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