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CHAPTER 5
DISCRETE
PROBABILITY
DISTRIBUTIONS
1
2
RANDOM VARIABLES
Tila College is a small private school with 2,500 students. According
to the registrar’s office, the frequency and relative frequency
distributions of the number of courses taken by all the students in
Fall 2010 are shown in the table below.
Table 1: Frequency and Relative Frequency Distribution of
the number of Courses Taken by Students
Number of Courses
Frequency
Relative Frequency
1
150
0.06
2
500
0.20
3
1000
0.40
4
600
0.24
5
250
0.10
 f = 2500
 = 1.00
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Random Variables
Now, suppose we want to conduct an experiment by randomly
selecting students from the population of 2,500 students.
Definition
1. The process of randomly selecting a
student from the population is called a
random or a chance experiment.
2. Let x = the number of courses taken by
the randomly selected student. Then x
depends on the outcome of the
experiment and can assume any values
(1, 2, 3, 4, or 5). x is also called a
random variable or a chance
variable.
3. A random variable is a variable whose
value is the outcome of a random or a
chance experiment.
4. A random variable x can be either
discrete random variable or
continuous random variable.


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Discrete Random Variable
Definition

A discrete random variable is a random variable that
assumes countable values.

In our example, the number of courses taken by
students in Fall 2010 is a discrete random variable
because it can assume any value, 1, 2, 3, 4, or 5.

Other examples of discrete random variables are:



The number of smokers in STA 12 class
The number of cars owned by families in your city
The number of fish caught on a fishing trip
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Continuous Random Variable
Definition

A continuous random variable is a random variable
that assumes any values within one or more intervals.

Suppose the volume of water used per month per
household in Victorville ranges from 2,000 to 4,000
gallons.




Let’s randomly select a household in the city of Victorville.
Let x =the volume of water used per month by the randomly
selected household.
Since there are infinitely many values between 2,000 and
4,000 gallons, the random variable, x, is a continuous random
variable and can assume infinite number of values.
Graphically,
2,000
x can assume any value on this number line 4,000
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Discrete and Continuous Random Variables
Example #1
Classify each of the following random
variables as discrete or continuous.
a.
b.
c.
d.
e.
f.
The time left on a parking meter
The number of bats broken by a major
league baseball team in a season
The number of cars in a parking lot
The total pounds of fish caught on a
fishing trip
The number of cars crossing a bridge on
a given day
The time spent by a physician
examining a patient
Solution
a.
b.
Continuous
Discrete
d.
Discrete
Continuous
e.
Discrete
f.
Continuous
c.
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PROBABLITY DISTRIBUTION OF A
DISCRETE RANDOM VARIABLE
Definition
A probability distribution of a discrete random variable lists
all possible values that the random variable can assume with their
corresponding probabilities.



As discussed in Section 5.1,
let x = number of courses taken by
a randomly selected student.
Then, we can write the probability
distribution of x as shown to the
right.
Note that the probabilities are actual
since the relative frequencies
represent population. As stated in
Chapter 4, if the relative frequencies
were for a sample, then the
probabilities would have been
approximate probabilities.



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Probability Distribution of a Discrete Random
Variable

A probability distribution of a discrete random
variable must satisfy the following conditions:
a.
The probability of each value of x is from 0 to 1.
0  P (x)  1
b.
The sum of the probabilities for all values of x is 1.
 P (x)  1


We can read the probability for any value of x as
P(x = 2) = P(2) = 0.20
P(x > 3) = P(4)+ P(5) =0.24 + 0.10 = 0.34
P(x =< 3) = P(1) + P(2) + P(3) = .06 + .20
+ .40 = 0.66
A probability distribution of a discrete random
variable can be represented by,
a.
b.
c.
Graph
Formula
Table

