, and standard deviation of a
discrete probability distribution
• Found the expected value of a discrete probability
distribution
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Section 5.2
Binomial Distributions
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Section 5.2 Objectives
• Determine if a probability experiment is a binomial
experiment
• Find binomial probabilities using the binomial
probability formula
• Find binomial probabilities using technology and a
binomial table
• Find the mean, variance, and standard deviation of a
binomial probability distribution
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Binomial Experiments
1. The experiment is repeated for a fixed number of
trials, where each trial is independent of other trials
2. There are only two possible outcomes of interest for
each trial. The outcomes can be classified as a
success (S) or as a failure (F)
3. The probability of a success P(S) is the same for
each trial
4. The random variable x counts the number of
successful trials
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Notation for Binomial Experiments
Symbol
n
p = P(S)
Description
The number of times a trial is repeated
The probability of success in a single trial
q = P(F)
The probability of failure in a single trial
q = (1 – p)
The random variable represents a count of
the number of successes in n trials:
r = 0, 1, 2, 3, … , n.
r
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Example: Binomial Experiments
Decide whether the experiment is a binomial
experiment. If it is, specify the values of n, p, and q, and
list the possible values of the random variable r.
1. A certain surgical procedure has an 85% chance of
success. A doctor performs the procedure on eight
patients. The random variable represents the number
of successful surgeries.
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Solution: Binomial Experiments
Binomial Experiment
1. Each surgery represents a trial. There are eight
surgeries, and each one is independent of the others.
2. There are only two possible outcomes of interest for
each surgery: a success (S) or a failure (F).
3. The probability of a success, P(S), is 0.85 for each
surgery.
4. The random variable r counts the number of
successful surgeries.
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Solution: Binomial Experiments
Binomial Experiment
• n = 8 (number of trials)
• p = 0.85 (probability of success)
• q = 1 – p = 1 – 0.85 = 0.15 (probability of failure)
• r = 0, 1, 2, 3, 4, 5, 6, 7, 8 (number of successful
surgeries)
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Example: Binomial Experiments
Decide whether the experiment is a binomial experiment.
If it is, specify the values of n, p, and q, and list the
possible values of the random variable r.
1. Flip a coin 6 times and count the number of times the
coin lands on heads.
Larson/Farber 4th ed
𝑛=
6
𝑝=
0.5
𝑟=
0,1,2,3,4,5,6
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Example: Binomial Experiments
Decide whether the experiment is a binomial
experiment. If it is, specify the values of n, p, and q, and
list the possible values of the random variable r.
2. A jar contains five red marbles, nine blue marbles, and
six green marbles. You randomly select three marbles
from the jar, without replacement. The random
variable represents the number of red marbles.
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Solution: Binomial Experiments
Not a Binomial Experiment
• The probability of selecting a red marble on the first
trial is 5/20.
• Because the marble is not replaced, the probability of
success (red) for subsequent trials is no longer 5/20.
• The trials are not independent and the probability of
a success is not the same for each trial.
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Example: Finding Binomial Probabilities
Microfracture knee surgery has a 75% chance of success
on patients with degenerative knees. The surgery is
performed on three patients. Find the probability of the
surgery being successful on exactly two patients.
Create a tree diagram and use it to compute this
probability.
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Finding Binomial Probabilities via Binary
Tree
Draw a tree diagram and use the Multiplication Rule
 9 
P(2 successful surgeries )  3    0.422
 64 
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Binomial Probability Formula
Binomial Probability Formula
• The probability of exactly r successes in n trials is
𝑃 𝑟 =
•
•
•
•
𝑛 𝐶𝑟 𝑝
𝑟 𝑛−𝑟
𝑞
𝑛!
=
𝑝𝑟 𝑞 𝑛−𝑟
𝑛 − 𝑟 ! 𝑟!
n = number of trials
p = probability of success
q = 1 – p probability of failure
r = number of successes in n trials
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Exercise 1: Finding Binomial
Probabilities via Binomial Formula
Repeat the previous exercise using the binomial
formula. .
