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Math 227 Elementary Statistics
Bluman 5th edition
2
CHAPTER 5
Discrete Probability Distribution
3
Objectives
• Construct a probability distribution for a
random variable.
• Find the mean, variance, and expected
value for a discrete random variable.
• Find the exact probability for X successes in
n trials of a binomial experiment.
4
Objectives (cont.)
• Find the mean, variance, and standard
deviation for the variable of a binomial
distribution.
5
Introduction
• Many decisions in business, insurance, and
other real-life situations are made by
assigning probabilities to all possible
outcomes pertaining to the situation and
then evaluating the results.
• This chapter explains the concepts and
applications of what is called a probability
distribution. In addition, special probability
distributions, the binomial distribution, is
explained.
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5.1 Probability Distribution
I. Random Variables
A random variable is a variable whose values are determined by chance.
• A random variable is discrete if it can potentially assume only a finite or countable number of values.
• A random variable is continuous if it potentially can take on any value on an interval.
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Example 1 : Identify the following random variables
as discrete or continuous.
(a) Weight of a package
Continuous
(b) Number of students in a first-grade classroom
Discrete
(c) Age of a cancer patient
Continuous
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Example 2 :
An experiment consists of tossing five fair coins. Let x be the random variable
that is the number of heads in the five tosses. Is x discrete or continuous? List
the sample space of values of x.
x is discrete.
Example 3 :
Let x be the random variable that gives the amount of time it takes a person to
drive to work. Is x discrete or continuous? List the sample space of values of x.
x is continuous.
Sample space for x = (0, ∞)
x
0
1
2
3
4
5
Possible # of heads
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II. Discrete Probability Distribution
• A discrete probability distribution consists of
the values a random variable can assume
and the corresponding probabilities of the
values. The probabilities are determined
theoretically or by observation.
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Example 1 :
The probabilities that a customer will purchase 0, 1 , 2, or 3 books are 0.45, 0.30,
0.15, and 0.10, respectively.
(a) Construct a probability distribution for the data.
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(b) Draw a graph for the distribution.
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0 1 2 3
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Example 2 :
Consider families with three children. Let x be the number of girls in a family. Find
the probability distribution for x and construct a graph for the probability
distribution.
x P (x)
0 1 / 8
1 3 / 8
2 3 / 8
3 1 / 8
S = { BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG} B
G
B
G
B
G
B
G
B
G
B
G
B
G
0
1/8
1/4
3/8
0 1 2 3
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III. Requirements for a Discrete
Probability Distribution
1. P (x) will always be a number between 0 and 1
inclusive:
2. The sum of the values of P (x) for each distinct
value of x is 1:
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Example 1 : A random variable x has this probability distribution :
(a) Find P (x =3)
(b) What is the probability that x is greater than 0?
(c) What value of x is most likely to occur?
(d) What is P (x = 8)?
0.2 + 0.3 + 0.1 + P (x = 3) = 1
0.6 + P (x = 3) = 1
P (x = 3) = 1 – 0.6 = 0.4
P (x > 0) = P (x = 1) + P (x = 2) + P (x = 3)
= 0.3 + 0.1 + 0.4 = 0.8
x = 3, because it has the highest probability. Therefore, it is more likely to
occur.
Since 8 is not in the domain of this distribution, therefore, P (x = 8) = 0.
15
Example 2 :
Determine whether the distribution represents a probability distribution. If it does
not, state why.
It does not represent a probability distribution because the sum of P(x) must
equal to 1, which in this example it exceeds 1.
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5.2 Mean, Variance, Standard
Deviation, and Expectation
I. The mean of a discrete probability distribution
Suppose two coins are tossed repeatedly, and the number of heads
that occurred is recorded. What will be the mean of the number of
heads?
1 1 1Mean of the number of heads would be 2 1 0 1
4 2 4
In the long run, you would expect
two heads ((HH) to occur approximately ¼ of the time,
one head to occur approximately ½ of the times (HT or TH), and
no heads (TT) to occur approximately ¼ of the time.
The sample space is {HH, HT, TH, TT}
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In order to find the mean for a probability distribution, one must
multiply each possible outcome by its corresponding probability
and find the sum of the products.
where X1, X2, X3, …,Xn are the outcomes and
P(X1), P(X2), P(X3),…P(Xn) are the corresponding probabilities.
1 1 2 2 3 3( ) ( ) ( ) . . . ( )
( )
n nX P X X P X X P X X P X
X P X
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II. Variance of a Probability
Distribution
• The variance of a probability distribution is
found by multiplying the square of each
outcome by its corresponding probability,
summing those products, and subtracting
the square of the mean.
• The formula for calculating the variance is:
• The formula for the standard deviation is:
2
2 2 2[ ( )]X P X
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Example 1 : For the probability distribution given below : find (a) the mean, (b) the
variance, and (c) the standard deviation. Also (d) construct a graph for the probability
distribution and describe the shape of the distribution.
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(a) The mean
(b) Variance
(c) Standard deviation
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(d) Construct a graph for the probability distribution and describe the shape of
the distribution.
0
1/10
2/10
3/10
4/10
5/10
0 1 2 3x
P(x
)
The shape of distribution is an almost bell-shaped curve.
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Example 2 :
Use the computational formula to find the standard deviation of probability distribution
given in example 1.
