Elementary Statistics
Eighth Edition
Chapter 4
Discrete
Probability
Distributions
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Slide - 1
Slide - 2
Chapter Outline
4.1 Probability Distributions
4.2 Binomial Distributions
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Slide - 3
Section 4.1 Probability Distributions
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Slide - 4
Section 4.1 Objectives
1. How to distinguish between discrete random variables
and continuous random variables
2. How to construct a discrete probability distribution and
its graph and how to determine if a distribution is a
probability distribution
3. How to find the mean, variance, and standard deviation
of a discrete probability distribution
4. How to find the expected value of a discrete probability
distribution
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Slide - 5
Random Variables (1 of 3)
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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Slide - 6
Random Variables (2 of 3)
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.
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Slide - 7
Random Variables (3 of 3)
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.
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Slide - 8
Example: Discrete and Continuous
Variables (1 of 2)
Determine whether each random variable x is discrete or
continuous. Explain your reasoning.
1. Let x represent the number of Fortune 500 companies
that lost money in the previous year.
Solution:
Discrete random variable (The number of companies that
lost money in the previous year can be counted.)
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Slide - 9
Example: Discrete and Continuous
Variables (2 of 2)
Determine whether each random variable x is discrete or
continuous. Explain your reasoning.
2. Let x represent the volume of gasoline in a 21-gallon
tank.
Solution:
Continuous random variable (The amount of gasoline in
the tank can be any volume between 0 gallons and 21
gallons.)
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Slide - 10
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)  
2. The sum of all the probabilities is 1.
 P( x) = 1
0 is less than or equal to P of x is less than or equal to 1
the sum of P of x = 1
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Slide - 11
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,
inclusive, and that the sum of all the probabilities is 1.
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Slide - 12
Example: Constructing and Graphing a
Discrete Probability Distribution
An industrial psychologist
administered a personality inventory
test for passive-aggressive traits to
150 employees. Each individual was
given a whole number score from 1 to
5, where 1 is extremely passive and 5
is extremely aggressive. A score of 3
indicated neither trait. The results are
shown. Construct a probability
distribution for the random variable x.
Then graph the distribution using a
histogram.
Score, x
Frequency, f
1
24
2
33
3
42
4
30
5
21
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Slide - 13
Solution: Constructing and Graphing a
Discrete Probability Distribution (1 of 3)
• 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.
24
P(1) =
= 0.16
150
33
P(2) =
= 0.22
150
30
P(4) =
= 0.20
150
P(5) =
42
P(3) =
= 0.28
150
21
= 0.14
150
• Discrete probability distribution:
x
P( x)
P of x
1
0.16
2
0.22
3
0.28
4
0.20
5
0.14
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Slide - 14
Solution: Constructing and Graphing a
Discrete Probability Distribution (2 of 3)
x
P of x
P( x)
1
0.16
2
0.22
3
0.28
4
0.20
5
0.14
This is a valid discrete probability distribution since
1. Each probability is between 0 and 1, inclusive,
0  P ( x )  
2. The sum of the probabilities equals 1,
 P ( x ) = 016 + 022 + 028 + 020 + 014 = 
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Slide - 15
Solution: Constructing and Graphing a
Discrete Probability Distribution (3 of 3)
x
P( x)
1
2
3
4
5
0.16
0.22
0.28
0.20
0.14
P of x
Because the width of each bar is
one, the area of each bar is equal
to the probability of a particular
outcome. Also, the probability of
an event corresponds to the sum
of the areas of the outcomes
included in the event.
Passive-Aggressive Traits
You can see that the distribution is approximately symmetric.
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Slide - 16
Example: Verifying a Probability
Distribution
Verify that the distribution for the three-day forecast and the
number of days of rain is a probability distribution.
Days of Rain, x
Probability, P ( x )
P of x
0
0.216
1
0.432
2
0.288
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3
0.064
Slide - 17
Solution: Verifying a Probability
Distribution
