copyright © 2013, 2010 and 2007 pearson education, inc. chapter discrete probability distributions...

Copyright © 2013, 2010 and 2007 Pearson Education, Inc.

Chapter

Discrete Probability Distributions

6


Section

Discrete Random Variables

6.1


Objectives1. Distinguish between discrete and continuous

random variables

2. Identify discrete probability distributions

3. Construct probability histograms

4. Compute and interpret the mean of a discrete random variable

5. Interpret the mean of a discrete random

6. Compute the standard deviation of a discrete random variable

5-3


Objective 1

• Distinguish between Discrete and Continuous Random Variables

5-4

Copyright © 2013, 2010 and 2007 Pearson Education, Inc.6-5

A random variable is a numerical measure of the outcome from a probability experiment, so its value is determined by chance. Random variables are denoted using letters such as X.


A discrete random variable has either a finite or countable number of values. The values of a discrete random variable can be plotted on a number line with space between each point.


A continuous random variable has infinitely many values. The values of a continuous random variable can be plotted on a line in an uninterrupted fashion.


Determine whether the following random variables are discrete or continuous. State possible values for the random variable.

(a)The number of light bulbs that burn out in a room of 10 light bulbs in the next year.

(b) The number of leaves on a randomly selected oak tree.

(c) The length of time between calls to 911.

EXAMPLE Distinguishing Between Discrete and Continuous Random Variables

EXAMPLE Distinguishing Between Discrete and Continuous Random Variables

Discrete; x = 0, 1, 2, …, 10

Discrete; x = 0, 1, 2, …

Continuous; t > 0


Objective 2

• Identify Discrete Probability Distributions

5-9


A probability distribution provides the possible values of the random variable X and their corresponding probabilities. A probability distribution can be in the form of a table, graph or mathematical formula.


The table to the right shows the probability distribution for the random variable X, where X represents the number of movies streamed on Netflix each month.

x P(x)

0 0.06

1 0.58

2 0.22

3 0.10

4 0.03

5 0.01

EXAMPLE A Discrete Probability DistributionEXAMPLE A Discrete Probability Distribution


Rules for a Discrete Probability Distribution

Let P(x) denote the probability that the random variable X equals x; then

• Σ P(x) = 1• 0 ≤ P(x) ≤ 1

Rules for a Discrete Probability Distribution

Let P(x) denote the probability that the random variable X equals x; then

• Σ P(x) = 1• 0 ≤ P(x) ≤ 1

5-12


EXAMPLE Identifying Probability DistributionsEXAMPLE Identifying Probability Distributions

x P(x)

0 0.16

1 0.18

2 0.22

3 0.10

4 0.30

5 0.01

Is the following a probability distribution?

No. Σ P(x) = 0.97



x P(x)

0 0.16

1 0.18

2 0.22

3 0.10

4 0.30

5 – 0.01


No. P(x = 5) = –0.01



x P(x)

0 0.16

1 0.18

2 0.22

3 0.10

4 0.30

5 0.04


Yes. Σ P(x) = 1


Objective 3

• Construct Probability Histograms

5-16


A probability histogram is a histogram in which the horizontal axis corresponds to the value of the random variable and the vertical axis represents the probability of that value of the random variable.


Draw a probability histogram of the probability distribution to the right, which represents the number of movies streamed on Netflix each month.

EXAMPLE Drawing a Probability HistogramEXAMPLE Drawing a Probability Histogram

x P(x)

0 0.06

1 0.58

2 0.22

3 0.10

4 0.03

5 0.01


Objective 4

• Compute and Interpret the Mean of a Discrete Random Variable

5-19


The Mean of a Discrete Random Variable

The mean of a discrete random variable is given by the formula

where x is the value of the random variable and P(x) is the probability of observing the value x.

The Mean of a Discrete Random Variable

The mean of a discrete random variable is given by the formula

where x is the value of the random variable and P(x) is the probability of observing the value x.

5-20

x x P x


Compute the mean of the probability distribution to the right, which represents the number of DVDs a person rents from a video store during a single visit.