3
𝑛 = 3, 𝑝 = , 𝑟 = 2
4
2
3 1
P(2 successful surgeries)  3 C2    
4 4
3 2
2
1
3!
3 1

   
(3  2)!2!  4   4 
 9  1  27
 3    
 0.422
 16  4  64
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Exercise 2: Finding Binomial Probabilities
Use TI-84 binompdf function
3
𝑛 = 3, 𝑝 = , 𝑟 = 2
4
TI-83/84
DISTR / 0: binompdf(n, p, r )
P(two successes ) = binompdf(3,0.75,2) =0.422
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Exercise 3: Binomial Probability
Distribution
• List the possible values of r with the corresponding
probability of each.
• Example: Binomial probability distribution for
Microfracture knee surgery: n = 3, p = 3
4
 Use binomial probability formula to find
probabilities.
r
0
1
2
3
P(r)
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Exercise 3: Binomial Probability
Distribution
Binomial Probability Distribution
• Solution: Binomial probability distribution for
Microfracture knee surgery: n = 3, p = 0.75
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r
0
1
2
3
P(r)
0.016
0.141
0.422
0.422
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Exercise 3: Binomial Probability
Distribution via TI-84
• Microfracture knee surgery: n = 3, p = 0.75
L1
r
L2 =binompdf(3,0.75, L1)
P(r)
0
0.016
1
0.141
2
0.422
3
0.422
TOTAL
1.000
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Example: Construct a Binomial Distribution
In a survey, workers in the U.S. were asked to name
their expected sources of retirement income. Seven
workers who participated in the survey are randomly
selected and asked whether they expect to rely on Social
Security for retirement
income. Create a binomial
probability distribution for
the number of workers who
respond yes.
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Solution: Constructing a Binomial
Distribution
• 25% of working Americans expect to rely on Social
Security for retirement income.
• n = 7, p = 0.25, q = 0.75, r = 0, 1, 2, 3, 4, 5, 6, 7
P(r = 0) = 7C0(0.25)0(0.75)7 = 1(0.25)0(0.75)7 ≈ 0.1335
P(r = 1) = 7C1(0.25)1(0.75)6 = 7(0.25)1(0.75)6 ≈ 0.3115
P(r = 2) = 7C2(0.25)2(0.75)5 = 21(0.25)2(0.75)5 ≈ 0.3115
P(r = 3) = 7C3(0.25)3(0.75)4 = 35(0.25)3(0.75)4 ≈ 0.1730
P(r = 4) = 7C4(0.25)4(0.75)3 = 35(0.25)4(0.75)3 ≈ 0.0577
P(r = 5) = 7C5(0.25)5(0.75)2 = 21(0.25)5(0.75)2 ≈ 0.0115
P(r = 6) = 7C6(0.25)6(0.75)1 = 7(0.25)6(0.75)1 ≈ 0.0013
P(r = 7) = 7C7(0.25)7(0.75)0 = 1(0.25)7(0.75)0 ≈ 0.0001
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Solution: Constructing a Binomial
Distribution
r
0
1
2
3
4
5
6
7
Larson/Farber 4th ed
P(r)
0.1335
0.3115
0.3115
0.1730
0.0577
0.0115
0.0013
0.0001
All of the probabilities are between
0 and 1 and the sum of the
probabilities is 1.00001 ≈ 1.
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Exercise 4: Finding Binomial
Probabilities
A survey indicates that 41% of women in the U.S.
consider reading their favorite leisure-time activity. You
randomly select four U.S. women and ask them if
reading is their favorite leisure-time activity. Find the
probability that at least two of them respond yes.
Solution:
• n = 4, p = 0.41, q = 0.59
• At least two means two or more.
• Find the sum of P(2), P(3), and P(4).
