Chapter 4
Discrete Random Variables
(离散型随机变量)
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Discrete Random Variables
4.1
4.2
4.3
4.4
Two Types of Random Variables
Discrete Probability Distributions
The Binomial Distribution
The Poisson Distribution
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Random Variables (随机变量)
A random variable is a variable that assumes numerical
values that are determined by the outcome of an experiment.
Example: 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.
The possible outcomes for such an experiment:
TTT, TTH, THT, THH,
HTT, HTH, HHT, HHH
Thus the possible values of X (number of heads) are 0,1,2,3.
From the definition of a random variable, X as defined in
this experiment, is a random variable.
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Two Types of Random Variables

Discrete random variable（离散型随机变量）:
Possible values can be counted or listed
 For example, the number of TV sets sold at the store
in one day. Here x could be 0, 1, 2, 3, 4 and so forth.

Continuous random variable （连续型随机变量）:
May assume any numerical value in one or more
intervals
 For example, the waiting time for a credit card
authorization, the interest rate charged on a business
loan
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Example: Two Types of Random Variables
Question
Random Variable x
Type
Family
size
x = Number of people in
family reported on tax return
Discrete(离散)
Distance from
home to store
x = Distance in miles from
home to a store
Continuous(连续)
Own dog
or cat
x = 1 if own no pet;
Discrete
= 2 if own dog(s) only;
= 3 if own cat(s) only;
= 4 if own dog(s) and cat(s)
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Discrete Probability Distributions
（离散概率分布）
The probability distribution of a discrete random variable is a
table, graph, or formula that gives the probability associated with
each possible value that the variable can assume
Notation: Denote the values of the random variable by x and the value’s
associated probability by p(x)
Properties
1. For any value x of the random variable, p(x)  0
2. The probabilities of all the events in the sample space
must sum to 1, that is,
 px   1
all x
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Example 4.3: Number of Radios
（Sold at Sound City in a Week）
 Let x be the random variable of the number of radios
sold per week
 x has values x = 0, 1, 2, 3, 4, 5

Given sales history over past 100 weeks
 Let f be the number of weeks (of the past 100) during
which x number of radios were sold
 Records tell us that







f(0)=3
No radios have been sold in 3 of the weeks
f(1)=20
One radios has been sold in 20 of the weeks
f(2)=50 Two radios have been sold in 50 of the weeks
f(3)=20 Three radios have been sold in 20 of the weeks
f(4)=5
Four radios have been sold in 4 of the weeks
f(5)=2
Five radios have been sold in 2 of the weeks
No more than five radios were sold in any of the past 100
weeks
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Frequency distribution of sales history over past 100 weeks
# Radios, x Frequency Relative Frequency Probability, p(x)
0
f(0) =3
3/100 = 0.03
p(0) = 0.03
1
f(1) =20
20/100 = 0.20
p(1) = 0.20
2
f(2) =50
0.50
p(2) = 0.50
3
f(3) =20
0.20
p(3) = 0.20
4
f(4) = 5
0.05
p(4) = 0.05
5
f(5) = 2
0.02
P(5) = 0.02
100
1.00
1.00

Interpret the relative frequencies
as probabilities
 So for any value x,
f(x)/n = p(x)
 Assuming that sales
remain stable over time
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Example 4.3 Continued



What is the chance that two radios will be sold in a
week?
 P(x = 2) = 0.50
What is the chance that fewer than 2 radios will be
sold in a week?
Using the addition
 p(x < 2) = p(x = 0 or x = 1)
rule for the mutually
exclusive values of
= p(x = 0) + p(x = 1)
the random variable.
= 0.03 + 0.20 = 0.23
What is the chance that three or more radios will be
sold in a week?
 p(x ≥ 3) = p(x = 3, 4, or 5)
= p(x = 3) + p(x = 4) + p(x = 5)
= 0.20 + 0.05 + 0.02 = 0.27
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Expected Value of a Discrete
Random Variable
The mean or expected value of a discrete random variable
X is:
 X   x p x 
All x
 is the value expected to occur in the long run and on average
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Example 4.3: Number of Radios
Sold at Sound City in a Week
How many radios should be expected to be sold in a
week?
 Calculate the expected value of the number of
radios sold, X
Radios, x
Probability, p(x)
x p(x)

