Transcript Chapter 7 Probability Distributions, Information about the Future

Chapter 7
Probability Distributions,
Information about the Future
.
7- 1
Random Processes
7- 2
Random Variable
• A random variable is the
numerical outcome of a
random (non-deterministic)
process.
• Intuitively, any numerically
measured variable that
possesses an uncertain
outcome is a random variable.
7- 3
Probability Distribution
• A probability distribution is a
model which describes a
specific kind of random
process.
• Specifically, a probability
distribution connects a
probability to each value the
random variable can assume.
7- 4
Probability Models
Probability models are excellent
descriptors of random processes.
The following is a probability
distribution (a model) for the
outcome of a coin toss.
Probability Distribution
Outcome Probability
Heads
.5
Tails
.5
7- 5
Types of Random
Variables
7- 6
Quantitative Random
Variables
Quantitative random variables
are divided into two classes.
1) Discrete
2) Continuous
7- 7
Discrete Random
Variables
• A discrete random variable
is a random variable which has
a countable number of
possible outcomes.
• The values that many discrete
random variables assume are
the counting numbers from 0
to N, where N depends upon
the nature of the variable.
• Example: The number of Statistics
pages in a standard math
textbook is a discrete random
variable.
7- 8
Continuous Random
Variables
• A continuous random
variable is a random variable
that can assume any value on
a continuous segment(s) of the
real number line.
• Heights, weights, volumes,
and time measurements are
usually measured on a
continuous scale.
• These measurements can take
on any value in some interval.
0
7- 9
Discrete or Continuous?
Classify the following as either a
discrete random variable or a
continuous random variable.
1. the speed of a train
2. the possible scores on the SAT
exam
3. the number of pizzas eaten on a
college campus each day
4. the daily takeoffs at Chicago’s
O’Hare Airport
5. the highest temperatures in Maine
and Florida tomorrow
7 - 10
Answers
1. the speed of a train
– continuous random variable
2. the possible scores on the SAT exam
– discrete random variable
3. the number of pizzas eaten on a
college campus each day
– discrete random variable
4. the daily takeoffs at Chicago’s
O’Hare Airport
– discrete random variable
5. the highest temperatures in Maine
and Florida tomorrow
– continuous random variable
7 - 11
Naming Convention
• Capital letters, such as X, will be
used to refer to the random
variable.
example:
X = number of cows in Texas
• Small letters, such as x, will refer to
a specific value of the random
variable.
example:
x = 1,498,000 cows in Texas
• Often the specific values will be
subscripted x1, x2, ..., xn.
7 - 12
Describing a Discrete
Random Variable
• State (Describe) the variable.
• List all of the possible values
of the variable.
• Determine the probabilities of
these values.
7 - 13
Example 1 (die tossing)
Random Phenomenon: Toss
a die and observe the outcome
of the toss.
•X=?
• What are the possible
values of X?
• What are the probabilities
of each value?
7 - 14
Example 1 - Solution
• Identify the Random Variable:
X = outcome of toss of die
• All possible Values: Integers
between 1 and 6. In this instance
x1 = 1, x2 = 2, ..., x6 = 6.
• Probability Distribution: The
outcomes of the toss of a die and
their probabilities are given in the
table. The probabilities are
deduced using the classical method
and the assumption of a fair die.
Value of X
1
2
3
4
5
6
Probability
1
6
1
6
1
6
1
6
1
6
1
6
7 - 15
Example 2
Random Phenomenon: The head
nurse of the pediatric division of the
Sisters of Mercy Hospital is trying to
determine the capacity requirement for
the nursery. She realizes that the
number of babies born at the hospital
each day is a random variable. And,
she will have to develop a description of
the randomness in order to develop her
plan.
• X=?
• What are the possible values of X?
• What are the probabilities of each
value?
Not all discrete random variables have
easily definable probability distributions.
7 - 16
Example 2 - Solution
• Identify the Random Variable:
X = # of babies born at hospital
each day
• Range of All possible Values:
Integers between 0 and some
large positive number.
• Probability Distribution:
Unknown, but could be
estimated using the relative
frequency idea in conjunction
with historical data on hospital
births.
7 - 17
Discrete Probability
Distributions
7 - 18
Discrete Probability
Distributions
• The random variable concept
is so general, that it is not very
useful by itself.
• What would be useful is to
determine what numerical
values the random variable
could assume and assess the
probabilities of each of these
values.
