chapter 5 discrete probability distributions copyright © the mcgraw-hill companies, inc. permission...

Chapter 5

Discrete Probability Distributions

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1

Chapter 5 Overview Introduction

• 5-1 Probability Distributions

• 5-2 Mean, Variance, Standard Deviation, and Expectation

• 5-3 The Binomial Distribution


Chapter 5 Objectives

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1. Construct a probability distribution for a random variable.

2. Find the mean, variance, standard deviation, and expected value for a discrete random variable.

3. Find the exact probability for X successes in n trials of a binomial experiment.

4. Find the mean, variance, and standard deviation for the variable of a binomial distribution.

5.1 Probability Distributions• A random variable is a variable whose values are

determined by chance. The variable includes only whole numbers ( no fractions or irrational numbers). Values of this variable include 0 , 1, 2, 3, …

• A discrete probability distribution consists of the values a random variable can assume and the corresponding probabilities of the values.

• The sum of the probabilities of all events in a sample space add up to 1. Each probability is between 0 and 1, inclusively.


Examples of random variables:

• 1. the number of speeding tickets issued on a certain stretch of I 95 S.

• 2. the number of heads which appear when 4 dimes are tossed.

• 3. The number of passes completed in a game by a quarterback

• 4. The number of defective parts in a lot of 1000.


Continuous random variables

• Temperature: Range 62-78 degrees • Can assume an infinite number of values • Examples: weight, height, temperature, time.

Chapter 5Discrete Probability Distributions

Section 5-1Example 5-1

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Example 5-1: Rolling a Die

Construct a probability distribution for rolling a single die.




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Discrete probability distribution

• -consists of a correspondence between the values of the random variable and its associated probability. The correspondence can be shown in a table.

• Properties:1. 0 ( ) 12. ( ) 1.0

p xp x

Example 5-2: Tossing CoinsRepresent graphically the probability distribution for the sample space for tossing three coins.

.


Discrete probability distribution

• A bag contains three red checkers and two black checkers. Two checkers are drawn in succession and without replacement from this bag. Let x represent the the number of red checkers drawn in two attempts.


Number of red checkers

• A) What are the values that x can take on? • Answer: x = 0, 1, 2• B) Find the probability distribution of x, the

number of red checkers.


Table

X P(x)0 0.11 0.62 0.3


Probability Distribution?

• C) Does the distribution satisfy the requirements of a discrete probability distribution?


Mean of a probability distribution

• Toss three coins and let x represent the number of heads that appear.

• Recall the sample space: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

• X = 0, 1, 1, 1, 2, 2, 2, 3


Mean

• Notice the the outcomes of x = 1 and x = 2 occur three times each, while the outcomes x = 0 and x = 3 occur once each:


Notice here we can find the mean by multiplying the values of x by the probability that x occurs.

Mean

• x p(x) xp(x)• 0 1/8 0(1/8)= 0• 1 3/8 1(3/8)=3/8• 2 3/8 2(3/8)=6/8• 3 1/8 3(1/8)=3/8• sum = 3/2 = 1.5 • Thus, mean number of heads = 1.5


Mean number of red checkers

• Recall the problem of selecting two checkers from a bag containing 2 black and 3 red checkers


X P(x) X*P(x)0 0.1 01 0.6 0.62 0.3 0.6

5-2 Mean, Variance, Standard Deviation, and Expectation


2 2 2

VARIANCE:

X P X

Rounding Rule

The mean, variance, and standard deviation should be rounded to one more decimal place than the outcome X.

When fractions are used, they should be reduced to lowest terms.


Mean, Variance, Standard Deviation, and Expectation



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Example 5-5: Rolling a DieFind the mean of the number of spots that appear when a die is tossed.

.


X P X 1 1 1 1 1 16 6 6 6 6 61 2 3 4 5 6

216 3.5



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Example 5-8: Trips of 5 Nights or MoreThe probability distribution shown represents the number of trips of five nights or more that American adults take per year. (That is, 6% do not take any trips lasting five nights or more, 70% take one trip lasting five nights or more per year, etc.) Find the mean.

.


Example 5-8: Trips of 5 Nights or More


X P X

0 0.06 1 0.70 2 0.20

3 0.03 4 0.01

1.2



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Example 5-9: Rolling a Die


2 2 2X P X

2 2 2 2 21 1 1 16 6 6 6

22 21 16 6

1 2 3 4

5 6 3.5

2 2.9 , 1.7

• .



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Example 5-11: On Hold for Talk Radio

A talk radio station has four telephone lines. If the host is unable to talk (i.e., during a commercial) or is talking to a person, the other callers are placed on hold. When all lines are in use, others who are trying to call in get a busy signal. The probability that 0, 1, 2, 3, or 4 people will get through is shown in the distribution. Find the variance and standard deviation for the distribution.




2 2 2 2

22 2

0 0.18 1 0.34 2 0.23

3 0.21 4 0.04 1.6

2 1.2 , 1.1

0 0.18 1 0.34 2 0.23

3 0.21 4 0.04 1.6


A talk radio station has four telephone lines. If the host is unable to talk (i.e., during a commercial) or is talking to a person, the other callers are placed on hold. When all lines are in use, others who are trying to call in get a busy signal.

Should the station have considered getting more phone lines installed?



No, the four phone lines should be sufficient.