EXAMPLE Computing the Mean of a Discrete Random Variable

EXAMPLE Computing the Mean of a Discrete Random Variable

x P(x)

0 0.06

1 0.58

2 0.22

3 0.10

4 0.03

5 0.01x x P x 0(0.06)1(0.58) 2(0.22) 3(0.10) 4(0.03) 5(0.01)

1.49


Interpretation of the Mean of a Discrete Random VariableSuppose an experiment is repeated n independent times and the value of the random variable X is recorded. As the number of repetitions of the experiment increases, the mean value of the n trials will approach μX, the mean of the random variable X. In other words, let x1 be the value of the random variable X after the first experiment, x2 be the value of the random variable X after the second experiment, and so on. Then

The difference between and μX gets closer to 0 as n increases.

Interpretation of the Mean of a Discrete Random VariableSuppose an experiment is repeated n independent times and the value of the random variable X is recorded. As the number of repetitions of the experiment increases, the mean value of the n trials will approach μX, the mean of the random variable X. In other words, let x1 be the value of the random variable X after the first experiment, x2 be the value of the random variable X after the second experiment, and so on. Then

The difference between and μX gets closer to 0 as n increases.

5-22

x

x1 x2 L xn

nx


The following data represent the number of DVDs rented by 100 randomly selected customers in a single visit. Compute the mean number of DVDs rented.


49.1100

... 10021

xxx

X


As the number of trials of the experiment increases, the mean number of rentals approaches the mean of the probability distribution.


Objective 5

• Compute and Interpret the Mean of a Discrete Random Variable

5-26


Because the mean of a random variable represents what we would expect to happen in the long run, it is also called the expected value, E(X), of the random variable.


EXAMPLE Computing the Expected Value of a Discrete Random VariableEXAMPLE Computing the Expected Value of a Discrete Random Variable

A term life insurance policy will pay a beneficiary a certain sum of money upon the death of the policy holder. These policies have premiums that must be paid annually. Suppose a life insurance company sells a $250,000 one year term life insurance policy to a 49-year-old female for $530. According to the National Vital Statistics Report, Vol. 47, No. 28, the probability the female will survive the year is 0.99791. Compute the expected value of this policy to the insurance company.

x P(x)

530 0.99791

530 – 250,000 = -249,470

0.00209

Survives

Does not survive

E(X) = 530(0.99791) + (-249,470)(0.00209)

= $7.50


Objective 6

• Compute the Standard Deviation of a Discrete Random Variable

5-29


Standard Deviation of a Discrete Random Variable

The standard deviation of a discrete random variable X is given by

where x is the value of the random variable, μX is the mean of the random variable, and P(x) is the probability of observing a value of the random variable.

Standard Deviation of a Discrete Random Variable

The standard deviation of a discrete random variable X is given by

where x is the value of the random variable, μX is the mean of the random variable, and P(x) is the probability of observing a value of the random variable.

5-30

X x x 2 P x

x2 P x X2


Compute the variance and standard deviation of the following probability distribution which represents the number of DVDs a person rents from a video store during a single visit.

EXAMPLE Computing the Variance and Standard Deviation of a Discrete Random VariableEXAMPLE Computing the Variance and Standard Deviation of a Discrete Random Variable

x P(x)

0 0.06

1 0.58

2 0.22

3 0.10

4 0.03

5 0.01


x P(x)0 0.06 –1.43 2.0449 0.1226941 0.58 –0.91 0.8281 0.4802982 0.22 –1.27 1.6129 0.3548383 0.1 –1.39 1.9321 0.193214 0.03 –1.46 2.1316 0.0639485 0.01 –1.48 2.1904 0.021904

EXAMPLE Computing the Variance and Standard Deviation of a Discrete Random Variable

EXAMPLE Computing the Variance and Standard Deviation of a Discrete Random Variable

X2 x X 2

P(x)1.236892

X 1.236892

1.11

x X 2x X

x X 2P(x)


Section

The Binomial Probability Distribution

6.2


Objectives

1. Determine whether a probability experiment is a binomial experiment

2. Compute probabilities of binomial experiments

3. Compute the mean and standard deviation of a binomial random variable

4. Construct binomial probability histograms

5-34


Criteria for a Binomial Probability ExperimentCriteria for a Binomial Probability Experiment