7 - 19
Discrete Probability
Distributions
• A discrete probability
distribution consists of (a list
of) all possible values of the
random variable with their
associated probabilities.
• The association of the possible
values of a random variable
with their respective
probabilities can be expressed
in three different forms: in a
table, in a graph, and in an
equation.
7 - 20
Characteristics
Discrete probability distributions
always have two
characteristics:
1. The sum of all of the
probabilities must equal 1.
2. The probability of any value
must be between 0 and 1,
inclusively.
[Relative frequencies also share
these properties]
7 - 21
Example 3 (Daily Sales)
K. J. Johnson is a computer
salesperson. During the last year he
has kept records on his computer
sales. He recognizes that his daily
sales constitute a random process
and wishes to determine the
probability distribution for daily sales.
The random variable is X = number of
computers sold each day.
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Example 3 - Solution
The probabilities for
this random variable
are computed in the
table based upon
200 days of sales
data obtained from
Mr. Johnson’s
records using the
relative frequency
concept.
Sales
Probability
x
P(X=x)
0
.2
1
.1
2
.3
3
.2
4
.2
P(x) = 1.0
The probability that Mr. Johnson will sell at least 2
computers each day is calculated as follows:
P(X  2) = P(X=2) + P(X=3) + P(X=4) = .3 + .2 + .2 = .7.
The probability that Mr. Johnson will sell at most 2
computers each day is calculated as follows:
P(X  2) = P(X=0) + P(X=1) + P(X=2) = .2 + .1 + .3 = .6.
7 - 23
Example 4, Is this a
prob. Dist’n?
• Tell whether or not the
following distribution is a
probability distribution.
X P(X)
-2 .25
0 .50
2 .25
• If the distribution is not a
probability distribution, give the
characteristic which is not
satisfied by the distribution.
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Example 4 - Solution
Yes. All probabilities are
between 0 and 1, and the sum of
the probabilities is 1.
7 - 25
Example 5 , Is this a
prob. Dist’n?
• Tell whether or not the
following distribution is a
probability distribution.
X P(X)
5 .46
10 .35
15 .25
• If the distribution is not a
probability distribution, give the
characteristic which is not
satisfied by the distribution.
7 - 26
Example 5 - Solution
No. The sum of the probabilities
is greater than one.
7 - 27
Example 6, Is this a
prob. Dist’n?
• Tell whether or not the
following distribution is a
probability distribution.
X
100
200
300
P(X)
-.10
.50
.50
• If the distribution is not a
probability distribution, give the
characteristic which is not
satisfied by the distribution.
7 - 28
Example 6 - Solution
No. You can't have negative
probabilities.
7 - 29
Example 7, Is this a
prob. Dist’n?
• Tell whether or not the
following distribution is a
probability distribution.
x
P(X=x) = 16 , for x = 1, 2, 3, 4, 5
• If the distribution is not a
probability distribution, give the
characteristic which is not
satisfied by the distribution.
7 - 30
Example 7 - Solution
No. See table.
The sum of the
probabilities is
15/16 which is
less than one.
P(X)=x/16 for x=1
to 5 only is NOT a
probability
distribution.
X
1
2
3
4
5
P(X)
1/16
2/16
3/16
4/16
5/16
7 - 31
Expected Value E(X) of
a random variable X
7 - 32
Importance of E(X)
• One of the most important
concepts in the analysis of
random phenomena is the
notion of expected value.
• Expected value is important
because it is a summary
statistic for a probability
distribution.
• It can also be used as a
criteria for comparing
alternative decisions in the
presence of uncertainty.
7 - 33
What is Expected Value?
• Conceptually, expected value is
closely allied with the notion of
mean or average.
• The expected value is a weighted
average, in which each possible
value of the random variable is
weighted by its probability.
• Definition:
The expected value of a random
variable X is the mean of the
random variable X. It is denoted by
E(X) and is given by computing the
following expression:
 = E(X) =  x* P(X=x)
=  x* P(x)
7 - 34
Digression on
Weighted Averages
• Weighted average of any
measurement (say prices Pt) is
always (t wtPt )/( t wt)
• weighted averages are
ubiquitous. Dow Jones Industrial
average is a weighted average;
• see
• http://www.indexarb.com/indexC
omponentWtsDJ.html
• S&P 500 index is similar with
weights available at
• http://www.indexarb.com/indexC
omponentWtsSP500.html
7 - 35
Average Value
• The expected value of a random
variable should be very close to
the average value of a large
number of observations from the
random process.
