4 - 1 © 2003 Pearson Prentice Hall Chapter 4 Discrete Random Variables

4 - 2 © 2003 Pearson Prentice Hall Learning Objectives 1.Distinguish Between the Two Types of Random Variables 2.Compute the Expected Value & Variance of Discrete Random Variables 3.Describe the Binomial Distribution and calculate probabilities for it 4.Understand derivation of formulas for mean and variance of binomial distribution

4 - 3 © 2003 Pearson Prentice Hall Data Types

4 - 4 © 2003 Pearson Prentice Hall Discrete Random Variables

4 - 5 © 2003 Pearson Prentice Hall Random Variable Assign numeric values to outcomes of an “experiment” Random Variable takes on different possible values corresponding to experiment outcomes Example: Number of Tails in 2 Coin Tosses Example: Number of Tails in 2 Coin Tosses

4 - 6 © 2003 Pearson Prentice Hall Discrete Random Variable Possible values are discrete Possible values are discrete E.g., Whole Number (0, 1, 2, 3 etc.) E.g., Whole Number (0, 1, 2, 3 etc.) Obtained by Counting Obtained by Counting Usually Finite Number of Values Usually Finite Number of Values But could be infinite (must be “countable”) But could be infinite (must be “countable”)

4 - 7 © 2003 Pearson Prentice Hall Exercise 4.3: Discrete or Continuous? Number of misspelled words Amount of water through Hoover dam in a day How late for class Number of bacteria in a water sample Amount of CO produced from burning a gallon of unleaded gas Your weight Number of checkout lanes at grocery store Amount of time waiting in line at grocery store

4 - 8 © 2003 Pearson Prentice Hall Discrete Probability Distribution 1.List of All possible [x, p(x)] pairs x = Value of Random Variable (Outcome) x = Value of Random Variable (Outcome) p(x) = Probability Associated with Value p(x) = Probability Associated with Value 2.Mutually Exclusive (No Overlap) 3.Collectively Exhaustive (Nothing Left Out) 4. 0  p(x)  1 5.  p(x) = 1

4 - 9 © 2003 Pearson Prentice Hall Discrete Probability Distribution Example Probability Distribution Values, x Probabilities, p(x) 01/4 =.25 12/4 =.50 21/4 =.25 Experiment: Toss 2 Coins. Count # Tails. © T/Maker Co.

© 2003 Pearson Prentice Hall Visualizing Discrete Probability Distributions ListingTable GraphEquation # Tails f(x) Count p(x) px n xnx pp xnx () ! !()! ()    x p(x)

© 2003 Pearson Prentice Hall Summary Measures 1.Expected Value Mean of Probability Distribution Mean of Probability Distribution Weighted Average of All Possible Values Weighted Average of All Possible Values May not equal any of the possible values May not equal any of the possible values  = E(X) =  x p(x)  = E(X) =  x p(x) 2.Variance Weighted Average Squared Deviation about Mean Weighted Average Squared Deviation about Mean  2 = E[ (x    (x    p(x)  2 = E[ (x    (x    p(x)

© 2003 Pearson Prentice Hall Thinking Challenge You toss 2 coins. You’re interested in the number of tails. What are the expected value & standard deviation of this random variable, number of tails? © T/Maker Co.

© 2003 Pearson Prentice Hall Expected Value & Variance Solution* xp(x)xp(x)x -  (x-  ) 2 ( x -  ) 2 p( p( x )

© 2003 Pearson Prentice Hall Expected Value & Variance Solution*  = 1.0 = 1.0 xp(x)xp(x)x -  (x-  ) 2 ( x -  ) 2 p( p( x )

© 2003 Pearson Prentice Hall Expected Value & Variance Solution*  = 1.0 = 1.0  2 =.50 =.50 xp(x)xp(x)x -  (x-  ) 2 ( x -  ) 2 p( p( x )

© 2003 Pearson Prentice Hall Exercise 4.25 Values 0, 1, 2 Dist X has probabilities.3,.4,.3 Dist Y has probabilities.1,.8,.1 Without doing any calculations: Which distribution has higher mean? Which distribution has higher mean? Which distribution has higher variance? Which distribution has higher variance?

