Chapter 4
Discrete Probability
Distributions
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Chapter Outline
• 4.1 Probability Distributions
• 4.2 Binomial Distributions
• 4.3 More Discrete Probability Distributions
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Section 4.1
Probability Distributions
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Section 4.1 Objectives
• How to distinguish between discrete random
variables and continuous random variables
• How to construct a discrete probability distribution
and its graph and how to determine if a distribution is
a probability distribution
• How to find the mean, variance, and standard
deviation of a discrete probability distribution
• How to 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 1 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.
0 1 2 3 … 24
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Example: Discrete and
Continuous Variables
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.)
x
0 1 2 3 … 500
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Example: Discrete and
Continuous Variables
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.)
x
0 1 2 3 … 21
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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 0 P(x) 1
the discrete random variable is
between 0 and 1, inclusive.
2. The sum of all the probabilities ΣP(x) = 1
is 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,
inclusive, and that the sum of all the probabilities
is 1.
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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 Score, x Frequency, f
indicated neither trait. The results 1 24
are shown. Construct a probability 2 33
distribution for the random variable 3 42
x. Then graph the distribution using 4 30
a histogram. 5 21
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Solution: Constructing and
Graphing 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.
24 33 42
P (1) 0.16 P (2) 0.22 P (3) 0.28
150 150 150
30 21
P (4) 0.20 P (5) 0.14
150 150
• 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 and
Graphing 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 and
Graphing a Discrete Probability
x 1 2
Distribution
3 4 5
P(x) 0.16 0.22 0.28 0.20 0.14
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.
You can see that the distribution is approximately
symmetric.
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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 0 1 2 3
Probability, P(x) 0.216 0.432 0.288 0.064
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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. = 0.216 + 0.432 + 0.288 + 0.064 = 1.
Days of Rain, x 0 1 2 3
Probability, P(x) 0.216 0.432 0.288 0.064
Because both conditions are met, the distribution is a
probability distribution.
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Example: Identifying
Probability Distributions
Determine whether each distribution is a probability
distribution. Explain your reasoning.
1. x 5 6 7 8
P(x) 0.28 0.21 0.43 0.15
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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Example: Identifying
Probability Distributions
Determine whether each distribution is a probability
distribution. Explain your reasoning.
2. x 1 2 3 4
P(x) 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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Mean
Mean of a discrete probability distribution
• μ = Σ xP(x)
• Each value of x is multiplied by its corresponding
probability and the products are added.
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Example: Finding the Mean
The probability distribution for the personality
inventory test for passive-aggressive traits is given. Find
the mean score.
x P(x) xP(x)
Solution: 1 0.16 1(0.16) = 0.16
2 0.22 2(0.22) = 0.44
3 0.28 3(0.28) = 0.84
4 0.20 4(0.20) = 0.80
5 0.14 5(0.14) = 0.70
μ = Σ 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: μ = Σ 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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Variance and Standard
Deviation
Variance of a discrete probability distribution
• 2 ( x ) 2 P ( x)
Standard deviation of a discrete probability
distribution
• 2 ( x ) 2 P ( x )
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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)
1 0.16
2 0.22
3 0.28
4 0.20
5 0.14
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Solution: Finding the Variance
and Standard Deviation
Recall μ = 2.94
x P(x) x–μ (x – μ)2 (x – μ)2P(x)
1 0.16 1 – 2.94 = –1.94 (–1.94)2 = 3.764 3.764(0.16) = 0.602
2 0.22 2 – 2.94 = –0.94 (–0.94)2 = 0.884 0.884(0.22) = 0.194
3 0.28 3 – 2.94 = 0.06 (0.06)2 = 0.004 0.004(0.28) = 0.001
4 0.20 4 – 2.94 = 1.06 (1.06)2 = 1.124 1.124(0.20) = 0.225
5 0.14 5 – 2.94 = 2.06 (2.06)2 = 4.244 4.244(0.14) = 0.594
Variance:σ2 = Σ(x – μ)2P(x) = 1.6164
Standard Deviation: = 1.3
Most of the data values differ from the mean by no
more than 1.3.
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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)
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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. Find the expected value and interpret its
meaning.
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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 $498 $248 $148 $73 –$2
1 1 1 1 1496
P(x)
1500 1500 1500 1500 1500
E ( x ) xP ( x )
1 1 1 1 1496
$498 $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.
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