An experiment is said to be a binomial experiment if

1. The experiment is performed a fixed number of times. Each repetition of the experiment is called a trial.

2. The trials are independent. This means the outcome of one trial will not affect the outcome of the other trials.

3. For each trial, there are two mutually exclusive (or disjoint) outcomes, success or failure.

4. The probability of success is fixed for each trial of the experiment.

Criteria for a Binomial Probability ExperimentCriteria for a Binomial Probability Experiment

An experiment is said to be a binomial experiment if

1. The experiment is performed a fixed number of times. Each repetition of the experiment is called a trial.

2. The trials are independent. This means the outcome of one trial will not affect the outcome of the other trials.

3. For each trial, there are two mutually exclusive (or disjoint) outcomes, success or failure.

4. The probability of success is fixed for each trial of the experiment.

5-35


Notation Used in the Binomial Probability Notation Used in the Binomial Probability DistributionDistribution

•There are n independent trials of the experiment.

•Let p denote the probability of success so that 1 – p is the probability of failure.

•Let X be a binomial random variable that denotes the number of successes in n independent trials of the experiment.So, 0 < x < n.

Notation Used in the Binomial Probability Notation Used in the Binomial Probability DistributionDistribution

•There are n independent trials of the experiment.

•Let p denote the probability of success so that 1 – p is the probability of failure.

•Let X be a binomial random variable that denotes the number of successes in n independent trials of the experiment.So, 0 < x < n.

5-36


Which of the following are binomial experiments?

(a) A player rolls a pair of fair die 10 times. The number X of 7’s rolled is recorded.

EXAMPLE Identifying Binomial ExperimentsEXAMPLE Identifying Binomial Experiments

Binomial experiment



(b) The 11 largest airlines had an on-time percentage of 84.7% in November, 2001 according to the Air

Travel Consumer Report. In order to assess reasons for delays, an official with the FAA randomly selects flights until she finds 10 that were not on time. The number of flights X that need to be selected is recorded.


Not a binomial experiment – not a fixed number of trials.



(c) In a class of 30 students, 55% are female. The instructor randomly selects 4 students. The

number X of females selected is recorded.


Not a binomial experiment – the trials are not independent.


Objective 2

• Compute Probabilities of Binomial Experiments

5-40


According to the Air Travel Consumer Report, the 11 largest air carriers had an on-time percentage of 79.0% in May, 2008. Suppose that 4 flights are randomly selected from May, 2008 and the number of on-time flights X is recorded.

Construct a probability distribution for the random variable X using a tree diagram.

EXAMPLE Constructing a Binomial Probability DistributionEXAMPLE Constructing a Binomial Probability Distribution


Binomial Probability Distribution Function

The probability of obtaining x successes in n independent trials of a binomial experiment is given by

where p is the probability of success.

Binomial Probability Distribution Function

The probability of obtaining x successes in n independent trials of a binomial experiment is given by

where p is the probability of success.

5-42

P x n Cx px 1 p n xx 0,1,2,...,n


Phrase Math Symbol

“at least” or “no less than”or “greater than or equal to” ≥

“more than” or “greater than” >

“fewer than” or “less than” <

“no more than” or “at most”or “less than or equal to ≤

“exactly” or “equals” or “is” =


EXAMPLE Using the Binomial Probability Distribution FunctionEXAMPLE Using the Binomial Probability Distribution Function

According to the Experian Automotive, 35% of all car-owning households have three or more cars.

(a)In a random sample of 20 car-owning households, what is the probability that exactly 5 have three or more cars?

P(5) 20 C5 (0.35)5 (1 0.35)20 5

0.1272




(b) In a random sample of 20 car-owning households, what is the probability that less than 4 have three or more cars?

P(X 4) P(X 3)

P(0) P(1) P(2) P(3)

0.0444




(c) In a random sample of 20 car-owning households, what is the probability that at least 4 have three or more cars?