• The larger the number of
observations collected the more
likely the expected value will be
close to the average of the
observations.
• For discrete random variables
the expected value is rarely one
of the possible outcomes of the
random variable.
7 - 36
E(X) for Daily Sales
The expected value of the probability
distribution given in Example 3 (daily
Sales) is computed in the table.
x
0
1
2
3
4
P(X=x)
.2
.1
.3
.2
.2
xP(X=x)
0
.1
.6
.6
.8
E(X) = 2.1
In the long run, data coming from a
random process with this distribution
should average about 2.1.
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Using Expected Values
to Compare Alternatives
Two Investment Opportunities
Profit
-2000
0
1000
2000
4000
Option A
Probability
xP(x) Profit
.2
-400 -3000
.1
0 -1000
.3
300 2000
.3
600 3000
.1
400 4000
E(XA) =xP(x) = $900
Option B
Probability
xP(x)
.2
-600
.1
-100
.2
400
.3
900
.2
800
E(XB) =xP(x) = $1400
• By calculating the expected values of the
two alternatives the information in each
distribution is condensed to a single point.
• This point characterizes the “center” of the
random process and facilitates comparison.
• In the long run, option B would be $500
more profitable.
• But on any one investment in option B, you
may lose as much as $3000 or make as
much as $4000.
7 - 38
Symbols
• The expected value, E(X), is
the “center point” for the
random process.
• The symbol x is often used to
represent E(X).
x = E(X)
7 - 39
Variance and
Standard Deviation
of a Discrete
Random Variable
7 - 40
Variance of a Discrete
Random Variable
• The expected value of a distribution
measures only one dimension of the
random variable (its central value).
• To gauge the variability of a random
variable we need another measure
similar to the variance measure
previously constructed but one
which accounts for the difference in
probabilities of the variable.
• The variance of a discrete
random variable X is given by
• The larger the variance the more
variability in the outcomes.
V ( X )  ( x  )2 p( x).
7 - 41
Standard Deviation as a
measure of risk
• The standard deviation is
computed by taking the square root
of the variance.
• In the “Investment Opportunity”
problem the variance and standard
deviation are as follows.
Option A
V(X) = 3,090,000
 = V(X) = 1,757.84
Option B
V(X) = 6,640,000
 = V(X) = 2,576.82
• The larger deviation reflects greater
variability in profits and increased
risk.
7 - 42
Sharpe Ratio
• Risk adjusted returns are compared
by computing the ratio
• Average return / std. Dev of returns
• Option A: 900/1,757.84
=0.5199199
• Option B: 1400/ 2,576.82
=0.5433053
Clearly Option B is slightly superior.
7 - 43
Example 9
Find the expected value, the variance,
and the standard deviation for a
random variable with the following
probability distribution.
X 400 420 440 460 480 500
P(X) 0 .1 .1 .2 .2 .4
7 - 44
Example 9 - Solution
X
P(X) XP(X)
400
0.0
0
420
0.1
42
440
0.1
44
460
0.2
92
480
0.2
96
500
0.4
200
Total 1.0
474
2
(X-) P(X)
0.0
291.6
115.6
39.2
7.2
270.4
724.0
Expected Value = 474
Variance = 724
Standard Deviation = 724  26.91
7 - 45
Probability
Distributions and
their Functions
7 - 46
Where do probability
distributions come from?
• In previous examples the
distribution is already given.
• In the “real world” there will be
very few instances in which
the probability distribution will
be conveniently available.
• Probabilities will have to be
determined using (i) classical,
(ii) relative frequency, or (iii)
subjective probability.
• Probability distributions can be
constructed from relative
frequency distributions(
depicted in histograms)
7 - 47
Probability Distribution
Functions (p.d.f.)
• Four well known discrete
distributions are: the discrete
uniform, binomial, Poisson,
and hypergeometric.
• Each of the discrete
distributions possesses a
probability distribution function.
• These math functions assign
probabilities to each value of
the random variable.
7 - 48
Discrete Probability
Distribution Function
Example of discrete p.d.f.:
P(X=x)=1/4, if x=1,2,3,4
P(X=x)=0, otherwise
This pdf does assign some value
to each possible discrete number
which can be the value of X.
All probability values need not be
positive. They can be zero!