0
1
2
3
4
5
p(0) = 0.03
p(1) = 0.20
p(2) = 0.50
p(3) = 0.20
p(4) = 0.05
p(5) = 0.02
1.00
0  0.03 = 0.00
1  0.20 = 0.20
2  0.50 = 1.00
3  0.20 = 0.60
4  0.05 = 0.20
5  0.02 = 0.10
2.10
• On average, expect to sell 2.1 radios per week
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Variance and Standard Deviation
The variance of a discrete random variable is:
 2X   x   X 2 px 
All x
• The variance is the average of the squared deviations of the
different values of the random variable from the expected
value
The standard deviation is the square root of the variance
X 
2
X
• The variance and standard deviation measure the spread of the
values of the random variable from their expected value
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Example 4.7: Number of Radios
Sold at Sound City in a Week
Calculate the variance and standard deviation of the number
of radios sold at Sound City in a week
Radios, x
0
1
2
3
4
5
Probability, p(x)
p(0) = 0.03
p(1) = 0.20
p(2) = 0.50
p(3) = 0.20
p(4) = 0.05
p(5) = 0.02
1.00
Standard deviation
 X  0.89  0.9434
(x - X)2 p(x)
(0 – 2.1)2 (0.03) = 0.1323
(1 – 2.1)2 (0.20) = 0.2420
(2 – 2.1)2 (0.50) = 0.0050
(3 – 2.1)2 (0.20) = 0.1620
(4 – 2.1)2 (0.05) = 0.1805
(5 – 2.1)2 (0.02) = 0.1682
0.8900
Variance
 X2  0.89
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3. Let X denote the number of service visits required to fix a problem
with a product purchased from a company.
X
1
2
3
4
P(x)
0.6
0.2
0.15
0.05
1. What is the expected number of visits required to fix the product?
2. The variance of the number of visits required to fix the product?
3. The probability that it will take more than 2 visits?
4. Given that it takes more than one visit, what is the probability
that it will take 4 visits?
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Ans: a.   E( X )  XP( X )  1* 0.6  2 * 0.2  3 * 0.15  4 * 0.05  1.65

b.
 2  Var(X)   ( X   ) 2 P( X )
 (1  1.65) 2 * 0.6  (2  1.65) 2 * 0.2  (3  1.65) 2 * 0.15  (4  1.65) 2 * 0.05
 0.8275
c.
d.
P( X  2)  P( X  3)  P( X  4)  0.15  0.05  0.2
P( X  4 | X  1) 
P( X  4, X  1)
0.05

 0.125
P( X  1)
0.2  0.15  0.05
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The Binomial Distribution (二项分布 )
The Binomial Experiment:
1. Experiment consists of n identical trials
2. Each trial results in either “success” or “failure”
3. Probability of success, p, is constant from trial to trial
4. Trials are independent
Note: The probability of failure, q, is 1 – p and is constant
from trial to trial
If x is the total number of successes in n trials of a
binomial experiment, then x is a binomial random
variable
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The Binomial Distribution #2
For a binomial random variable x, the probability
of x successes in n trials is given by the binomial
distribution:
n!
px  =
p x q n- x
x!n - x !
• Note: n! is read as “n factorial” and n! = n × (n-1) × (n-2) × ... × 1
– For example, 5! = 5  4  3  2  1 = 120
• Also, 0! =1
• Factorials are not defined for negative numbers or fractions
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The Binomial Distribution #3
• What does the equation mean?
– The equation for the binomial distribution consists of the
product of two factors
n!
px  =
x!n - x !