Solution
If the distribution is a probability distribution, then (1) each
probability is between 0 and 1, inclusive, and (2) the sum of
all the probabilities equals 1.
1. Each probability is between 0 and 1.
2.
 P ( x ) = 0216 + 0432 + 0288 + 0064 = 
Days of Rain, x
Probability, P ( x )
P of x
0
1
2
3
0.216
0.432
0.288
0.064
Because both conditions are met, the distribution is a
probability distribution.
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Slide - 18
Example: Identifying Probability
Distributions (1 of 2)
Determine whether each distribution is a probability
distribution. Explain your reasoning.
1.
x
5
6
7
8
0.28
0.21
0.43
0.15
P of x
P( x)
Solution
• Each probability is between 0 and 1, but the sum of
all the probabilities is 1.07, which is greater than 1.
• The sum of all the probabilities in a probability
distribution always equals 1. So, this distribution is
not a probability distribution.
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Slide - 19
Example: Identifying Probability
Distributions (2 of 2)
Determine whether each distribution is a probability
distribution. Explain your reasoning.
2.
x
P( x)
P of x
1
2
3
4
1
2
1
4
5
4
−1
half
1 fourth
5 fourths
negative 1
Solution
• The sum of all the probabilities is equal to 1, but
P ( 3) and P ( 4 ) are not between 0 and 1.
• Probabilities can never be negative or greater than 1.
So, this distribution is not a probability distribution.
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Slide - 20
Mean
Mean of a discrete probability distribution
•  =  xP ( x )
• Each value of x is multiplied by its corresponding
probability and the products are added.
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Slide - 21
Example: Finding the Mean
The probability distribution for the personality inventory test
for passive-aggressive traits is given. Find the mean score.
Solution:
x
1
2
3
4
5
P( x)
xP ( x )
0.16
0.22
0.28
0.20
0.14
1(0.16) = 0.16
P of x
x P of x
1 left parenthesis 0.16 right parenthesis, = 0.16
2(0.22) = 0.44
2 left parenthesis 0.22 right parenthesis, = 0.44
3(0.28) = 0.84
3 left parenthesis 0.28 right parenthesis, = 0.84
4(0.20) = 0.80
4 left parenthesis 0.20 right parenthesis, = 0.80
5(0.14) = 0.70
5 left parenthesis 0.14 right parenthesis, = 0.70
 =  xP ( x ) = 294
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Solution: Finding the Mean
The probability distribution for the personality inventory test
for passive-aggressive traits is given. Find the mean score.
Solution:
 =  xP ( x ) = 294
Recall that a score of 3 represents an individual who
exhibits neither passive nor aggressive traits and the
mean is slightly less than 3. So, the mean personality
trait is neither extremely passive nor extremely
aggressive, but is slightly closer to passive.
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Slide - 23
Variance and Standard Deviation
Variance of a discrete probability distribution
•  2 = ( x −  ) P( x)
2
Standard deviation of a discrete probability distribution
•  =  = ( x −  ) P( x)
2
2
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Slide - 24
Example: Finding the Variance and
Standard Deviation
The probability distribution for the personality inventory test
for passive-aggressive traits is given. Find the variance and
standard deviation.
Score, x Probability, P ( x )
P of x
1
2
0.16
0.22
3
0.28
4
5
0.20
0.14
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Slide - 25
Solution: Finding the Variance and
Standard Deviation
 = 2.94
Recall
( x –  )2
x–
x
P( x)
1
0.16
1 – 2.94 = –1.94
( –1.94 )
2
2
0.22
2 – 2.94 = –0.94
( –0.94 )
2
3
0.28
3 – 2.94 = 0.06
( 0.06 )
4
0.20
4 – 2.94 = 1.06
(1.06 )
5
0.14
5 – 2.94 = 2.06
( 2.06 )
P of x
x minus mu
1 minus 2.94 = negative 1.94
2 minus 2.94 = negative 0.94
3 minus 2.94 = 0.06
4 negative 2.94 = 1.06
5 minus 2.94 = 2.06
( x –  )2 P ( x )
left parenthesis x minus mu right parenthesis squared
left parenthesis x minus mu right parenthesis squared P of x
= 3.7636
3.7636 ( 0.16 ) = 0.60218
= 0.8836
0.8836 ( 0.22 ) = 0.19439
left parenthesis negative 1.94 right parenthesis squared = 3.7636
3.7636 left parenthesis 0.16 right parenthesis, = 0.60218
left parenthesis negative 0.94 right parenthesis squared = 0.8836
2
= 0.0036
0.0036 ( 0.28 ) = 0.00101
= 1.1236
1.1236 ( 0.20 ) = 0.22472
= 4.2436
4.2436 ( 0.14 ) = 0.59410
left parenthesis 0.06 right parenthesis squared = 0.0036
2
0.0036 left parenthesis 0.28 right parenthesis = 0.00101
left parenthesis 1.06 right parenthesis squared = 1.1236
2
1.1236 left parenthesis 0.20 right parenthesis, = 0.22472
left parenthesis 2.06 right parenthesis squared = 4.2436
Variance:
Standard Deviation:
0.8836 left parenthesis 0.22 right parenthesis = 0.19439
4.2436 left parenthesis 0.14 right parenthesis, = 0.59410
 =  ( x −  ) P ( x ) = 16164