7 - 49
Determining Probabilities
for a Specific Value (just
plug into the formula)
To determine the probability for a
specific value, use the value as the
argument to the function. Pdf is
P(X=x)=x2/30. [sum is unity]
To determine the probability that X = 3,
P(X=3) = 32 = 9 .
30 30
To determine the probability that X = 4,
P(X=4) = 42 = 16 .
30 30
7 - 50
The Discrete Uniform
Distribution
7 - 51
Definition
• In the discrete uniform
distribution each value of the
random variable is assigned
identical probabilities.
• This distribution is one of the
simplest probability
distributions.
• There are many situations in
which the discrete uniform
distribution arises.
7 - 52
Example 10
• The outcome of the throw of a
single die.
• If the die is “fair”, then each of the
outcomes is equally likely.
• Resulting probability
distribution:
Throwing a
Die
x P(X=x)
1
2
3
4
5
6
1
6
1
6
1
6
1
6
1
6
1
6
7 - 53
The Binomial
Distribution
7 - 54
Definition
• A binomial experiment is a
random experiment which
satisfies all of the following
conditions:
1 There are only two outcomes
on each trial of the experiment.
– One of the outcomes is usually
referred to as a success, and the
other as a failure.
2 The experiment consists of n
identical trials as described in
(1).
7 - 55
Definition Continued
3 The probability of success on
any one trial is denoted by p
and does not change from
trial to trial.
– Note that the probability of a
failure is 1- p and also does not
change from trial to trial.
4 The trials are independent.
5 The binomial random variable
is the count of the number of
successes in n trials.
7 - 56
Example 11 (r.v. is # of
heads in 4 tosses)
• Toss a coin 4 times and record the
number of heads as the random
variable.
• The number of heads in 4 tosses is
a binomial random variable.
Events
TTTT
HTTT
THTT
TTHT
TTTH
HHTT
HTHT
HTTH
THHT
THTH
TTHH
THHH
HTHH
HHTH
HHHT
HHHH
Number of
heads
0 heads
Probability
1
16
1 head
4
16
2 heads
6
16
3 heads
4
16
4 heads
1
16
7 - 57
Pascal Triangle to
compute nCx Binomial
coefficients
•
1
•
1
1
•
1
2
1
•
1 3
3
1
•
1
4
6
4
1
•
1
5 10 10
5 1
•
1
6
15 20 15 6 1
• 4 coin x=0,1,2,3,4 and corresp.
P(x): 1/16,4/16,6/16,4/16, 1/16
• Note each row is created from
previous row by always starting and
ending with ones, computing sums
of numbers from the previous row
7 - 58
Ex. 11 Continued (r.v. is
# of heads in 4 tosses)
1 There are only 2 outcomes, heads
or not heads.
2 The experiment will consist of 4
tosses of a coin.
3 The probability of a getting a head
(success) is .5 and does not change
from trial to trial.
4 The outcome of one toss will not
affect other tosses.
5 The variable of interest is the
number of heads in 4 tosses.
7 - 59
The Binomial
Probability Distribution
Function
n
x
x
P(X = x) = C p (1- p)
n-x
n
x
where C represents the number of
possible combinations of n objects taken
x at a time (without replacement) and is
given by
n!
Cx 
where n! = n(n
x!(n - x)!
n
n!
where n! = n(n -1)(n - 2) ... 1, and 0! = 1;
n - x)!
n = the number of trials, and
p = the probability of a success.
7 - 60
Calculating a Binomial
Probability by plugging in
the Binomial formula
• The parameters of the distribution
(n and p) as well as the value of the
random variable must be specified.
• For example, to determine the
probability that x=3, given that n=4
and p=0.5, substitute those values
in the probability distribution
function as follows:
P(X=3) = C34(21)3(1 21)4  3.
4
4!
4  3 2 1
C
Since, 3  3!(4  3)!  321(1)  4, then
4 .25.
P(X  3) = 4(21)3(21)  16
7 - 61
Example 12
Calculate Cnx for the
following combinations of x
and n.
A. n = 4 and x = 2
B. n = 12 and x =8
7 - 62
Example 12 - Solution
A.
4!
4(3)(2!)
C 

6
2!(4  2)! 2!(2)(1)
4
2
B.
12!
12(11)(10)(9)(8!)
C 

 495
8!(12  8)!