Number of ways to
get x successes and
(n–x) failures in n
trials
p x q n- x
The chance of getting x
successes and (n–x)
failures in a particular
arrangement
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Example 4.10: Incidence of Nausea
 The
company claims that, at most, 10 percentage of all
patients treated with Phe-Mycin would experience nausea
as a side effect of taking the drug.

x = number of patients who will experience nausea
following treatment with Phe-Mycin out of the 4 patients
tested

Find the probability that 2 of the 4 patients treated will
experience nausea Given: n = 4, p = 0.1, with x = 2
Then: q = 1 – p = 1 – 0.1 = 0.9 and
4!
0.1 20.9 42
px  2 
2!4  2!
 60.1 2 0.9 2  0.0486
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Example: Binomial Distribution
n = 4, p = 0.1
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Binomial Probability Table
(Appendix Table A.1, P817)
Table 4.7(a) for n = 4, with x = 2 and p = 0.1
p = 0.1
values of p (.05 to .50)
x
0
1
2
3
4
0.05
0.8145
0.1715
0.0135
0.0005
0.0000
0.95
0.1
0.6561
0.2916
0.0486
0.0036
0.0001
0.9
0.15
0.5220
0.3685
0.0975
0.0115
0.0005
0.85
…
…
…
…
…
…
…
0.50
0.0625
0.2500
0.3750
0.2500
0.0625
0.50
4
3
2
1
0
x
values of p (.05 to .95)
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P(x = 2) = 0.0486
21
Example 4.11: Incidence of Nausea
after Treatment
x = number of patients who will experience nausea
following treatment with Phe-Mycin out of the 4
patients tested
Find the probability that at least 3 of the 4 patients
treated will experience nausea
Set x = 3, n = 4, p = 0.1, so q = 1 – p = 1 – 0.1 = 0.9
Then:
p x  3  p  x  3 or 4 
 p  x  3  p x  4 
 0.0036  .0001  0.0037
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Using the addition rule
for the mutually
exclusive values of the
binomial random
variable
22
Example 4.11: Rare Events

Suppose at least three of four sampled patients actually
did experience nausea following treatment
 If p = 0.1 is believed, then there is a chance of only 37 in
10,000 of observing this result
 So this is very unlikely!
 But it actually occurred
 So, this is very strong evidence that p does not equal
0.1
 There is very strong evidence that p is actually
greater than 0.1
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Several Binomial Distributions
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Mean and Variance of a Binomial Random
Variable
If x is a binomial random variable with parameters (
参数) n and p (so q = 1 – p), then
mean X  np
variance  X2  npq
standard deviation  X 
npq
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Back to Example 4.11

Of 4 randomly selected patients, how many
should be expected to experience nausea after
treatment?

Given: n = 4, p = 0.1

Then X = np = 4  0.1 = 0.4

So expect 0.4 of the 4 patients to experience
nausea

If at least three of four patients experienced nausea,
this would be many more than the 0.4 that are
expected
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Binomial Distribution EXAMPLE:




Pat Statsdud is registered in a statistics course and
intends to rely on luck to pass the next quiz.
The quiz consists on 10 multiple choice questions with 5
possible choices for each question, only one of which is
the correct answer.
Pat will guess the answer to each question
Find the following probabilities
 Pat gets no answer correct
 Pat gets two answer correct?
 Pat fails the quiz
 If all the students in Pat’s class intend to guess the
answers to the quiz, what is the mean and the standard
deviation of the quiz mark?
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
Solution
 Checking the conditions
 An answer can be either correct or incorrect.
 There is a fixed finite number of trials (n=10)
 Each answer is independent of the others.
 The probability p of a correct answer (.20) does not
change from question to question.
Determining the binomial probabilities:
Let X = the number of correct answers
10!
P( X  0) 
(.20 ) 0 (.80 )100  .1074
0! (10  0)!
10!
P( X  2) 
(.20 ) 2 (.80 )10 2  .3020
2! (10  2)!
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Determining the binomial probabilities:
Pat fails the test if the number of correct answers is less
than 5, which means less than or equal to 4.
P(X4 = p(0) + p(1) + p(2) + p(3) + p(4)
= .1074 + .2684 + .3020 + .2013 + .0881
=.9672
The mean and the standard deviation of the quiz mark?
μ= np = 10(.2) = 2.
σ= [np(1-p)]1/2 = [10(.2)(.8)]1/2 = 1.26
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The Poisson Distribution
Consider the number of times an event occurs over an
interval of time or space, and assume that
1. The probability of occurrence is the same for any
intervals of equal length
2. The occurrence in any interval is independent of
an occurrence in any non-overlapping interval
If x = the number of occurrences in a specified
interval, then x is a Poisson random variable
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The Poisson Distribution