2
 =  2 = 16164  1.27  13
Most of the data values differ from the mean by no more than 1.3.
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Slide - 26
Expected Value
Expected value of a discrete random variable
• Equal to the mean of the random variable.
• E ( x ) =  =  xP ( x )
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Slide - 27
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. Find the
expected value and interpret its meaning.
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Slide - 28
Solution: Finding an Expected Value (1 of 2)
• 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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Slide - 29
Solution: Finding an Expected Value (2 of 2)
• Probability distribution for the possible gains
(outcomes)
Gain, x
$498
$248
$148
P( x)
1
1500
1
1500
1
1500
P of x
start fraction 1 over 1500 end fraction
start fraction 1 over 1500 end fraction
start fraction 1 over 1500 end fraction
$73
1
1500
start fraction 1 over 1500 end fraction
−$2
negative $2
1496
1500
start fraction 1496 over 1500 end fraction
E ( x ) =  xP ( x )
1
1
1
1
1496
+ $248 
+ $148 
+ $73 
+ ( −$2 ) 
1500
1500
1500
1500
1500
= −$1.35
You can expect to lose an average of $1.35 for each ticket
you buy.
= $498 
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Section 4.2 Binomial Distributions
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Section 4.2 Objectives
1. How to determine whether a probability experiment is a
binomial experiment
2. How to find binomial probabilities using the binomial
probability formula
3. How to find binomial probabilities using technology,
formulas, and a binomial probability table
4. How to construct and graph a binomial distribution
5. How to find the mean, variance, and standard deviation
of a binomial probability distribution
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Slide - 32
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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Slide - 33
Notation for Binomial Experiments
Symbol
Description
n
The number of times a trial is repeated
p
The probability of success in a single trial
q
The probability of failure in a single trial
(q = 1 − p)
left parenthesis q = 1 minus p right parenthesis
x
The random variable represents a count of the
number of successes in n trials:
x = 0, 1, 2, 3,  , n.
x = 0, 1, 2, 3, ellipsis, n.
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Slide - 34
Example: Identifying and Understanding
Binomial Experiments (1 of 2)
Decide whether each 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 x. If it is
not, explain why.
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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Slide - 35
Solution: Identifying and Understanding
Binomial Experiments (1 of 3)
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 x counts the number of successful
surgeries.
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Slide - 36
Solution: Identifying and Understanding
Binomial Experiments (2 of 3)
Binomial Experiment
• n = 8 (number of trials)
• p = 0.85 (probability of success)
• q = 1 − p = 1 − 0.85 = 0.15 (probability of failure)
• x = 0, 1, 2, 3, 4, 5, 6, 7, 8 (number of successful
surgeries)
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Slide - 37
Example: Identifying and Understanding
Binomial Experiments (2 of 2)
Decide whether each 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 x.
If it is not, explain why.
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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Slide - 38
Solution: Identifying and Understanding
Binomial Experiments (3 of 3)
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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Slide - 39
Binomial Probability Formula
Binomial Probability Formula
• The probability of exactly x successes in n trials is
P( x ) = n C x p q
x
n− x
n
=
p x q n− x
( n − x ) x
• n = number of trials
• p = probability of success
• q = 1 − p probability of failure
• x = number of successes in n trials
• Note: number of failures is
n−x
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Slide - 40
Example: Finding a Binomial
Probability
Rotator cuff surgery has a 90% chance of success. The
surgery is performed on three patients. Find the probability
of the surgery being successful on exactly two patients.
(Source: The Orthopedic Center of St. Louis)
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Slide - 41
Solution: Finding a Binomial
Probability (1 of 2)
Method 1: Draw a tree diagram and use the Multiplication
Rule.
 81 
3

