1. 6- 1
Chapter Six
Discrete Probability Distributions
GOALS
When you have completed this chapter, you will be able to:
ONE
Define the terms random variable and probability distribution.
TWO
Distinguish between a discrete and continuous probability
distributions.
THREE
Calculate the mean, variance, and standard deviation of a discrete
probability distribution.
FOUR
Describe the characteristics and compute probabilities using the
binomial probability distribution.
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2. 6- 2
Chapter Six continued
A Survey of Probability Concepts
GOALS
When you have completed this chapter, you will be able to:
FIVE
Describe the characteristics and compute probabilities using the
hypergeometric distribution.
SIX
Describe the characteristics and compute the probabilities using
the Poisson distribution.
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3. 6- 3
Random Variables
A random variable is a numerical value
determined by the outcome of an
experiment.
A probability distribution is the listing of all
possible outcomes of an experiment and the
corresponding probability.
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4. 6- 4
Types of Probability Distributions
A discrete probability distribution can
assume only certain outcomes.
A continuous probability distribution
can assume an infinite number of values
within a given range.
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5. 6- 5
Types of Probability Distributions
Examples of a discrete distribution are:
The number of students in a class.
The number of children in a family.
The number of cars entering a carwash in a hour.
Number of home mortgages approved by Coastal
Federal Bank last week.
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6. 6- 6
Types of Probability Distributions
Examples of a continuous distribution include:
The distance students travel to class.
The time it takes an executive to drive to work.
The length of an afternoon nap.
The length of time of a particular phone call.
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7. 6- 7
Features of a Discrete Distribution
The main features of a discrete probability
distribution are:
The sum of the probabilities of the various
outcomes is 1.00.
The probability of a particular outcome is
between 0 and 1.00.
The outcomes are mutually exclusive.
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8. 6- 8
Example 1
EXAMPLE 1: Consider a random experiment in
which a coin is tossed three times. Let x be the
number of heads. Let H represent the outcome of
a head and T the outcome of a tail.
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9. 6- 9
EXAMPLE 1 continued
The possible outcomes for such an experiment
will be:
TTT, TTH, THT, THH,
HTT, HTH, HHT, HHH.
Thus the possible values of x (number of
heads) are 0,1,2,3.
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10. 6- 10
EXAMPLE 1 continued
 The outcome of zero heads occurred once.
 The outcome of one head occurred three times.
 The outcome of two heads occurred three times.
 The outcome of three heads occurred once.
 From the definition of a random variable, x as
defined in this experiment, is a random variable.
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11. 6- 11
The Mean of a Discrete Probability
Distribution
 The mean:
reports the central location of the data.
is the long-run average value of the random variable.
is also referred to as its expected value, E(X), in a
probability distribution.
is a weighted average.
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12. 6- 12
The Mean of a Discrete Probability
Distribution
The mean is computed by the formula:
µ = Σ[ xP( x)]
where w represents the mean and P(x) is the
probability of the various outcomes x.
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13. 6- 13
The Variance of a Discrete
Probability Distribution
The variance measures the amount of spread
(variation) of a distribution.
The variance of a discrete distribution is
denoted by the Greek letter d2 (sigma
squared).
The standard deviation is the square root of
2.
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14. 6- 14
The Variance of a Discrete
Probability Distribution
The variance of a discrete probability distribution
is computed from the formula:
2 2
σ = Σ[( x − µ ) P ( x)]
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15. 6- 15
EXAMPLE 2
Dan Desch, owner of
College Painters, # of Houses Weeks
studied his records for Painted
10 5
the past 20 weeks and
reports the following 11 6
number of houses
painted per week: 12 7
13 2
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16. 6- 16
EXAMPLE 2 continued
Probability Distribution:
Number of houses Probability, P(x)
painted, x
10 .25
11 .30
12 .35
13 .10
Total 1.00
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17. 6- 17
EXAMPLE 2 continued
Compute the mean number of houses painted per
week:
µ = E ( x) = Σ[ xP( x)]
= (10)(.25) + (11)(.30) + (12)(.35) + (13)(.10)
= 11.3
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18. 6- 18
EXAMPLE 2 continued
Compute the variance of the number of
houses painted per week:
2 2
σ = Σ[( x − µ ) P( x)]
= (10 − 11.3) 2 (.25) + ... + (13 − 11.3) 2 (.10)
= 0.4225 + 0.0270 + 0.1715 + 0.2890
= 0.91
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19. 6- 19
Binomial Probability Distribution
 The binomial distribution has the following
characteristics:
An outcome of an experiment is classified into
one of two mutually exclusive categories, such
as a success or failure.