P(X 4) 1 P(X 3)

1 0.0444

0.9556


Objective 3

• Compute the Mean and Standard Deviation of a Binomial Random Variable

5-47


Mean (or Expected Value) and Standard Deviation of a Binomial Random Variable

A binomial experiment with n independent trials and probability of success p has a mean and standard deviation given by the formulas

Mean (or Expected Value) and Standard Deviation of a Binomial Random Variable

A binomial experiment with n independent trials and probability of success p has a mean and standard deviation given by the formulas

5-48

X np and X np 1 p


According to the Experian Automotive, 35% of all car-owning households have three or more cars. In a simple random sample of 400 car-owning households, determine the mean and standard deviation number of car-owning households that will have three or more cars.

EXAMPLE Finding the Mean and Standard Deviation of a Binomial Random Variable

EXAMPLE Finding the Mean and Standard Deviation of a Binomial Random Variable

X np

(400)(0.35)

140

X np(1 p)

(400)(0.35)(1 0.35)

9.54


Objective 4

• Construct Binomial Probability Histograms

5-50


(a) Construct a binomial probability histogram with n = 8 and p = 0.15.

(b) Construct a binomial probability histogram with n = 8 and p = 0. 5.

(c) Construct a binomial probability histogram with n = 8 and p = 0.85.

For each histogram, comment on the shape of the distribution.

EXAMPLE Constructing Binomial Probability Histograms



Construct a binomial probability histogram with n = 25 and p = 0.8. Comment on the shape of the distribution.




For a fixed probability of success, p, as the number of trials n in a binomial experiment increase, the probability distribution of the random variable X becomes bell-shaped.As a general rule of thumb, if np(1 – p) > 10, then the probability distribution will be approximately bell-shaped.

For a fixed probability of success, p, as the number of trials n in a binomial experiment increase, the probability distribution of the random variable X becomes bell-shaped.As a general rule of thumb, if np(1 – p) > 10, then the probability distribution will be approximately bell-shaped.

5-61


Use the Empirical Rule to identify unusual observations in a binomial experiment.

The Empirical Rule states that in a bell-shaped distribution about 95% of all observations lie within two standard deviations of the mean.


The Empirical Rule states that in a bell-shaped distribution about 95% of all observations lie within two standard deviations of the mean.

5-62



95% of the observations lie between μ – 2σ and μ + 2σ.

Any observation that lies outside this interval may be considered unusual because the observation occurs less than 5% of the time.


95% of the observations lie between μ – 2σ and μ + 2σ.

Any observation that lies outside this interval may be considered unusual because the observation occurs less than 5% of the time.

5-63


According to the Experian Automotive, 35% of all car-owning households have three or more cars. A researcher believes this percentage is higher than the percentage reported by Experian Automotive. He conducts a simple random sample of 400 car-owning households and found that 162 had three or more cars. Is this result unusual ?

EXAMPLE Using the Mean, Standard Deviation and Empirical Rule to Check for Unusual Results in a Binomial Experiment


X np

(400)(0.35)

140

X np(1 p)

(400)(0.35)(1 0.35)

9.54




The result is unusual since 162 > 159.1

X np

(400)(0.35)

140

X np(1 p)

(400)(0.35)(1 0.35)

9.54

X 2 X 140 2(9.54)

120.9

X 2 X 140 2(9.54)

159.1


Section

The Poisson Probability Distribution

6.3


Objectives

1. Determine whether a probability experiment follows a Poisson process

2. Compute probabilities of a Poisson random variable

3. Find the mean and standard deviation of a Poisson random variable

5-67


Objective 1

• Determine if a Probability Experiment Follows a Poisson Process

5-68


A random variable X, the number of successes in a fixed interval, follows a Poisson process provided the following conditions are met.

1.The probability of two or more successes in any sufficiently small subinterval is 0.

2.The probability of success is the same for any two intervals of equal length.

3.The number of successes in any interval is independent of the number of successes in any other interval provided the intervals are not overlapping.


The Food and Drug Administration sets a Food Defect Action Level (FDAL) for various foreign substances that inevitably end up in the food we eat and liquids we drink.

For example, the FDAL level for insect filth in chocolate is 0.6 insect fragments (larvae, eggs, body parts, and so on) per 1 gram.

