Chapter 5:Probability Distribution

Types of Variables

Chapter 1: Variable definitionA characteristic or attribute that can assume

different values.

Chapter 5: Random variableA variable whose value are determined by

chance.

Random Variables

Variables whose values are determined by chance.

Two Types of Variables1. Discrete

• Finite number of possible values2. Continuous

• Assumes all values between two values

Discrete Probability Distribution

Consists of the values a random variable can assume and the corresponding probabilities of the values. The probabilities are determined theoretically or by observations

What does this mean?

Example: Constructing a probability distribution

for rolling a single dieSolution:

Sample Space: 1, 2, 3, 4, 5, 6Probability: each has 1/6 of a

chance

Construction a Probability Distribution

First, make a tableThe Outcomes are placed on topThe probabilities are placed on the bottom

Outcome X 1 2 3 4 5 6

Probability

P(X)16

16

16

16

16

16

Construction a Probability Distribution

Second, make a chartP(X)

X

1

12

16

1 2 3 4 5 6 7

Rules of Probability Distribution

Rule 1:The sum of the probabilities of all the events

in the sample space must equal 1∑P(X)=1

Rule 2:The probability of each event in the sample

space must be between or equal to 0 and 10≤ P(X) ≤1

Practice

Page 258 #’s 1-25

Chapter 5 Section 2

Finding the Mean of Probability Distribution Formula μ= ∑X*P(X)

1. Mu(μ)= mean

2. ∑ = sum of

3. X= outcomes

4. P(X)= probability of outcomes

Example of Probability Distribution Mean

Find the average number of spots that appear when a die is tossed.

Probability

P(X)

654321Outcome X

16

16

16

16

16

16

Example continued

μ= ∑X*P(X)μ= X •P(X ) + X •P(X ) + X •P(X ) + … + X •P(X )

μ= 1• + 2• + 3• + 4• + 5• + 6•

μ= 21 = 3.5

μ= 3.5*

1 1 2 2 3 3 n

n

16

16

16

16

16

16

6

* Theoretically mean because there cannot be a 3.5 rolled with a die

Rounding Rule

The rounding rule for Mean, Standard Deviation, and Variance is:

Formula for Variance of Probability Distribution

Formula: σ²= ∑[X²•P(X)] - μ²

1. σ = sigma = sum

2. Mu(μ)= mean

3. ∑ = sum of

4. X= outcomes

5. P(X)= probability of outcomes

Variance of Probability Distribution

Probability

P(X)

654321Outcome X

16

16

16

16

16

16

σ²= ∑[X²•P(X)] - μ²1²•1/6+2²•1/6+3²•1/6+4²•1/6+5²•1/6+6²•1/615.1715.17-12.25 2.9 = σ²

- μ²- 3.5²

Finding Standard Deviation of a Probability Distribution

Formula:σ = √σ²σ = √2.9σ = 1.7

Assignment:

Page 267#’s 1-10

Expected Value Example

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

Table Set Up for Expected Value

Probability

-$1$349Gain

LoseWin

11,000

9991,000

E(X) = 349 • + (-1) • = 1

1,000999

1,000E(X) = -$0.65

This does not mean that you will lose $.65 if you participate. It means the that average lose of every person who plays will be $.65

Expected Value

Formulaμ= ∑X*P(X)

1. E(X) = expected value

2. ∑ = sum of

3. X= outcomes

4. P(X)= probability of outcomes

E(X)= ∑X*P(X)

Is it fair?

How is a gambling game fair?The expected value of the game is zero.

Who does it favor?Expected value

Positive: The playerNegative: The house

Example:Roulette: House wins $0.90 on every $1 betCraps: House wins $0.88 on every $1 bet

Your turn

One thousand tickets are sold at $1 each for four prizes ($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?

Table

Gain

Prob

Practice

Page # 26812-18 even

Chapter 5 Section 3 The Binomial Distribution

The Binomial Experiment

1. There must be a fixed number of Trials

2. Each Trial can have only two outcomes

1. Successful

2. Failure

3. Outcomes must be independent of each other

4. The probability must remain the same for each trial

Binomial Probability Formula

!( )

( )! !x n Xn

P X p qn X X

•P(S) The symbol for the probability of success

•P(F) The symbol for the probability of failure

•p The numerical probability of success

•q The numerical probability of failure

•n The number of trials

•X The number of successes in n trials

Example:

A coin is tossed 3 times. Find the probability of getting exactly two heads.

!( )

( )! !x n Xn

P X p qn X X

n = 3

X = 2

p = 1/2

q = 1/2

2 3 23! 1 1(2 )

(3 2)!2! 2 2P heads

(2 ) 0.375P heads

Page 277 # 4

A burglar alarm system has six fail-safe components. The probability of each failing is 0.05. Find these probabilities.

Exactly three will failFewer than two will failNone will fail

Exactly three will fail

n = 6X = 3p = 0.05q = 0.95

Use chart

0.002

Fewer than two will fail

n = 6X = 1 or 0p = 0.05q = 0.95

1 0or

0.232 0.735+0.967

None will fail

n = 6X = 0p = 0.05q = 0.95

0.735

Your turn

Page 277-278#’s 11-12

Binomial Distribution

Mean Formula:μ = n • p

Variance Formula:σ²= n • p • q

Standard Deviation Formula:σ=√n • p • q

Examples:

No Examples today. I think you can handle it.

Page 278#’s 14-27

Multinomial Distribution

31 21 2 3

1 2 3

!( )

! ! ! !KXX X X

KK

nP x p p p p

X X X X

1

2

3

5

3

1

1

n

X

X

X

1

2

3

.50

.30

.20

p

p

p

3 1 15!( ) .5 .3 .2

3!1!1!P x

( ) .15P x

Page 284 Example 5-25

In a music store, a manager found that the probabilities that a person buys 0, 1, or 2 or more CDs are 0.3,.6, and .1 respectively. If 6 customers enter the store, find the probability that 1 won’t buy any CD’s, 3 will buy 1 CD, and 2 will buy 2 or more CDs.

1

2

3

6

1

3

2

n

X

X

X

1

2

3

.30

.60

.10

p

p

p

( ) 0.03888P X

Practice

Page 290 #’s 1-6

The Poisson Distribution

A discrete probability distribution that is useful when n is large and p is small and when the independent variables occur over a period of time.Ex: area, volume, time

Formula

( ; )!

xeP X

X

The letter e is a constant approximately equal to 2.7183

Answers are rounded to four decimal places

Problem

If there are 200 typographical errors randomly distributed in a 500 page novel, find the probability that a given page contains exactly three errors.

Hypergeometric Distribution

Given a population two typesEx: male and femaleSuccess and failure

Without replacementMore accurate than Binomial Distribution

Formula

a x b n X

a b n

C C

C

a = population 1b = population 2n = total sectionX = selection wanted

Example

Ten people apply for a job. Five have completed college and five have not. If a manager selects three applicants at random, find the probability that all three are graduates.

a = college graduates = 5b = nongraduates = 5n = total section = 3X = selection wanted = 3

Page 287 Example 5-30

a x b n X

a b n

C C

C

5 3 5 3 3

5 5 3

C C

C

a = college graduates = 5b = nongraduates = 5n = total section = 3X = selection wanted = 3

Practice

Page 291#’s 17-21