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Probability Distribution of a Discrete Random
Variable
Example #2
Each of the following tables lists
certain values of x and their
probabilities. Determine whether
or not each one satisfies the two
conditions required for a valid
probability distribution.
(1)
(2)
(3)
x
P(x)
x
P(x)
x
P(x)
5
-.36
1
.27
0
.15
6
.48
2
.24
1
.08
7
.62
3
.49
2
.20
8
.26
3
.50
Solution
Although the sum of
probabilities for all values of x
is equal to 1, the probability of
x = 5 is negative. Thus, this
table does not represent a valid
probability distribution.
2. Each probability ranges from 0
to 1, and the sum of the
probabilities is 1.00. Thus, this
table represents a valid
probability distribution.
3. Although each probability lies
between 0 and 1, the sum of
the probabilities is less than
1.0. Thus, this table does not
represent a valid probability10
distribution.
1.
Probability Distribution of a Discrete Random
Variable
Solution
Example #3
The following table gives the
From the given table,
probability distribution of a discrete
a. P(x=1) = 0.17
random variable x.
b. P(x  1) = P(x = 0) + P(x = 1)
= .03 + .17 = .20
P(x) .03 .17 .22 .31 .15 .12
c. P(x  3) = P(x = 3) + P(x = 4) + P(x = 5)
Find the following probabilities:
= .31 + .15 + .12 = .58
d. P(0  x  2) = P(x = 0) + P(x = 1) + P( x = 2)
a. P(x=1)
b. P(x  1)
= .03 + .17 + .22 = .42
c. P(x  3) d. P(0  x  2)
e. Probability that x assumes a value < 3 e. P(x < 3) = P(0) + P(1) + P(2) = 0.42
f. P(x > 3) = P(x = 4) + P(x = 5) = 0.27
f. Probability that x assumes a value > 3
g. P(2  x  4) = P(x = 2) + P(x = 3) + P(x = 4)
g. Probability that x assumes a value
= .22 + .31 + .15 = .68
x
0
1
2
in the interval 2 to 4
3
4
5
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Probability Distribution of a Discrete Random
Variable
Example #4
The following table shows the probability distribution for the sum, denoted by x,
of the faces on a pair of Nathan’s dice:
x
2
3
4
5
6
7
8
9
10
11
12
P(x)
.065
.065
.08
.095
.11
.17
.11
.095
.08
.065
.065
a. Draw a bar graph for this probability
b. Determine the probability that the sum of the faces on a single roll of Nathan’s
dice is: (i) an even number (ii) 7 or 11 (iii) 4 to 6 (iv) no less than 9
Solution
0 . 18
0 . 16
0 . 14
0 . 12
0.1
0.08
0.06
0.04
0.02
0
2
3
4
5
6
7
8
9
10
11
12
i. P(x=even number) = P(x=2)+P(x=4)+P(x=6)+P(x=8)
+P(x=10)+P(x=12)= .51
ii. P(x=7 or x=11) = P(x=7) + P(x=11)
= .17 + .065 = .235
iii. P(x=4 to x=6) = P(x=4)+P(x=5)+P(x=6)
= .08 + .095 + .11 = .285
iv. P(x  9) = P(x=9) + P(x=10) + P(x=11) + P(x=12)
12
= .095 + .08 + .065 + .065 = .305
Probability Distribution of a Discrete Random
Variable
Example #5
According to a survey, 30% of
adults are against using animals for
research. Assume that this result
holds true for the current
population of all adults. Let x be
the number of adults who are
against using animals for research
in a random sample of two adults.
Obtain the probability distribution
for x. Draw a tree diagram for this
problem.
Let A =
F=
against animal research
for animal research
Solution
x can assume 0, 1, or 2. Thus,
the probability distribution is
x
P(x)
0
P(F) P(F)=.7 (.7) = 0.49
1
P(F)P(A)+P(A)P(F)= .7(.3)+.3(.7)
=.42
2
P(A) P(A)=.3(.3) = .09
A
.30
F
.70
A
.30
F
.70
P(AA) = .09
P(AF) = .21
A
.30
F
P(FA) = .21
.70
P(FF) = .49
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MEAN OF A DISCRETE RANDOM
VARIABLE
The mean of a discrete variable
is denoted by µ and is define as
the sum of the product of each
value of the discrete variable and
the corresponding probabilities. It
is also called an expected value
and denoted by E(x)
Formal Definition:
The mean of a discrete random
variable x is the value that is
expected to occur per repetition,
on average, if an experiment is
repeated a large number of times.
It is calculated as
µ = E(x) = Σ x P(x)
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STANDARD DEVIATION OF A DISCRETE
RANDOM VARIABLE
Definition
The standard deviation of a discrete random variable is denoted by 
and is define as the measure of the spread of its probability distribution.
It is computed as,
Basic Formula
Variance
 2    x    P ( x ) 
 2   x 2P ( x )   2
    x    P ( x ) 