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Solution: Finding Binomial Probabilities
P(r = 2) = 4C2(0.41)2(0.59)2 = 6(0.41)2(0.59)2 ≈ 0.351094
P(r = 3) = 4C3(0.41)3(0.59)1 = 4(0.41)3(0.59)1 ≈ 0.162654
P(r = 4) = 4C4(0.41)4(0.59)0 = 1(0.41)4(0.59)0 ≈ 0.028258
P(r ≥ 2) = P(2) + P(3) + P(4)
≈ 0.351094 + 0.162654 + 0.028258
≈ 0.542
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Section 5.3 Objectives
•
•
•
•
Use the binomial table to find P(r )
Make histograms for binomial distributions
Compute µ and 𝜎 for binomial distributions
Compute the minimum number of trials n needed to
achieve a given probability of success P(r).
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Exercise 1: Finding Binomial
Probabilities Using a Table
About thirty percent of working adults spend less than
15 minutes each way commuting to their jobs. You
randomly select six working adults. What is the
probability that exactly three of them spend less than 15
minutes each way commuting to work? Use a table to
find the probability. (Source: U.S. Census Bureau)
Solution:
• Binomial with n = 6, p = 0.30, r = 3
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Solution: Finding Binomial Probabilities
Using a Table
• A portion of Table 2 is shown
The probability that exactly three of the six workers
spend less than 15 minutes each way commuting to
work is 0.185.
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Exercise 2: Graphing a Binomial
Distribution
Fifty-nine percent of households in the U.S. subscribe to
cable TV. You randomly select six households and ask
each if they subscribe to cable TV. Construct a
probability distribution for the random variable x. Then
graph the distribution. (Source: Kagan Research, LLC)
Solution:
• n = 6, p = 0.59, q = 0.41
• Find the probability for each value of r
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Graphing a Binomial
Distribution
Solution: Graphing a Binomial
Distribution
x
0
1
2
3
4
5
6
P(x)
0.005
0.041
0.148
0.283
0.306
0.176
0.042
Histogram:
Subscribing to Cable TV
0.35
Probability
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
Households
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Mean, Variance, and Standard Deviation
• Mean: μ = np
• Variance: σ2 = npq
• Standard Deviation:   npq
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Exercise 3: Finding the Mean, Variance,
and Standard Deviation
In Pittsburgh, Pennsylvania, about 56% of the days in a
year are cloudy. Find the mean, variance, and standard
deviation for the number of cloudy days during the
month of June. Interpret the results and determine any
unusual values. (Source: National Climatic Data Center)
Solution: n = 30, p = 0.56, q = 0.44
Mean: μ = np = 30∙0.56 = 16.8
Variance: σ2 = npq = 30∙0.56∙0.44 ≈ 7.4
Standard Deviation:   npq  30  0.56  0.44  2.7
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Solution: Finding the Mean, Variance, and
Standard Deviation
μ = 16.8 σ2 ≈ 7.4
σ ≈ 2.7
• On average, there are 16.8 cloudy days during the
month of June.
• The standard deviation is about 2.7 days.
• Values that are more than 2.5 standard deviations
from the mean are considered unusual:
 16.8 – 2.5(2.7) =10.05, A June with 10 cloudy
days would be unusual.
 16.8 + 2.5 (2.7) =23.55 A June with 24 cloudy
days would also be unusual.
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Binomial Probabilities: Equivalent
Formulas
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Exercise 4: Compute Binomial
Probabilities
If 22% of U.S. households have a Nintendo game,
compute the probability that
a. exactly 5 of 12 randomly chosen families will have
Nintendo games.
b. at most, 5 randomly chosen families will have
Nintendo games.