The mean number of people calling at any one time is 1.6.

Since the standard deviation is 1.1, most callers would be accommodated by having four phone lines because µ + 2σ would be

1.6 + 2(1.1) = 1.6 + 2.2 = 3.8.

Very few callers would get a busy signal since at least 75% of the callers would either get through or be put on hold. (See Chebyshev’s theorem in Section 3–2.)



Expectation• The expected value, or expectation, of a

discrete random variable of a probability distribution is the theoretical average of the variable.

• The expected value is, by definition, the mean of the probability distribution.

E X X P X

Chapter 5Discrete Probability

Distributions


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Gain X

Probability P(X)

Example 5-12: Winning Tickets

One thousand tickets are sold at $1 each for a color television valued at $350. What is the expected value of the gain if you purchase one ticket?


$349 -$11

1000999

1000

99911000 1000$349 $1 $0.65E X

Win Lose

Gain X

Probability P(X)

Example 5-13: Winning Tickets

One thousand tickets are sold at $1 each for four prizes of $100, $50, $25, and $10. After each prize drawing, the winning ticket is then returned to the pool of tickets. What is the expected value if you purchase two tickets?


$100 $50 $25 $10 $02

10002

10002

10002

1000992

1000

2 2 21000 1000 1000

99221000 1000

$100 $50 $25

$10 $0 $1.63$2

E X

Alternate Approach

Expected value

• Life insurance. A person pays $100 for a one year term life insurance policy. The probability that this person will live one more year is 0.97. The policy pays his beneficiaries $3000 upon the death of the policy holder. What is the expected value of x, the amount of money the insurance company expects to make?


Expected Value

X P(x) X*P(x)

100 0.97 97

-2900 0.03 -87

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39

There are two outcomes: The insured lives for one more year or dies within one year. If the person lives for one more year, the insurance company gains $100. If the person dies within a year, the beneficiaries receive $2900. (The premium of $100 is subtracted from the benefit.

5-3 The Binomial Distribution


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.

The Binomial Distribution


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.

Notation for the Binomial Distribution


The symbol for the probability of success

The symbol for the probability of failure

The numerical probability of success

The numerical probability of failure

and P(F) = 1 – p = q

The number of trials

The number of successes

P(S)

P(F)

p

q

P(S) = p

n

X

Note that X = 0, 1, 2, 3,...,n

Introduction to Binomial Probability

• Suppose there is a 0.30 probability that a customer walks in a store and buys at least one product. Find the probability that two out of three customers will buy at least one product.

• This is a binomial experiment. Why? • 1) Fixed number of trials (3) • 2) Two outcomes • 3) Trials independent


Binomial experiment

• There are C(3,2) that any two of the three customers will purchase at least one product.

• (Check this by using a tree diagram). • C(3,2) =

• For each of the three outcomes, the probability is 0.3(0.3)(0.7) =


Binomial probability

• Thus P(x = 2) =

• This result is generalized on the next slide.




!

- ! ! X n Xn

P X p qn X X

In a binomial experiment, the probability of exactly X successes in n trials is



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Example 5-16: Survey on Doctor Visits

A survey found that one out of five Americans say he or she has visited a doctor in any given month. If 10 people are selected at random, find the probability that exactly 3 will have visited a doctor last month.


!

- ! ! X n Xn

P X p qn X X

3 7

10! 1 43

7!3! 5 5

P

1510,"one out of five" , 3 n p X

0.201



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Example 5-17: Survey on EmploymentA survey from Teenage Research Unlimited (Northbrook, Illinois) found that 30% of teenage consumers receive their spending money from part-time jobs. If 5 teenagers are selected at random, find the probability that at least 3 of them will have part-time jobs.


3 25!3 0.30 0.70

2!3! P

5, 0.30,"at least 3" 3,4,5 n p X

0.132

4 15!4 0.30 0.70

1!4! P 0.028

5 05!5 0.30 0.70

0!5! P 0.002

3 0.132

0.028

0.002

0.162

P X



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Example 5-18: Tossing CoinsA coin is tossed 3 times. Find the probability of getting exactly two heads, using Table B.


123, 0.5, 2 n p X 2 0.375 P



Mean: np2Variance: npq

The mean, variance, and standard deviation of a variable that has the binomial distribution can be found by using the following formulas.

Standard Deviation: npq



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Example 5-23: Likelihood of TwinsThe Statistical Bulletin published by Metropolitan Life Insurance Co. reported that 2% of all American births result in twins. If a random sample of 8000 births is taken, find the mean, variance, and standard deviation of the number of births that would result in twins.


8000 0.02 160 np

2 8000 0.02 0.98 156.8 157 npq

8000 0.02 0.98 12.5 13 npq

Using the TI83+

• Suppose n = 10 , p = 0.3 and x = 3. • We can calculate the binomial probability

using the TI 83+ calculator by following the steps below:

• 1) Enter 2nd • 2) DISTR • 3) Option 10 (binompdf( • 4. enter binompdf(10, 0.3, 3) =0.2668


Using the TI 83+ to calculate cumulative binomial probability

• Let n = 10, p = 0.3 and x is less than or equal to 3. So what is calculated is

• P(x =0) + P(x = 1) + P(x = 2) + p(x = 3). • 1. 2nd • 2. DISTR

• 3) Option A (binomcdf(10, 0.3, 3))

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