The data collected are the results of counts.
The probability of success stays the same for
each trial.
The trials are independent.
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20. 6- 20
Binomial Probability Distribution
To construct a binomial distribution, let
 n be the number of trials
 x be the number of observed successes
 be the probability of success on each trial
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21. 6- 21
Binomial Probability Distribution
The formula for the binomial probability
distribution is:
x n− x
P( x)= n C xπ (1 − π )
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22. 6- 22
EXAMPLE 3
The Alabama Department of Labor reports that 20% of
the workforce in Mobile is unemployed. From a sample
of 14 workers, calculate the following probabilities:
Exactly three are unemployed.
At least three are unemployed.
At least none are unemployed.
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23. 6- 23
EXAMPLE 3 continued
 The probability of exactly 3:
P (3)=14 C 3 (.20) 3 (1 − .20)11
= (364)(.0080)(.0859)
= .2501
 The probability of at least 3 is:
3 11 14 0
P( x ≥ 3)= 14 C3 (.20) (.80) + ...+ 14 C14 (.20) (.80)
= .250 + .172 + ... + .000 = .551
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24. 6- 24
Example 3 continued
The probability of at least one being unemployed.
P( x ≥ 1) = 1 − P(0)
0 14
= 1− 14 C0 (.20) (1 − .20)
= 1 − .044 = .956
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25. 6- 25
Mean & Variance of the Binomial
Distribution
The mean is found by:
µ = nπ
The variance is found by:
σ = nπ (1 − π )
2
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26. 6- 26
EXAMPLE 4
From EXAMPLE 3, recall that , =.2 and n=14.
π
Hence, the mean is:
i = n = 14(.2) = 2.8.
The variance is:
i 2 = n (1- ( ) = (14)(.2)(.8) =2.24.
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27. 6- 27
Finite Population
A finite population is a population consisting of a
fixed number of known individuals, objects, or
measurements. Examples include:
The number of students in this class.
The number of cars in the parking lot.
The number of homes built in Blackmoor.
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28. 6- 28
Hypergeometric Distribution
The hypergeometric distribution has the following
characteristics:
 There are only 2 possible outcomes.
 The probability of a success is not the same on each trial.
 It results from a count of the number of successes in a
fixed number of trials.
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29. 6- 29
Hypergeometric Distribution
 The formula for finding a probability using the
hypergeometric distribution is:
( S Cx )( N −S Cn −x )
P( x ) =
N Cn
where N is the size of the population, S is the
number of successes in the population, x is the
number of successes in a sample of n
observations.
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30. 6- 30
Hypergeometric Distribution
 Use the hypergeometric distribution to find the
probability of a specified number of successes or
failures if:
the sample is selected from a finite population
without replacement (recall that a criteria for the
binomial distribution is that the probability of
success remains the same from trial to trial).
the size of the sample n is greater than 5% of the
size of the population N .
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31. 6- 31
EXAMPLE 5
 The National Air Safety Board has a list of 10
reported safety violations. Suppose only 4 of the
reported violations are actual violations and the
Safety Board will only be able to investigate five of
the violations. What is the probability that three of
five violations randomly selected to be investigated
are actually violations?
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32. 6- 32
EXAMPLE 5 continued
( 4 C 3 )(10− 4 C 5− 2
P(3) =
10 C 5
( 4 C 3 )( 6 C 2 ) 4(15)
= = = .238
10 C 5 252
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33. 6- 33
Poisson Probability Distribution
The binomial distribution becomes more skewed
to the right (positive) as the probability of success
become smaller.
The limiting form of the binomial distribution
where the probability of success w is small and n
is large is called the Poisson probability
distribution.
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34. 6- 34
Poisson Probability Distribution
The Poisson distribution can be described
mathematically using the formula:
µe x −u
P( x) =
x!
where wis the mean number of successes in a
particular interval of time, e is the constant
2.71828, and x is the number of successes.
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35. 6- 35
Poisson Probability Distribution
The mean number of successes T can be
determined in binomial situations by n , where
n is the number of trials and the probability of a
success.
The variance of the Poisson distribution is also
equal to n .
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36. 6- 36
EXAMPLE 6
 The Sylvania Urgent Care facility specializes in caring
for minor injuries, colds, and flu. For the evening hours
of 6-10 PM the mean number of arrivals is 4.0 per hour.
What is the probability of 4 arrivals in an hour?
x −u −4
µ e 2
4 e
P( x) = = = .1465
x! 2!
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