1000


= 0.243
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Slide - 42
Solution: Finding a Binomial
Probability (2 of 2)
Method 2: Use the binomial probability formula.
9
1
n = 3, p = , q = , and x = 2.
10
10
3
9
P ( 2) =
 
3
−
2

( )  10 
2
1
1
 
 10 
 81  1 
= 3
 
100

 10 
 81 
= 3

1000


= 0.243
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Slide - 43
Example: Constructing a Binomial
Distribution
In a survey, U.S. adults were
asked how often they go
online. The results are shown
in the figure. Six adults who
participated in the survey are
randomly selected and asked
whether they go online several
times a day. Construct a
binomial probability distribution
for the number who respond
that they go online several
times a day. (Source: Pew
Research)
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Slide - 44
Solution: Constructing a Binomial
Distribution (1 of 2)
p = 0.48 and q = 0.52
n = 6, possible values for x are 0, 1, 2, 3, 4, 5 and 6
P ( 0 ) = 6C0 (0.48)0 (0.52)6 = 1(0.48)0 (0.52)6  0.020
P (1) = 6C1 (0.48)1 (0.52)5 = 6(0.48)1 (0.52)5  0.109
P ( 2 ) = 6C2 (0.48) 2 (0.52) 4 = 15(0.48) 2 (0.52) 4  0.253
P ( 3) = 6C3 (0.48)3 (0.52)3 = 20(0.48)3 (0.52)3  0.311
P ( 4 ) = 6C4 (0.48) 4 (0.52) 2 = 15(0.48) 4 (0.52) 2  0.215
P ( 5 ) = 6C5 (0.48)5 (0.52)1 = 6(0.48)5 (0.52)1  0.079
P ( 6 ) = 6C6 (0.48)6 (0.52)0 = 1(0.48)6 (0.52)0  0.012
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Slide - 45
Solution: Constructing a Binomial
Distribution (2 of 2)
Notice in the table that all the probabilities are between 0
and 1 and that the sum of the probabilities is about 1.
x
P( x)
0
0.020
1
0.109
2
0.253
3
0.311
4
0.215
5
0.079
6
0.012
Blank
P of x
 P( x) = 0.999
sum of P of x = 0.999
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Slide - 46
Example: Finding a Binomial
Probabilities Using Technology
A survey found that 75% of U.S. adults are experiencing
digital device fatigue. (This is the state of mental
exhaustion caused by concurrent and excessive use of
digital devices such as smartphones and tablet
computers.) You randomly select 100 adults. What is the
probability that exactly 66 are experiencing digital device
fatigue? Use technology to find the probability. (Source:
The Harris Poll)
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Slide - 47
Solution: Finding a Binomial
Probabilities Using Technology (1 of 2)
Solution
Minitab, Excel,
StatCrunch, and the T I-84
Plus each have features
that allow you to find
binomial probabilities. Try
using these technologies.
You should obtain results
similar to these displays.
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Slide - 48
Solution: Finding a Binomial
Probabilities Using Technology (2 of 2)
Solution
From these displays, you can see that the probability that
exactly 66 adults are experiencing digital device fatigue is
about 0.011. Because 0.011 is less than 0.05, this can be
considered an unusual event.
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Slide - 49
Example: Finding Binomial
Probabilities Using Formulas
A survey found that 22% of U.S. adults say that the
economy, unemployment, and jobs are the most important
problems facing the U.S. today. You randomly select four
adults and ask them whether the economy, unemployment,
and jobs are the most important problems facing the U.S.
today. Find the probability that (1) exactly two of them
respond yes, (2) at least two of them respond yes, and (3)
fewer than two of them respond yes. (Source: Ipsos)
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Slide - 50
Solution: Finding Binomial Probabilities
Using Formulas (1 of 4)
Solution
1. Using n = 4, p = 0.22, q = 0.78, and x = 2, the probability
that exactly two adults will respond yes is
P ( 2 ) = 4 C2 ( 0.22 ) ( 0.78 ) = 6 ( 0.22 ) ( 0.78 )  0.177.
2
2
2
2
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Slide - 51
Solution: Finding Binomial Probabilities
Using Formulas (2 of 4)
Solution
2. To find the probability that at least two adults will respond
yes, find the sum of P(2), P(3), and P(4).
Begin by using the binomial probability formula to write
an expression for each probability.
P ( 2 ) = 4 C2 ( 0.22 ) ( 0.78 ) = 6 ( 0.22 ) ( 0.78 )
2
2
2
2
P ( 3) = 4 C3 ( 0.22 ) ( 0.78 ) =  ( 0.22 ) ( 0.78 )
3
1
3
1
P ( 4 ) = 4 C4 ( 0.22 ) ( 0.78 ) = ( 0.22 ) ( 0.78 )
4
0
4
0
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Slide - 52
Solution: Finding Binomial Probabilities
Using Formulas (3 of 4)
Solution
2. So, the probability that at least two will respond yes is