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Solution: Compute Binomial Probabilities
L1
L2 =binompdf(12,0.22,L1)
r
TOTAL
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P(r)
0
0.0507
1
0.1717
2
3
0.2663
0.2503
4
0.1589
5
0.0717
0.9696
65
Solution: Compute Binomial Probabilities
r
5
P(r)
binompdf(12, 0.22, 5) = 0.0717
a) P(exactly 5 of 12 randomly chosen families will have Nintendo games)
P(5) = binompdf(12,0.22,5) = 0.0717
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Solution: Compute Binomial Probabilities
r
P(r)
0
binompdf(12, 0.22, 0) = 0.0507
1
binompdf(12, 0.22, 1) = 0.1717
2
binompdf(12, 0.22, 2) = 0.2663
3
binompdf(12, 0.22, 3) = 0.2503
4
binompdf(12, 0.22, 4) = 0.1589
5
binompdf(12, 0.22, 5) = 0.0717
sum = 0.9696
b) P(at most five randomly chosen families will have Nintendo games)
P(𝑟 ≤ 5 = 𝑃(0 + 𝑃(1 + 𝑃(2 + 𝑃(3 + 𝑃(4 + 𝑃(5 = 0.9696
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Exercise 5: Use Binomial Cumulative
Density Function
If 22% of U.S. households have a Nintendo game,
compute the probability that at most, 5 of 12 randomly
chosen families will have Nintendo games.
Solution:
P(𝑟 ≤ 5 =
𝑃(0 + 𝑃(1 + 𝑃(2 + 𝑃(3 + 𝑃(4 + 𝑃(5 = binomcdf( 12,0.22,5) =
0.9696
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Quota Problems
• We can use the binomial distribution table
“backwards” to solve for a minimum number of
trials.
• Given: r and p
• Find: value of n that satisfies requirement for
expected value, or reliability specification
• Use calculator technology or tables together with
guess/check strategy
• Show minimum of three guesses and three checks
Example: Satisfy Quota for Expected
Value of Distribution
Suppose a satellite requires 3 solar cells for its power,
the probability that any one of these cells will fail is
0.15, and the cells operate and fail independently.
We want to find the least number of cells each satellite
should have so that the expected number of working
cells (the mean) is no smaller that 3. That is, n is the
total number of cells, and r is the number of working
cells, p is probability that a cell will work.
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Example: Satisfy Quota for Expected
Value of Distribution
cell 1
cell 2
cell 3
…
cell n
We want to find the least number of cells each satellite
should have so that the expected number of working
cells (the mean) is no smaller that 3. That is, n is the
total number of cells, and r is the number of working
cells, p is probability that a cell will work, and 𝜇 ≥ 3.
Since 𝜇 = 𝑛𝑝, 𝑛𝑝 ≥ 3 𝑎𝑛𝑑 3 ≤ 𝑛𝑝
Let S = “cell works”
Larson/Farber 4th ed
p = 0.85, q = 0.15
71
Example: Satisfy Quota for Expected
Value of Distribution
cell 1
Larson/Farber 4th ed
cell 2
cell 3
…
cell n
72
Example: Satisfy Quota for Expected
Value of Distribution
Probability Distribution: n = 4, p = 0.85
0.85
0.15
4
P(r)
0.001
0.011
0.098
0.368
0.522
1.000
0.600
0.500
P(r)
p=
q=
n=
r
0
1
2
3
4
TOTAL
0.400
0.300
0.200
0.100
0.000
0
1
2
r
3
4
𝑛 ≥ 3.53 ==> choose 𝑛 = 4
𝜇 = 𝑛𝑝 = 4 0.85 = 3.4
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Example: Satisfy Quota for Expected
Value of Distribution
𝑛 ≥ 3.53 ==> choose 𝑛 = 4
𝜇 = 𝑛𝑝 = 4 0.85 = 3.4
Interpretive Statement: If we put up lots of
satellites with four cells, we can expect that an
average of 3.4 cells will work in each satellite.
Note that each additional solar satellite will
produce four new cells (e.g. trials) of the
binomial experiment.
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Exercise 6: Satisfy Reliability Quota
Suppose a satellite requires 3 solar cells for its power,
the probability that any one of these cells will fail is
0.15, and the cells operate and fail independently.
We want to find the smallest number of cells the
satellite should have to be 99% sure that there will be
adequate power (i.e. at least three cells work).
How many cells do we need to meet the specification
P(𝑟 ≥ 3 ≥ 99%?