   x 2P ( x )   2
2

Standard
deviation
Short-Cut Formula

2
Note
 A higher value of  signifies that x can assume values over a larger range
about the mean

A smaller value of  signifies that x can assume values that are closely
clustered about the mean.
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Standard Deviation of a Discrete Random Variable
Example #7
The following table gives the probability distribution of the
number of camcorder sold on a given day at an electronic store
Camcorder sold
0
1
2
3
4
5
6
Probability
.05
.12
.19
.30
.20
.10
.04
Calculate the standard deviation for this probability distribution.
Solution
X
0
1
2
3
4
5
6
P(x)
.05
.12
.19
.30
.20
.10
.04
xP(x)
0
.12
.38
.90
.80
.50
.24
X2
0
1
4
9
16
25
36
x2P(x)
0
.12
.76
2.7
3.2
2.5
1.44
   x 2P ( x )   2  10.72  (2.94)2  1.4410
   xP(x)=2.94
2
 x P(x)=10.72
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Standard Deviation of a Discrete Random Variable
Example #8
An instant lottery ticket costs $2. Out of a total of 10,000 tickets printed for this
lottery, 1000 tickets contain a price of $5 each, 100 tickets have a price of $10
each, and 5 tickets have a price of $1000 each, and 1 ticket has a price of $5000.
Let x be the random variable that denotes the net amount a player wins by playing
this lottery. Write the probability distribution of x. Determine the mean and
standard deviation of x. How will you interpret the values of the mean and
standard deviation.
  0.40  E ( x )
Solution
x
f
P(x)
xP(x)
X2
x2P(x)
-2
8894
.8894
1.7788
4
3.5576
3
1000
.1000
0.3000
9
0.9000
8
100
0.0100 0.0800
64
0.6400
998
5
0.0005 0.4990
996004
498.002
4998
1
0.0001 0.4998 24980004 2498.0004
 xP(x)=-0.40
2
 x P(x)=3001.1
   x 2P ( x )   2
 3001.1  ( .40)2
 $54.78
Players are expected
to lose, on an
average, 40 cents per
ticket with a standard
17
deviation of $54.78.
Identifying Unusual Results
Range Rule of Thumb
According to the range rule of thumb, most
values should lie within 2 standard deviations
of the mean.
We can therefore identify “unusual” values
by determining if they lie outside these limits:
  2
Maximum usual value =
  2
Minimum usual value =
Example – continued
We found for families with two children, the mean number of
girls is 1.0 and the standard deviation is 0.7 girls.
Use those values to find the maximum and minimum usual
values for the number of girls.
maximum usual value    2  1.0  2  0.7   2.4
minimum usual value    2  1.0  2  0.7   0.4
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Identifying Unusual Results
Probabilities
Rare Event Rule for Inferential Statistics
If, under a given assumption (such as the
assumption that a coin is fair), the probability
of a particular observed event (such as 992
heads in 1000 tosses of a coin) is extremely
small, we conclude that the assumption is
probably not correct.
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Identifying Unusual Results
Probabilities
Using Probabilities to Determine When
Results Are Unusual
Unusually high: x successes among n
trials is an unusually high number of
P( x or more)  0.05
successes if
.
Unusually low: x successes among n trials
is an unusually low number of successes if
P( x or fewer)  0.05
.
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Identifying Unusual Results
Probabilities
Suppose you were tossing a coin to determine
whether it favors heads, an supposed 1000
tosses resulted in 501 heads.
a) P(exactly 501 heads in 1000 tosses)=0.0252
Notice 0.0252<0.05
so, is very unlikely.
b) p(501 or more heads in 1000 tosses)=0.487
Since 0.487 in not greater than 0.05 then is not unusually high
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