© 2003 Pearson Prentice Hall Definition of Independence X and Y are independent if (and only if) For any sets of outcome values A and B For any sets of outcome values A and B P(X in A and Y in B) = P(X in A)P(Y in B) P(X in A and Y in B) = P(X in A)P(Y in B) That is, any pair of events defined by outcome sets for X and Y are independent events That is, any pair of events defined by outcome sets for X and Y are independent events

© 2003 Pearson Prentice Hall Binomial Distribution 1.Number of ‘Successes’ in a Sample of n Observations (Trials) # Reds in 15 Spins of Roulette Wheel # Reds in 15 Spins of Roulette Wheel # Defective Items in a Batch of 5 Items # Defective Items in a Batch of 5 Items # Correct on a 33 Question Exam # Correct on a 33 Question Exam # Customers Who Purchase Out of 100 Customers Who Enter Store # Customers Who Purchase Out of 100 Customers Who Enter Store # of Bush-Cheney supporters in survey of 100 people # of Bush-Cheney supporters in survey of 100 people

© 2003 Pearson Prentice Hall Binomial Distribution Properties 1.Sequence of n Identical Trials 2.Each Trial Has 2 Outcomes ‘Success’ (Desired/specified Outcome) or ‘Failure’ ‘Success’ (Desired/specified Outcome) or ‘Failure’ 3.Constant Trial Probability 4.Trials Are Independent 5.# of successes in n trials is a binomial random variable

© 2003 Pearson Prentice Hall Exercise: Binomial? Pick 6 students from this class Each flips a coin Each flips a coin Count # of heads Count # of heads Pick 6 students from this class X= # of 1st year students selected X= # of 1st year students selected Random digit dialing of 100 numbers # of Bush-Cheney supporters # of Bush-Cheney supporters Random digit dialing of 100 numbers Sum of ages of respondents Sum of ages of respondents

© 2003 Pearson Prentice Hall Binomial Probability Distribution Function p(x) = Probability of x ‘Successes’ n=Sample Size p=Probability of ‘Success’ x=Number of ‘Successes’ in Sample (x = 0, 1, 2,..., n)

© 2003 Pearson Prentice Hall Binomial Probability Distribution Example Experiment: Toss 1 Coin 5 Times in a Row. Note # Tails. What’s the Probability of 3 Tails?

© 2003 Pearson Prentice Hall Cumulative Distribution Function

© 2003 Pearson Prentice Hall Binomial Probability Table (Portion) Cumulative Probabilities F X (k)

© 2003 Pearson Prentice Hall Binomial Distribution Characteristics n = 5 p = 0.1 n = 5 p = 0.5 Mean Standard Deviation

© 2003 Pearson Prentice Hall Exercise 4.37 N=4, x=2, q=.4– compute p(x)

© 2003 Pearson Prentice Hall Exercise 4.39 N=25, p=.5 Compute E(X), Var(X), sigma Compute E(X), Var(X), sigma

© 2003 Pearson Prentice Hall Exercise 4.51 Pr(domestic abuse) = 1/3, or maybe 1/10 Sample of 15 women; 4 have been abused If p=1/3, what is Pr(X>=4)? If p=1/10, what is pr(X>=4)? Given evidence from the sample, which abuse rate seems more plausible? Note: this is a preview of thinking about sampling distributions

© 2003 Pearson Prentice Hall Exercise Response rate p=.4 If mail 20, what is Pr(X>12)? If want 100 responses with Pr >.95, how many should you mail?

© 2003 Pearson Prentice Hall Solution By empirical rule, 95% of observations are within 2sigma Find n such that mu-2sigma > 100 Find n such that mu-2sigma > 100 Mu =.4n Mu =.4n Sigma squared = npq=.24n Sigma squared = npq=.24n Solving for n  292 Solving for n  292 Note that for n=292 E(X) =292*.4=116.8 In order to be sure to get 100, need to have a higher average number of returns In order to be sure to get 100, need to have a higher average number of returns

© 2003 Pearson Prentice Hall Useful Observation 1 For any X and Y

© 2003 Pearson Prentice Hall One Binary Outcome Random variable X, one binary outcome Code success as 1, failure as 0 P(success)=p, p(failure)=(1-p)=q E(X)= p

© 2003 Pearson Prentice Hall Independent, identically distributed X 1, …, X n ; E(X i )=p; Binomial X = X 1, …, X n ; E(X i )=p; Binomial X = Mean of a Binomial By useful observation 1

© 2003 Pearson Prentice Hall Useful Observation 2 For independent X and Y

Useful Observation 3 For independent X and Y cancels by obs. 2

© 2003 Pearson Prentice Hall Variance of Binomial Independent, identically distributed X 1, …, X n ; E(X i )=p; Binomial X = X 1, …, X n ; E(X i )=p; Binomial X =

© 2003 Pearson Prentice Hall

End of Chapter Any blank slides that follow are blank intentionally.

© 2003 Pearson Prentice Hall Useful Observation 4 (We’ll use this later) For any X