P(fail) =
0.15
Larson/Farber 4th ed
P(fail) =
0.15
P(fail) =
0.15
...
P(fail) =
0.15
75
Exercise 6: Satisfy Reliability Quota
n
P(r >=3 ) = 1 - P( r <=2)
x Y1 = 1 - binomcdf(x,0.85,2)
Expected Successes
µ = np
3
0.6141
2.55
4
0.8905
3.40
5
0.9734
4.25
6
0.9941
5.10
7
0.9988
5.95
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Exercise 6: Satisfy Reliability Quota
Conclusion: We need at least 6 cells to be 99%
sure that all satellites will be operational.
Interpretive Statement: In order to meet the 99%
reliability specification, we will need to install at
least 6 cells in each satellite to ensure that the
satellite will be fully operational.
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Exercise 6: Interpretive Statement
• If ___ cells are installed in a satellite, then there is a
99% probability that the satellite will be operational.
OR
• 99% of all satellites with ____ cells in them will be
operational.
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Exercise 6: Interpretive Statement(s)
• If 6 cells are installed in a satellite, then there is a
99% probability that the satellite will be operational.
OR
• 99% of all satellites with 6 cells in them will be fully
operational.
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Section 5.3 Summary
• Determined if a probability experiment is a binomial
experiment
• Found binomial probabilities using the binomial
probability formula
• Graphed a binomial distribution
• Found the mean, variance, and standard deviation of
a binomial probability distribution
• Solved quota problems
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Section 5.4
More Discrete Probability
Distributions
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Section 5.4 Objectives
• Find probabilities using the geometric distribution
• Find probabilities using the Poisson distribution
• Use a Poisson Approximation to the Binomial
Distribution
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Exercise 1: Geometric Distribution
Susan is taking Western Civilization on a pass/fail basis.
Historically, the passing rate for this course has been
77% each term. Let n = 1, 2, 3, . . . represent the
number of times a student takes this course until a
passing grade is received. Assume the attempts are
independent.
Construct a tree diagram for n = 1, 2, and 3 attempts.
List the outcomes and their associated probabilities.
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Exercise 1: Geometric Distribution
Susan is taking Western Civilization on a pass/fail basis.
Historically, the passing rate for this course has been
77% each term. Use your tree diagram to compute:
a) What is the probability that Susan passes on the first
try?
b) What is the probability that Susan passes on the
second try?
c) What is the probability that Susan passes on the
third try?
d) What is the probability that Susan passes on the nth
try?
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Exercise 1: Geometric Distribution
n
Factors
1
0.77
2
0.23
0.77
3
0.23
0.23
P(n)
0.770
0.177
0.77
0.041
𝑃 𝑛 = 𝑝𝑞 𝑛−1
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Geometric Distribution
A discrete probability distribution. It satisfies the
following conditions:
 A trial is repeated until a success occurs
 The repeated trials are independent of each other
 The probability of success p is constant for each
trial
• The probability that the first success will occur on
trial n is P(n) = p(q)n – 1, where q = 1 – p.
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86
Geometric Distribution
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Geometric Distribution – TI-84 Calculator
The probability of success on exactly the nth try is given
by 𝑃 𝑛 = 𝑝𝑞𝑛−1 = 𝑔𝑒𝑜𝑚𝑒𝑡𝑝𝑑𝑓(𝑝, 𝑛
1
Larson/Farber 4th ed
2
3
...
n
88
Geometric Distribution – TI-84 Calculator
The probability that a success will occur on or before
the nth try is given by
• 𝑃 𝑛 ≤ 𝑛𝑚𝑎𝑥 = 𝑃 1 + 𝑃 2 + 𝑃 3 + . . . +𝑃(𝑛max =
𝑔𝑒𝑜𝑚𝑒𝑡𝑐𝑑𝑓(𝑝, 𝑛max
1
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2
3
...