8!4(3)(2)(1)
12
8
7 - 63
Binomial Tables
• In order to avoid tedious
calculations, binomial tables
containing a large collection of
binomial distributions have
been constructed.
• These tables are found in the
Appendix (pp. 523 -527).
7 - 64
Example 13
The random variable X is a binomial
random variable with n = 12 and p = .8.
Using the tables, find the following:
A. the probability that X is at most 4
B. the probability that X is at least 1
C. the probability that X is more than 10
7 - 65
Example 13 - Solution
A.
P(X  4)
=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)
=.0000+.0000+.0000+.0001+.0005
=.0006
B.
P(X 1)=1-P(X=0)=1-.0000=1
C.
P(X>10)=P(X=11)+P(X=12)
=.2062+.0687=.2749
7 - 66
The Shape of a
Binomial
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
9 10 11 12
If p is small, the distribution
tends to be skewed with a tail
on the right.
7 - 67
The Shape of a
Binomial
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
9
If p is near .5, the distribution
is symmetrical.
7 - 68
The Shape of a
Binomial
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
9
10
11
If p is large, the distribution
tends to be skewed with a tail
on the left.
7 - 69
The Expected Value and
Variance of a Binomial
Random Variable
• The expected value of a
binomial random variable
can be computed using the
simple analytic expression
E(X) = np.
• The variance of a binomial
random variable can be
computed using the analytic
expression
V(X) = np(1-p)= n p q,
Where q=(1-p) by definition.
7 - 70
Example 14
The random variable X is a binomial
random variable with n=12 and p=.8.
A. Find the expected value of X.
B. Find the variance of X.
C. Find the standard deviation of X.
7 - 71
Example 14 - Solution
A.  = E(X) = np = (12)(.8) = 9.6
B. V(X) = np(1-p) = (12)(.8)(1-.8)
= 1.92
C.  = V(X) = 1.92 = 1.386
7 - 72
The Poisson
Distribution
7 - 73
Poisson vs. Binomial
• The Poisson distribution is
similar to the binomial in that
the random variable
represents a count of the total
number of “successes”.
• The major difference between
the two distributions is that the
Poisson does not have a fixed
number of trials.
• Instead, the Poisson uses a
fixed interval of time or space
in which the number of
“successes” are recorded.
7 - 74
Definition
In order to qualify as a Poisson
random variable an experiment
must meet two conditions:
1 “Successes” occur one at a
time. That is, two or more
“successes” cannot occur at
exactly the same point in time or
exactly at the same point in
space.
2 The occurrence of a “success”
in any interval is independent of
the occurrence of a “success” in
any other interval.
7 - 75
The Poisson
Probability Distribution
Function
-l
e l
P(X = x) =
, for x = 0,1,2,...
x!
x
where the transcendental constant
e is the limit of (1+[1/n])n as n
becomes large without bound
e = 2.71828..., and
l = average number of “successes”
Note: The variance of the Poisson
distribution is equal to the mean (l).
7 - 76
The Shape of the
Poisson Distribution
l = .3
0.8
0.6
0.4
0.2
0
0
1
2
0.25
3
4
5
l=3
0.2
0.15
0.1
0.05
0
0 1 2 3 4 5 6 7 8 9 10 11 12
As l
increases,
the shape of
the Poisson
distribution
begins to
resemble a
bell shaped
distribution.
0.12
0.1
l = 12
0.08
0.06
0.04
0.02
0
1
3
5
7
9 11 13 15 17 19
7 - 77
Poisson Random
Variables for Time
• The majority of Poisson
applications are related to the
number of occurrences of
some event in a specific
duration of time.
• The average number of
“successes” that occur within
the duration of time will define
the one and only parameter l
of the Poisson random
variable.
7 - 78
Example 15 (morning
calls)
The number of calls received by
an office on Monday morning
between 8:00 AM and 9:00 AM
has a Poisson distribution with
l equal to 4.0.
X = the number of calls received by
an office on Monday morning between
8:00 AM and 9:00 AM
l = 4.0
7 - 79
Example 15 - A (No
morning calls)
A. Determine the probability of
getting no calls between
eight and nine in the
morning.
l
4
e l
e 4
P(X=0) =

.0183
x!
0!
x
0
7 - 80
Example 15 - B (exactly
5 morning calls)
B. Calculate the probability of
getting exactly five calls
between eight and nine in
the morning.
e  l l x e 4 4 5
P(X=5)=

.1563
x!