P ( x   ) = P ( 2 ) + P ( 3) + P ( 4 ) = 6 ( 0.22 ) ( 0.78 )
2
+4 ( 0.22 ) ( 0.78 ) + 1( 0.22 ) ( 0.78 )
3
1
4
2
0
 0.212.
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Slide - 53
Solution: Finding Binomial Probabilities
Using Formulas (4 of 4)
Solution
3. To find the probability that fewer than two adults will
respond yes, find the sum of P(0) and P(1).
P ( 0 ) = 4 C0 ( 0.22 ) ( 0.78 ) = ( 0.22 ) ( 0.78 )
0
4
0
P (1) = 4 C1 ( 0.22 ) ( 0.78 ) =  ( 0.22 ) ( 0.78 )
1
3
1
4
3
So, the probability that fewer than two will respond yes is
P ( x  2 ) = P ( 0 ) + P (1)
= 1( 0.22 ) ( 0.78 ) + 4 ( 0.22 ) ( 0.78 )  0.788.
0
4
1
3
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Slide - 54
Example: Finding a Binomial
Probability Using a Table
About 5% of workers (ages 16 years and older) in the United
States commute to their jobs by using public transportation
(excluding taxicabs). You randomly select eight workers.
What is the probability that exactly three of them use public
transportation to get to work? Use a table to find the
probability. (Source: American Community Survey)
Solution:
• Binomial with n = 8, p = 0.05, x = 3
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Slide - 55
Solution: Finding Binomial Probabilities
Using a Table (1 of 2)
• A portion of Table 2 is shown
According to the table, the probability is 0.005.
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Slide - 56
Solution: Finding Binomial Probabilities
Using a Table (2 of 2)
• You can check the result using technology.
So, the probability that exactly four of the eight workers
carpool to work is 0.005. Because 0.005 is less than 0.05,
this can be considered an unusual event.
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Slide - 57
Example: Graphing a Binomial
Distribution
Sixty-four percent of people who have survived cancer are
ages 65 years or older. You randomly select six people who
have survived cancer and ask them whether they are 65
years of age or older. Construct a probability distribution for
the random variable x. Then graph the distribution. (Source:
American Cancer Society)
Solution:
• n = 6, p = 0.64, q = 0.36
• Find the probability for each value of x
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Slide - 58
Solution: Graphing a Binomial
Distribution (1 of 2)
x
P( x)
P of x
0
1
2
3
4
5
6
0.002
0.023
0.103
0.245
0.326
0.232
0.069
Notice in the table that all the probabilities are between 0
and 1 and that the sum of the probabilities is 1.
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Slide - 59
Solution: Graphing a Binomial
Distribution (2 of 2)
Histogram:
People Who Have Survived
Cancer, 65 Years of Age or Older
From the histogram, you can see that it would be unusual for
none or only one of the survivors to be age 65 years or older
because both probabilities are less than 0.05.
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Slide - 60
Mean, Variance, and Standard Deviation
• Mean:
 = np
• Variance:
 = npq
2
• Standard Deviation:
 = npq
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Slide - 61
Example: 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 Oceanic and Atmospheric Administration)
Solution: n = 30, p = 0.56, q = 0.44
Mean:  = np = 30  0.56 = 16.8
2
Variance:  = npq = 30  0.56  0.44  7.4
Standard Deviation:  =
npq = 30  0.56  0.44  2.7
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Slide - 62
Solution: Mean, Variance, and Standard
Deviation
 = 16.8  2  7.4   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 two standard
deviations from the mean are considered
unusual.
– 16.8 − 2 ( 2.7 ) = 11.4; A June with 11 cloudy days or
less would be unusual.
– 16.8 + 2 ( 2.7 ) = 22.2; A June with 23 cloudy days or
more would also be unusual.
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Slide - 63
Copyright
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provided solely for the use of instructors in teaching their
courses and assessing student learning. Dissemination or sale of
any part of this work (including on the World Wide Web) will
destroy the integrity of the work and is not permitted. The work
and materials from it should never be made available to students
except by instructors using the accompanying text in their
classes. All recipients of this work are expected to abide by these
restrictions and to honor the intended pedagogical purposes and
the needs of other instructors who rely on these materials.
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Slide - 64