𝑛max
89
Geometric Distribution – TI-84 Calculator
The probability that a success will occur after try 𝑛𝑚𝑖𝑛 is given
by 𝑃 𝑛 > 𝑛𝑚𝑖𝑛 =
1 − 𝑃 𝑛 ≤ 𝑛𝑚𝑖𝑛 = 1 − 𝑔𝑒𝑜𝑚𝑒𝑡𝑐𝑑𝑓(𝑝, 𝑛m𝑖𝑛
1
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2
𝑛m𝑖𝑛
> 𝑛m𝑖𝑛
90
Exercise 2 Solution: Geometric
Distribution
Susan is taking Western Civilization on a pass/fail basis.
Historically, the passing rate for this course has been
77% each term. Complete the following table:
Probability Description Probability
Expression
success on exactly the 3rd
try
success on or before the
3rd try
success after the 3rd try
Larson/Farber 4th ed
TI-84 Command
91
Exercise 2 Solution: Geometric
Distribution
Susan is taking Western Civilization on a pass/fail basis.
Historically, the passing rate for this course has been
77% each term. Complete the following table:
Probability
Description
success on exactly
the 3rd try
success on or before
the 3rd try
success after the 3rd
try
Larson/Farber 4th ed
Probability Expression
𝑃 3 = (0.77 (0.23
𝑃 𝑛 ≤ 3
𝑃 𝑛> 3 =
1 −𝑃 𝑛 ≤3
2
TI-84 Command
𝑔𝑒𝑜𝑚𝑒𝑡𝑝𝑑𝑓(0.77,3 =
0.041
𝑔𝑒𝑜𝑚𝑒𝑡𝑐𝑑𝑓(0.77,3 =
0.988
1 −
𝑔𝑒𝑜𝑚𝑒𝑡𝑐𝑑𝑓(0.77,3 =
0.012
92
Exercise 3: Geometric Distribution
From experience, you know that the probability that you
will make a sale on any given telephone call is 0.23.
Find the probability that your first sale on any given day
will occur on your fourth or fifth sales call.
Solution:
• P(sale on fourth or fifth call) = P(4) + P(5)
• Geometric with p = 0.23, q = 0.77, n = 4, 5
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Solution: Geometric Distribution
• P(4) = 0.23(0.77)4–1 ≈ 0.105003
• P(5) = 0.23(0.77)5–1 ≈ 0.080852
P(sale on fourth or fifth call) = P(4) + P(5)
≈ 0.105003 + 0.080852
≈ 0.186
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Poisson Distribution
A discrete probability distribution. It satisfies the following
conditions:
 The experiment consists of counting the number of
times an event occurs in a given interval
 The interval can be an interval of time, area, or
volume
 The probability of the event occurring is the same for
each interval
 The number of occurrences in one interval is
independent of the number of occurrences in other
intervals
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The Poisson Distribution
• This distribution is used to model the number of
“rare” events that occur in a time interval, volume,
area, length, etc…
• Examples:
 Number of auto accidents during a month
 Number of diseased trees in an acre
 Number of customers arriving at a bank
Poisson Distribution
• The probability of exactly r occurrences in an interval is
𝑒 −𝜆 ∙ 𝜆𝑟
𝑃 𝑟 =
= 𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑝𝑑𝑓(𝜆, 𝑟
𝑟!
where e  2.71828 and λ is the
mean number of occurrences in
the interval
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The Poisson Distribution
Exercise 4: Poisson Distribution
The mean number of accidents per month at a certain
intersection is 3. What is the probability that in any
given month four accidents will occur at this
intersection?
𝑢𝑛𝑖𝑡 𝑟𝑎𝑡𝑒 =
𝜆 = 3, r =4
𝑃 4 =
Larson/Farber 4th ed
3 𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡𝑠
𝑚𝑜𝑛𝑡ℎ
𝑒 −3 ∙34
4!
