Probability Distribution Chapter 5: Probability Distribution

Types of Variables Chapter 1: Variable definition A characteristic or attribute that can assume different values. Chapter 5: Random variable A variable whose value are determined by chance.

Random Variables Variables whose values are determined by chance. Two Types of Variables Discrete Finite number of possible values 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 die Solution: Sample Space: 1, 2, 3, 4, 5, 6 Probability: each has 1/6 of a chance

Construction a Probability Distribution First, make a table The Outcomes are placed on top The probabilities are placed on the bottom Outcome X 1 2 3 4 5 6 Probability P(X) 1 6 1 6 1 6 1 6 1 6 1 6

Construction a Probability Distribution Second, make a chart P(X) 1 1 2 1 6 X 1 2 3 4 5 6 7

Rules of Probability Distribution 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 1 0≤ P(X) ≤1

Practice Page 258 #’s 1-25

Chapter 5 Section 2 Finding the Mean of Probability Distribution Formula μ= ∑X*P(X) Mu(μ)= mean ∑ = sum of X= outcomes 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) 6 5 4 3 2 1 Outcome X

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 1 6 1 6 1 6 1 6 1 6 1 6 6 * Theoretically mean because there cannot be a 3.5 rolled with a die

Rounding Rule ONE DECIMAL to the RIGHT OF the GIVEN The rounding rule for Mean, Standard Deviation, and Variance is: to the RIGHT OF the GIVEN

Formula for Variance of Probability Distribution σ²= ∑[X²•P(X)] - μ² σ = sigma = sum Mu(μ)= mean ∑ = sum of X= outcomes P(X)= probability of outcomes

Variance of Probability Distribution Outcome X 1 2 3 4 5 6 Probability P(X) 1 1 1 1 1 1 6 6 6 6 6 6 σ²= ∑[X²•P(X)] - μ² 1²•1/6+2²•1/6+3²•1/6+4²•1/6+5²•1/6+6²•1/6 15.17 15.17-12.25 2.9 = σ² - 3.5² - μ²

Finding Standard Deviation of a Probability Distribution Formula: σ = √σ² σ = √2.9 σ = 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 Win Lose Gain $349 -$1 Probability 1 1,000 999 1,000 E(X) = 349 • + (-1) • = 1 1,000 999 E(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 Formula μ= ∑X*P(X) E(X)= ∑X*P(X) E(X) = expected value ∑ = sum of X= outcomes P(X)= probability of outcomes E(X)= ∑X*P(X)

Is it fair? How is a gambling game fair? Who does it favor? Example: The expected value of the game is zero. Who does it favor? Expected value Positive: The player Negative: The house Example: Roulette: House wins $0.90 on every $1 bet Craps: 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 # 268 12-18 even

Chapter 5 Section 3 The Binomial Distribution The Binomial Experiment There must be a fixed number of Trials Each Trial can have only two outcomes Successful Failure Outcomes must be independent of each other The probability must remain the same for each trial

Binomial Probability Formula 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. n = 3 X = 2 p = 1/2 q = 1/2

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 fail Fewer than two will fail None will fail

Exactly three will fail n = 6 X = 3 p = 0.05 q = 0.95 Use chart 0.002

Fewer than two will fail X = 1 or 0 p = 0.05 q = 0.95 1 or 0.967 0.232 + 0.735

None will fail n = 6 X = 0 p = 0.05 q = 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

Multinomial Distribution

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.

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 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 types Ex: male and female Success and failure Without replacement More accurate than Binomial Distribution

Formula a = population 1 b = population 2 n = total section X = 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 = 5 b = nongraduates = 5 n = total section = 3 X = selection wanted = 3

Page 287 Example 5-30 a = college graduates = 5 b = nongraduates = 5 n = total section = 3 X = selection wanted = 3

Practice Page 291 #’s 17-21