EXAMPLE Illustrating a Poisson ProcessEXAMPLE Illustrating a Poisson Process


Objective 2

• Compute Probabilities of a Poisson Random Variable

5-71


Poisson Probability Distribution Function

If X is the number of successes in an interval of fixed length t, then the probability of obtaining x successes in the interval is

where λ (the Greek letter lambda) represents the average number of occurrences of the event in some interval of length 1 and e ≈ 2.71828.


If X is the number of successes in an interval of fixed length t, then the probability of obtaining x successes in the interval is

where λ (the Greek letter lambda) represents the average number of occurrences of the event in some interval of length 1 and e ≈ 2.71828.

5-72

P x t x

x!e t x 0,1,2,3,...


The Food and Drug Administration sets a Food Defect Action Level (FDAL) for various foreign substances that inevitably end up in the food we eat and liquids we drink. For example, the FDAL level for insect filth in chocolate is 0.6 insect fragments (larvae, eggs, body parts, and so on) per 1 gram. Suppose that a chocolate bar has 0.6 insect fragments per gram. Compute the probability that the number of insect fragments in a 10-gram sample of chocolate is(a) exactly three. Interpret the result.(b) fewer than three. Interpret the result.(c) at least three. Interpret the result.

EXAMPLE Illustrating a Poisson ProcessEXAMPLE Illustrating a Poisson Process


(a) λ = 0.6; t = 10

(b) P(X < 3) = P(X < 2) = P(0) + P(1) + P(2) = 0.0620

(c) P(X > 3) = 1 – P(X < 2) = 1 – 0.0620 = 0.938

P(3) (0.610)3

3!e 0.6(10)

0.0892


Objective 3

• Find the Mean and Standard Deviation of a Poisson Random Variable

5-75


Mean and Standard Deviation of a Poisson Random Variable

A random variable X that follows a Poisson process with parameter λ has mean (or expected value) and standard deviation given by the formulas

where t is the length of the interval.

Mean and Standard Deviation of a Poisson Random Variable

A random variable X that follows a Poisson process with parameter λ has mean (or expected value) and standard deviation given by the formulas

where t is the length of the interval.

5-76

X t and X t X



If X is the number of successes in an interval of fixed length and X follows a Poisson process with mean μ, the probability distribution function for X is


If X is the number of successes in an interval of fixed length and X follows a Poisson process with mean μ, the probability distribution function for X is

5-77

P x x

x!e x 0,1,2, 3,...


The Food and Drug Administration sets a Food Defect Action Level (FDAL) for various foreign substances that inevitably end up in the food we eat and liquids we drink. For example, the FDAL level for insect filth in chocolate is 0.6 insect fragments (larvae, eggs, body parts, and so on) per 1 gram.

(a)Determine the mean number of insect fragments in a 5 gram sample of chocolate.

(b) What is the standard deviation?

EXAMPLE Mean and Standard Deviation of a Poisson Random Variable

EXAMPLE Mean and Standard Deviation of a Poisson Random Variable


EXAMPLE Mean and Standard Deviation of a Poisson Random Variable EXAMPLE Mean and Standard Deviation of a Poisson Random Variable

We would expect 3 insect fragments in a 5-gram sample of chocolate.

(a) X t

(0.6)(5)

3

(b) X t

(0.6)(5)

3


In 1910, Ernest Rutherford and Hans Geiger recorded the number of α-particles emitted from a polonium source in eighth-minute (7.5 second) intervals. The results are reported in the table to the right. Does a Poisson probability function accurately describe the number of α-particles emitted?

EXAMPLE A Poisson Process? EXAMPLE A Poisson Process?

Source: Rutherford, Sir Ernest; Chadwick, James; and Ellis, C.D.. Radiations from Radioactive Substances. London, Cambridge University Press, 1951, p. 172.