=
𝜆 𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡𝑠
𝑚𝑜𝑛𝑡ℎ
= 𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑝𝑑𝑓 3,4 ≈ 0.168
99
Exercise 5: Poisson Distribution
On average, boat fishermen on Pyramid Lake catch
0.667 fish per hour. Suppose you decide to fish the lake
on a boat for 7 hours.
a) What is S ?
b) What is the expected number of fish caught over the
7-hour period?
c) Construct a table for r = 0, 1, 2, . . ., 8 fish
r
0
1
2
3
4
5
6
7
8
P(r)
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Exercise 5 Solution: Poisson Distribution
On average boat fishermen on Pyramid Lake catch
0.667 fish per hour. Suppose you decide to fish the lake
on a boat for 7 hours.
a) What is S? S = fish is caught
b) What is the expected number of fish caught over the
7-hour period?
• 𝑢𝑛𝑖𝑡 𝑟𝑎𝑡𝑒 =
0.667 𝑓𝑖𝑠ℎ
ℎ𝑜𝑢𝑟
=
𝜆 𝑓𝑖𝑠ℎ
7 ℎ𝑜𝑢𝑟𝑠
• 𝜆 = 0.667 7 ≈ 4.7 // Expected number of fish
caught in a 7 hour period
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Exercise 5 Solution: Poisson Distribution
On average boat fishermen on Pyramid Lake catch
0.667 fish per hour. Suppose you decide to fish the lake
on a boat for 7 hours.
c) Construct a table for r = 0,1,2,3 … 8 fish
λ
4.7
L1 r
P(r) =
poissonpdf(
L2 4.7, L1)
Larson/Farber 4th ed
0
1
2
3
4
5
6
0.009
0.043
0.100
0.157
0.185
0.174
0.136
7
8
0.091 0.054
102
Exercise 5 Solution: Poisson Distribution
Probability Expression
Probability
Description
At least 4 fish?
More than 7
fish?
𝑃 𝑟 ≥ 4 = 1 − 𝑃(𝑟 ≤ 3
1 − 𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑐𝑑𝑓 4.7,3
= 0.690
𝑃 𝑟>7 =𝑃 𝑟≥8
= 1 − 𝑃(𝑟 ≤ 7
1 − 𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑐𝑑𝑓 4.7,7
= 0.104
Between 2 and 𝑃 2 ≤ 𝑟 ≤ 5
5 fish inclusive = 𝑃 𝑟 ≤ 5 − 𝑃 𝑟 ≤ 1
Larson/Farber 4th ed
TI-84 Command
𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑐𝑑𝑓 4.7,5
− 𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑐𝑑𝑓 4.7,1
= 0.617
103
Finding Poisson Probabilities
Using the Table
• We can use Page A-16 instead of the formula:
1) Find λ at the top of the table
2) Find r along the left margin of the table
Using the Poisson Table Page A17
Recall, λ = 4
r=0
r=4
r=7
Poisson Approximation to the Binomial
Exercise 6: Poisson Approximation to the
Binomial
In Kentucky’s Pick 4 game, you pay $1 to select a
sequence of four digits, such as 2283. If you play this
game once every day, find the probability of winning
exactly once in 365 days.
Solution: Consider a binomial experiment:
1. 𝑛 = 365
2. Because there is one winning set of numbers
among the 10,000 that are possible
1
𝑝=
10,000
Example: Poisson Approximation to the
Binomial
1. 𝑛 ≥ 100
2. 𝑛𝑝 < 10
[n = 365]
[ np = 0.0365]
Conclusion: We may use the Poisson distribution as
an approximation to the binomial distribution.
Winning is a “rare” event.
λ = 𝜇 = 𝑛𝑝 = 365 ⋅
1
10,000
= 0.0365
On average, we can expect to win 0.0365 times in a
365-day interval.
Example: Poisson Approximation to the
Binomial
𝑒 −0.0365 ∙ 0.03651
𝑃 1 =
1!
= 𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑝𝑑𝑓 0.0365,1 = 0.035
Interpretive Statement: The probability of
winning exactly once in a 365-day interval is
about 0.035.
Section 5.4 Summary
• Found probabilities using the geometric distribution
• Found probabilities using the Poisson distribution
• Used a Poisson Approximation to the Binomial
Distribution
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