5!
7 - 81
Example 15 - C (E(X) of
morning calls)
C. What will be the expected
number of calls received by
the office during this time
period? What is the
variance?
Remember that l = 4.0, and that l
is the mean, or average
number of “successes”.
Also, remember that the variance
of a Poisson distribution is
equal to the mean.
Thus, the expected number of calls
is 4.0, and the variance is also
4.0.
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Example 15 - D (Plot of
morning calls)
D. Graph the probability
distribution of the number of
calls using values from the
Poisson distribution tables in
the Appendix (pp. 528-532).
0.2
0.18
0.16
0.14
0.12
p(x) 0.1
0.08
0.06
0.04
0.02
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
x
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Poisson Random
Variables for Length
and Space
• There are a number of
Poisson applications that
measure the number of
“successes” in some area or
length.
• The average number of
“successes” in the area or
length will define the l
parameter of the Poisson
random variable.
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Example 16 (carpet
weaving errors)
The number of weaving errors in
a twenty foot by ten foot roll of
carpet has a Poisson distribution
with l equal to 0.1.
X = the number of weaving errors in a
20x10 foot roll of carpet
l = 0.1
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Example 16 - A (carpet
weaving errors p.d.f.)
A. Using the distribution tables ,
construct the probability
distribution for the carpet. l =
0.1
X
0
1
2
3
P(X)
.9048
.0905
.0045
.0002
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Example 16 - B (<2
carpet weaving errors)
B. What is the probability of
observing less than 2 errors
in the carpet? l = 0.1
P(X<2) = P(X=0) + P(X=1)
= .9048 + .0905 = .9953
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Example 16 - C (>5
carpet weaving errors)
C. What is the probability of
observing more than 5 errors
in the carpet? l = 0.1
P(X>5) = 0
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The Hypergeometric
Distribution
7 - 89
Hypergeometric vs.
Binomial
• Similarities
–Both random variables have
only two outcomes on each
trial of the experiment.
–They both count the number
of successes in n trials of an
experiment.
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Hypergeometric vs.
Binomial
• Differences
The hypergeometric distribution
differs from the binomial
distribution in the lack of
independence between trials.
[the probability of “success”
will vary between trials for the
hypergeometic pdf ].
In addition, hypergeometric
distributions have finite
populations in which the
TOTAL number of “successes”
and “failures’ are known.
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CN-A
n-x
N
The Hypergeometric
Probability Distribution
Function
C Ax CN-A
n-x
P(X = x) =
where 0
N
Cn
, where 0  x  minimum of (A,n)
A = the largest number of “successes” possible in population
N = the size of the total population
n = size of the sample drawn
7 - 92
Example 17 (dist’n of
memory chips)
Suppose that a shipment from
Matsua Semiconductor
contains 30 memory chips of
which two are bad.
If a memory board requires 16
chips, what is the probability
distribution for the number of
defective chips on the memory
board?
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Example 17 - Solution
• The random variable under
consideration is given as
X = number of defective chips on
the memory board.
The three parameters of the
distribution are
A = 2 (a “success” in this case is
a defective chip).
N = 30, and
n = 16.
• The maximum value of X in this
case is 2 =min(2,16).
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Example 17 - Solution
30-2
C20 C16-0
P( X = 0) =
30
C16
30, 421, 755
=
= .209 ,
145, 422 , 675
30-2
C12 C16-1
P(X = 1) =
30
C16
(2) (37, 442,160)
=
= .515, and
145, 422, 675
30-2
C 22 C16-2
P(X = 2) =
30
C16
(1) (40,116, 600)
=
= .276.
145, 422, 675
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The Expected Value and
Variance of a Hypergeometric
Random Variable
• The expected value of a
hypergeometric random
variable can be obtained
using the expression
A
E(X) = n
.
N
• The variance of a
hypergeometric random
variable is
A
A (N-n)
V(X) = [ n
(1 )]
.
N
N (N-1)
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Example 18, E(X) and
V(X) for Hypergeom’c
Compute the expected value and
variance for the random variable for
memorychips defined in Example 17.
A = 2, N = 30, and n =16
2 ) = 1.067
E(X) = 16 ( 30
V(X) = [ 16 ( 2 ) (1 - 2 ) ] (30-16)/(30-1) = .481
30
30
If the experiment were repeated many
times, the average number of defective
chips per board would be slightly greater
than 1.
7 - 97