Chapter 7 Probability Distributions, Information about the Future

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Chapter 7

Probability Distributions, Information about the Future

.

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Random Processes

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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.

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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.

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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

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Types of Random Variables

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Quantitative Random Variables

Quantitative random variables are divided into two classes.

1) Discrete

2) Continuous

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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 pages in a standard math textbook is a discrete random variable.

Statistics

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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

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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

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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


4. the daily takeoffs at Chicago’s O’Hare Airport


5. the highest temperatures in Maine and Florida tomorrow

– continuous random variable

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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.

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Describing a Discrete Random Variable

• State (Describe) the variable.

• List all of the possible values of the variable.

• Determine the probabilities of these values.

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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?

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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

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Example 2

Not all discrete random variables have easily definable probability distributions.

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?

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• 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.

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Discrete Probability Distributions

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• 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.

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• 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.

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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]

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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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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.

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.

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Example 4, Is this a prob. Dist’n?

• Tell whether or not the following distribution is a probability distribution.

• If the distribution is not a probability distribution, give the characteristic which is not satisfied by the distribution.

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Yes. All probabilities are between 0 and 1, and the sum of the probabilities is 1.

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Example 5 , Is this a prob. Dist’n?



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No. The sum of the probabilities is greater than one.

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No. You can't have negative probabilities.

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P(X=x) = , for x = 1, 2, 3, 4, 5


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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.

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Expected Value E(X) of a random variable X

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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.

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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)

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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/indexComponentWtsDJ.html

• S&P 500 index is similar with weights available at

• http://www.indexarb.com/indexComponentWtsSP500.html

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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.

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E(X) for Daily Sales

The expected value of the probability distribution given in Example 3 (daily Sales) is computed in the table.

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

• 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.

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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)

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Variance and Standard Deviation

of a Discrete Random Variable

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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 p x( ) ( ) ( ). 2

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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

= 1,757.84

Option B

V(X) = 6,640,000

= 2,576.82

• The larger deviation reflects greater variability in profits and increased risk.

= V(X)

= V(X)

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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.

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Example 9

Find the expected value, the variance, and the standard deviation for a random variable with the following probability distribution.

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Probability Distributions and their Functions

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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)

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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.

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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!

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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) = .

To determine the probability that X = 4,

P(X=4) = .

330

= 930

2

430

=1630

2

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The Discrete Uniform Distribution

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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.

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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:

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The Binomial Distribution

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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).

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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.

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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.

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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

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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.

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The Binomial Probability Distribution

Function

where represents the number of possible combinations of n objects taken x at a time (without replacement) and is given by

, and 0! = 1;

n = the number of trials, and

p = the probability of a success.

P(X = x) = C p (1- p)xn x n-x

xn

C

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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:

Since,

P(X=3) = C ( ) (1 )34 3 4 31

212

.

C 4, then34 4!

3!(4 3)!4 3 2 13 2 1(1)

P(X 3) = 4( ) ( ) .25.12

12

416

3

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Example 12

Calculate for the following combinations of x and n.

A. n = 4 and x = 2

B. n = 12 and x =8

Cxn

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A.

B.

C12!

8!(12 8)!

12(11)(10)(9)(8!)

8!4(3)(2)(1)4958

12

C4!

2!(4 2)!

4(3)(2!)

2!(2)(1)62

4

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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).

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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

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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

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The Shape of a Binomial

If p is small, the distribution tends to be skewed with a tail on the right.

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 1 2 3 4 5 6 7 8 9 10 11 12

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If p is near .5, the distribution is symmetrical.

0

0.05

0.1

0.15

0.2

0.25

0 1 2 3 4 5 6 7 8 9

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If p is large, the distribution tends to be skewed with a tail on the left.

0

0.05

0.1

0.15

0.2

0.25

0.3

0 1 2 3 4 5 6 7 8 9 10 11

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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.

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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.

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A. = E(X) = np = (12)(.8) = 9.6

B. V(X) = np(1-p) = (12)(.8)(1-.8)

= 1.92

C. = = = 1.3861.92V(X)

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The Poisson Distribution

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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.

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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.

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The Poisson Probability Distribution

Function

where the transcendental constant e is the limit of (1+[1/n])n as n becomes large without bound

e = 2.71828..., and

= average number of “successes”

Note: The variance of the Poisson distribution is equal to the mean ().

P(X = x) = e

x!, for x = 0,1,2,...

- x

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The Shape of the Poisson Distribution

As increases, the shape of the Poisson distribution begins to resemble a bell shaped distribution.

0

0.2

0.4

0.6

0.8

0 1 2 3 4 5

0

0.05

0.1

0.15

0.2

0.25

0 1 2 3 4 5 6 7 8 9 10 11 12

0

0.02

0.04

0.06

0.08

0.1

0.12

1 3 5 7 9 11 13 15 17 19

= .3

= 3

= 12

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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 of the Poisson random variable.

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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 equal to 4.0.

X = the number of calls received by an office on Monday morning between 8:00 AM and 9:00 AM

= 4.0

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Example 15 - A (No morning calls)

A. Determine the probability of getting no calls between eight and nine in the morning.

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Example 15 - B (exactly 5 morning calls)

B. Calculate the probability of getting exactly five calls between eight and nine in the morning.

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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 = 4.0, and that 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).

x

p(x)

00.020.040.060.08

0.10.120.140.160.18

0.2

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

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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 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 equal to 0.1.

X = the number of weaving errors in a 20x10 foot roll of carpet

= 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. = 0.1

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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? = 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? = 0.1

P(X>5) = 0

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The Hypergeometric Distribution

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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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The Hypergeometric Probability Distribution

Function

A = the largest number of “succ-esses” possible in population

N = the size of the total population

n = size of the sample drawn

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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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P(X = 0) = C C

C

= 30,421,755

145,422,675 = .209,

02

16-030-2

1630

P(X = 1) = C C

C

= (2) (37, 442,160)

145, 422,675 = .515, and

P(X = 2) = C C

C

= (1) (40,116,600)

145, 422,675 = .276.

12

16-130-2

1630

22

16-230-2

1630

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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

• The variance of a hypergeometric random variable is

E(X) = n A

N .

V(X) = [ n A

N (1 -

A

N )]

(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

E(X) = 16 ( ) = 1.067

If the experiment were repeated many times, the average number of defective chips per board would be slightly greater than 1.

V(X) = [ 16 ( 230

) (1 - 230

) ] (30-16)/(30-1) = .481

230

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Chapter 7 Probability Distributions, Information about the Future

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