Section

The Hypergeometric Probability Distribution

6.4


Objectives

1. Determine whether a probability experiment is a hypergeometric experiment

2. Compute probabilities of hypergeometric experiments

3. Find the mean and standard deviation of a hypergeometric random variable

5-83


Objective 1

• Determine if a Probability Experiment is a Hypergeometric Experiment

5-84


Criteria for a Hypergeometric Probability ExperimentA probability experiment is said to be a hypergeometric experiment provided

•The finite population to be sampled has n elements.•For each trial of the experiment, there are two possible outcomes, success or failure. There are exactly k successes in the population.•A sample size of n is obtained from the population of size N without replacement.

Criteria for a Hypergeometric Probability ExperimentA probability experiment is said to be a hypergeometric experiment provided

•The finite population to be sampled has n elements.•For each trial of the experiment, there are two possible outcomes, success or failure. There are exactly k successes in the population.•A sample size of n is obtained from the population of size N without replacement.

5-85


Notation Used in the Hypergeometric Probability Distribution

•The population is size N. •The sample is size n.•There are k successes in the population.•Let the random variable X denote the number of successes in the sample of size n, so x must be greater than or equal to the larger of 0 or n – (N – k), and x must be less than or equal to the smaller of n or k.

Notation Used in the Hypergeometric Probability Distribution

•The population is size N. •The sample is size n.•There are k successes in the population.•Let the random variable X denote the number of successes in the sample of size n, so x must be greater than or equal to the larger of 0 or n – (N – k), and x must be less than or equal to the smaller of n or k.5-86


EXAMPLE A Hypergeometric Probability DistributionEXAMPLE A Hypergeometric Probability Distribution

The Dow Jones Industrial Average (DJIA) is a collection of thirty publicly traded companies that are meant to be representative of the United States economy. In one certain month 18 of the 30 stocks in the DJIA increased in value. If an investor randomly invests in four stocks at the beginning of this month and records X, the number of stocks that increased in value during the month, is this a hypergeometric probability experiment? List the possible values of the random variable X.



This is a hypergeometric probability experiment because

1.The population consists of N = 30 stocks.2.Two outcomes are possible: either the stock increases in value, or it does not increase in value. There are k = 18 successes.3.The sample size is n = 4.

The possible values of the random variable arex = 0, 1, 2, 3, 4.


Objective 2

• Construct the Probabilities of Hypergeometric Experiments

5-89


Hypergeometric Probability Distribution

The probability of obtaining x success based on a random sample of size n from a population of size N is given by

where k is the number of successes in the population.

Hypergeometric Probability Distribution

The probability of obtaining x success based on a random sample of size n from a population of size N is given by

where k is the number of successes in the population.

5-90

P x k Cx N k Cn x N Cn



The Dow Jones Industrial Average (DJIA) is a collection of thirty publicly traded companies that are meant to be representative of the United States economy. In one certain month 18 of the 30 stocks in the DJIA increased in value.

What is the probability that an investor randomly invests in four stocks at the beginning of this month and three of the stocks increased in value?



N = 30; n = 4; k = 18; x = 3

P(x) k Cx N k Cn x N Cn

18 C3 30 18 C4 3 30 C4

816 12 27,405

0.3573


Objective 3

• Compute the Mean and Standard Deviation of a Hypergeometric Random Variable

5-93


Mean and Standard Deviation of a Hypergeometric Random Variable

A hypergeometric random variable has mean and standard deviation given by the formulas

where n is the sample sizek is the number of successes in the populationN is the size of the population

Mean and Standard Deviation of a Hypergeometric Random Variable

A hypergeometric random variable has mean and standard deviation given by the formulas

where n is the sample sizek is the number of successes in the populationN is the size of the population

5-94

X nk

N ; X

N n

N 1

n

k

NN k

N



The Dow Jones Industrial Average (DJIA) is a collection of thirty publicly traded companies that are meant to be representative of the United States economy. In one certain month 18 of the 30 stocks in the DJIA increased in value. In a random sample of four stocks, determine the mean and standard deviation of the number of stocks that will increase in value.



N = 30; n = 4; k = 18

We would expect 2.4 stocks out of 4 stocks to increase in value during the month.

X 4 18

302.4

X 30 4

30 14

18

3030 18

300.9277

copyright © 2013, 2010 and 2007 pearson education, inc. chapter discrete probability distributions...

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