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III. Expected Value
• Expected value or expectation is used in
various types of games of chance, in
insurance, and in other areas, such as
decision theory.
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Expected Value (cont.)
• The expected value of a discrete random variable
of a probability distribution is the theoretical
average of the variable. The formula is:
• For a discrete random variable x, the expected
value of x is the mean of the random variable x.
• The symbol E(X) is used for the expected value.
E X X P X
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Example 1 : (Ref: General Statistics by Chase/Bown, 4th Ed.)
A high school class decides to raise some money by conducting a raffle. The
students plan to sell 2000 tickets at $1 apiece. They will give one prize of $100, two
prizes of $50, and three prizes of $25. If you plan to buy one ticket, what are your
expected net winnings?
Expected net winnings ≈ $ - 0.8625
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Example 2 : (Ref: Elementary Statistics by Triola, 9th edition)
In New Jersey’s Pick 4 lottery game, you pay 50 cents to select a sequence of
four digits, such as 7273. If you win by selecting the same sequence of four
digits that are drawn, you collect $2,788.
(a) How many different selections are possible?
(b) What is the probability of winning?
10 · 10 · 10 · 10 = 10,000
P (winning in one ticket) = 1 / 10,000
(c) If you win, what is your net profit?
Net Profit = $2,788 – $0.50 = $2,787.50
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(d) Find the expected value.
In the long run, you will lose $0.22 per ticket.
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5.3 The Binomial Distribution
I. Binomial Experiment
• Many types of probability problems have only
two possible outcomes or they can be reduced
to two outcomes.
• Examples include: when a coin is tossed it can
land on heads or tails, when a baby is born it is
either a boy or girl, etc
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The Binomial Experiment
• The binomial experiment is a probability experiment that
satisfies these requirements:
1. Each trial can have only two possible
outcomes—success or failure.
2. There must be a fixed number of trials.
3. The outcomes of each trial must be independent of
each other.
4. The probability of success must remain the same for
each trial.
• The outcomes of a binomial experiment and the
corresponding probabilities of these outcomes are called a
binomial distribution.
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Example 1: (Ref: Exploring Statistics by Kitchens, 2nd ed.)
Which of the following are binomial random variables?
(a) The number of successful heart transplants out of five patients.
(b) The length of a prison term for possession of marijuana.
(c) The name of each student in Math 227.
(d) The number of approved food stamp recipients out of 50 applications.
2 outcomes – S / F, n = 5, independent → Binomial random variable.
More than 2 outcomes → NOT a binomial random variable.
More than 2 outcomes → NOT a binomial random variable.
2 outcomes, n = 50, independent → Binomial random variable.
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• The symbol for the probability of success
• The symbol for the probability of failure
• The numerical probability of success
• The numerical probability of failure
• The number of trials
• The number of successes
II. Notation for the Binomial Distribution
P S( )
P F( )
p
q
P S p( ) P F p q( ) 1
n
X
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III. Binomial Probability Formula
A binomial experiment consists of n identical trials
with probability of success p on each trial. The
probability of x successes in n trials is
OR
!( )
( )! !
X n XnP X p q
n X X
33
Example 1 :
(a) Find 8C3
(b) Find 12C7
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Example 2 :
Consider a binomial experiment with n = 15, p = 0.3, and x = number of
success. Use the Binomial Formula for P(x) to find the probability that
(a) x = 11
(b) x is less than 2
35
Example 3 :
It was found that 68% of American victims of health care fraud are senior citizens.
If 10 victims are randomly selected, find the probability that exactly 3 are senior
citizens.
36
Example 4 :
Consider a binomial experiment with n = 12, p = 0.4, and x = number of success.
Use the Binomial Table to find the probability that
(a) x is greater than 5 but less than 8
(b) x is greater than 7
(c) x equals 6
P (x > 5 but x < 8)
= P (6 ≤ x ≤ 7) = P (x = 6) + P (x = 7)
Look at the Binomial Table for values
= 0.177 + 0.101 = 0.278
Choose values on the table greater than 7.
P (x = 8) + P (x = 9) + P (x = 10) + P (x = 11) + P (x = 12)
= 0.042 + 0.012 + 0.002 + 0+ + 0+ = 0.056
P (x = 6) = 0.177
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IV. Mean and Variance of a Binomial
Distribution
For a binomial experiment consisting of n trials with the probability of
success p, the mean or expected of x is
For a binomial experiment consisting of n trials with the probability of
success p, the variance of x is
The standard deviation of x is
pn
qpn 2
instead of using
instead of using
n p q
38
Example 1 :
Assume that 60% of a college’s student loan applications are approved. Ten
applications are chosen at random.
(a) What is the probability that eight or more are approved?
(b) How many applications are expected to be approved?
(Find the mean of the number approved out of ten applications.)
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(c) What is the standard deviation of the number approved out of the application?
40
Summary
• A probability distribution can be graphed, and the mean,
variance, and standard deviation can be found.
• The mathematical expectation can also be calculated for
a probability distribution.
• Expectation is used in insurance and games of chance.
• The binomial distribution is used when there are only two
outcomes for an experiment, a fixed number of trials, the
probability is the same for each trial, and the outcomes
are independent of each other.
41
Conclusion
• Many decisions in business, insurance,
and other real-life situations are made by
assigning probabilities to all possible
outcomes pertaining to the situation and
then evaluating the results.
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