Example

The number of failures in a large computer
system during a given day.

The number of customers who arrive at the
checkout counter of a store in one hour.
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The Poisson Distribution Continued
Suppose μ is the mean or expected number of
occurrences during a specified interval
The probability of x occurrences in the interval
when m are expected is described by the
Poisson distribution:
px  
e


x!
x
where x can take any of the values x = 0, 1, 2, 3, …
and e = 2.71828… (e is the base of the natural logs)
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Example 4.13: ATC Center Errors
Suppose that an air traffic control (ATC) center has been
averaging 20.8 errors per year and lately the center
experiences 3 errors in a week.
Let x be the number of errors made by the ATC
center during one week
Given: μ = 20.8 errors per year
Then:  = 0.4 errors per week
• Because there are 52 weeks per year, μfor a week is:
μ = (20.8 errors/year) / (52 weeks/year) = 0.4 errors/week
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Example 4.13: ATC Center Errors Continued
Find the probability that 3 errors (x =3) will occur in a
week
– Want p(x = 3) when μ= 0.4
px  3 
e
0.4
0.4
3
3!
 0.0072
Find the probability that no errors (x = 0) will occur in a
week
– Want p(x = 0) when  = 0.4
e 0.4 0.40
px  0 
 0.6703
0!
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Poisson Probability Table
(Appendix Table A.2, P821)
=0.4
Table 4.9
x
0
1
2
3
4
5
, Mean number of Occurrences
0.1
0.9048
0.0905
0.0045
0.0002
0.0000
0.0000
0.2
0.8187
0.1637
0.0164
0.0011
0.0001
0.0000
…
…
…
…
…
…
…
0.4
0.6703
0.2681
0.0536
0.0072
0.0007
0.0001
e 0.4 0.43
px  3 
 0.0072
3!
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…
…
…
…
…
…
…
1.00
0.3679
0.3679
0.1839
0.0613
0.0153
0.0031
35
Example: Poisson Distribution
 = 0.4
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Mean and Variance of a
Poisson Random Variable
If x is a Poisson random variable with parameter , then
mean X  
variance X2  
standard deviation X 
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
37
Several Poisson Distributions
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Back to Example 4.13
In the ATC center situation, 28.0 errors occurred on
average per year
Assume that the number x of errors during any span of
time follows a Poisson distribution for that time span
Per week, the parameters of the Poisson distribution are:
• mean  = 0.4 errors/week
• Because there are 52 weeks per year,  for a week is
 = (20.8 errors/year) / (52 weeks/year) = 0.4
errors/week
• standard deviation  = 0.6325 errors/week.
• Because  = √0.4 = 0.6325
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Poisson Distribution Example
Customers arrive at a rate
of 72 per hour. What is
the probability of 4
customers arriving in 3
minutes?
Solution:
72 per hr. = 1.2 per
min.
= 3.6 per 3
min.
px  
e


x!
x
e 3.6 3.6 
px  4 
 0.1912
4!
4
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Discrete probability distribution
Discrete
Binomial
P( x) n C x (1   )
x
Poisson
n x
 xe 
P( x) 
x!
Probability
distribution
P (x )
Mean of
Probability
distribution
  [ xP( x)]
  n

 2  [( x   )2 P( x)]
 2  n (1   )
2 
Variance of
Probability
distribution
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Binomial distribution
P( x) n C x (1   )
x
n x
  n
  n (1   )
2
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Poisson Distribution

 
2
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