Transcript Chapter 6

Probability Distributions
Chapter 6
McGraw-Hill/Irwin
©The McGraw-Hill Companies, Inc. 2008
GOALS
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Define the terms probability distribution and
random variable.
Distinguish between discrete and continuous
probability distributions.
Calculate the mean, variance, and standard
deviation of a discrete probability distribution.
Describe the characteristics of and compute
probabilities using the binomial probability
distribution.
What is a Probability Distribution?
Experiment: Toss a
coin three times.
Observe the number of
heads. The possible
results are: zero
heads, one head, two
heads, and three
heads.
What is the probability
distribution for the
number of heads?
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Probability Distribution of Number of
Heads Observed in 3 Tosses of a Coin
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Characteristics of a Probability
Distribution
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Random Variables
Random variable - a quantity resulting from an
experiment that, by chance, can assume different
values.
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Types of Random Variables
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
Discrete Random Variable can assume only
certain clearly separated values. It is usually
the result of counting something
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Continuous Random Variable can assume an
infinite number of values within a given
range. It is usually the result of some type of
measurement
Discrete Random Variables - Examples
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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.
Continuous Random Variables Examples
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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.
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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The Mean of a Probability Distribution
MEAN
•The mean is a typical value used to represent the
central location of a probability distribution.
•The mean of a probability distribution is also
referred to as its expected value.
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The Variance, and Standard
Deviation of a Probability Distribution
Variance and Standard Deviation
• Measures the amount of spread in a distribution
• The computational steps are:
1. Subtract the mean from each value, and square this difference.
2. Multiply each squared difference by its probability.
3. Sum the resulting products to arrive at the variance.
The standard deviation is found by taking the positive square root
of the variance.
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Mean, Variance, and Standard
Deviation of a Probability Distribution - Example
John Ragsdale sells new cars for Pelican Ford.
John usually sells the largest number of cars
on Saturday. He has developed the following
probability distribution for the number of cars
he expects to sell on a particular Saturday.
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Mean of a Probability Distribution - Example
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Computational Formula for Variance
An equivalent formula for the variance of a discrete
distribution (and one that is simpler to use) is the
following:
   x P x   
2
2
2
all x
After we have calculated the mean, , we calculate the
above sum and subtract the square of the mean to get
the variance.
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Variance and Standard Deviation of a Probability
Distribution - Example
In the previous example, we already found that
the mean was 2.10. We calculate the following
sum:
2
2
2
2
2
2


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









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
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


0.1  5.7
x
P
x

0
0
.
1

1
0
.
2

2
0
.
3

3
0
.
3

4

4
x 0
and subtract the square of the mean to get
  5.7  2.10   1.29
2
2
the variance of the probability distribution.
The standard deviation is then   1.29  1.14
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A Special Distribution
There are certain types of probability
distributions that are encountered often in
many real-world situations. In this chapter, we
concentrate on a single particular probability
distribution that occurs in many real-world
situations, the binomial distribution.
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Binomial Probability Distribution
Characteristics of a Binomial Experiment
• The experiment consists of a fixed number of trials,
• The trials are identical – they are performed the
same way.
• The trials are independent of each other,
• Each trial has two possible outcomes: Success or
Failure,
• The probability of Success is the same for each trial.
• The random variable is the number of Successes.
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Example 1: Presidential polls
Suppose that today is October 15, 2008, and I
want to predict the outcome of the 2008
Presidential election. I select a random sample
of n = 1067 voters from the population of all
registered and likely voters in the United
States, and I ask each voter the following
question: “Do you intend to vote for Senator
John McCain in the Presidential election?”
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Example 1: (continued)
The experiment consists of 1067 trials,
2) The trials are identical, since each one consists of
randomly selecting a registered and likely voter and
asking the same question.
3) The trials are independent of each other, due to
random sampling,
4) Each trial results in one of two possible outcomes:
Success = {voter says, “Yes”} or Failure = {voter
says, “No”},
5) P(Success) is the same for each trial, due to
random sampling.
Therefore this is a binomial experiment.
1)
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Example 1: (continued)
If I let X = number of voters in the sample who
answer “Yes”, then X is a random variable that
has a binomial distribution with n = 1067 and
 = Senator McCain’s level of support in the
voting population, expressed as a fraction.
Using this probability distribution, I can
calculate probabilities of various outcomes of
the poll.
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Example 1: (continued)
For example, a recent poll say that the Senator’s level
of support in the voting population is  = 0.441, with
Senator Obama’s level of support being 0.496. In our
experiment, then, X has a binomial distribution with
n = 1067 and  = 0.441. How likely is it that a majority
of voters in the sample will be supporters of Senator
McCain? We will answer this question later, after we
acquire some more tools for calculating binomial
probabilities.
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Binomial Probability Formula
The probability that a binomial random variable
X will be found to take on the value x when we
perform the binomial experiment can be found
using the following equation:
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Binomial Probability – Example 2
There are five flights
daily from Pittsburgh via
US Airways into the
Bradford, Pennsylvania,
Regional Airport.
Suppose the probability
that any flight arrives
late is .20.
What is the probability
that none of the flights
is late today?
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Binomial Dist. – Mean and Variance
The mean, or expectation, of a binomial Random variable X gives
the expected number of Successes out of the n trials of the
experiment. The variance and standard deviation give us an idea
about how spread out the probability distribution is.
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Mean and Variance, Example 1:
Let’s go back to our poll example. We have
n = 1067 and  = 0.441. Then the expected
number of “Yes” answers in our poll is
 = n = (1067)(0.441) = 470.547. The
variance of the distribution is
2 = n(1-) = 263.0358, and the standard
deviation is  = 16.2184.
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Binomial Dist. – Mean and Variance:
Example 2
For the example
regarding the number of
late flights, recall that
 =.20 and n = 5.
What is the average
number of late flights?
What is the variance of
the number of late
flights?
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Binomial Distribution - Table
Five percent of the worm gears produced by an automatic, highspeed Carter-Bell milling machine are defective. What is the
probability that out of six gears selected at random none will be
defective? Exactly one? Exactly two? Exactly three? Exactly four?
Exactly five? Exactly six out of six?
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Binomial Distribution - MegaStat
Five percent of the worm
gears produced by an
automatic, high-speed
Carter-Bell milling
machine are defective.
What is the probability
that out of six gears
selected at random none
will be defective?
Exactly one? Exactly
two? Exactly three?
Exactly four? Exactly
five? Exactly six out of
six?
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Binomial – Shapes for Varying 
(n constant)
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Binomial – Shapes for Varying n
( constant)
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Cumulative Binomial Probability
Distributions
A study in June 2003 by the Illinois Department of Transportation
concluded that 76.2 percent of front seat occupants used seat
belts. A sample of 12 vehicles is selected. What is the probability
the front seat occupants in at least 7 of the 12 vehicles are wearing
seat belts?
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Example 1: (continued)
We will use MegaStat to calculate the probability that a
majority of those polled support Sen. McCain. We
want the probability that X is at least 534. By the
complement rule, this is 1 minus the probability that X
is no more than 533. From the table given by
MegaStat, we find that P(X  533) = 0.99995, so that
P(X > 534) = 0.00005. It is very unlikely that a majority
of the sample will say, “Yes.” (Of course, the latest poll
also includes voters who are still undecided.)
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End of Chapter 6
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