1. 1 Random Variables and Discrete probability Distributions SESSION 2

2. 2 A random variable is discrete if it can assume a countable number of values. A random variable is continuous if it can assume an uncountable number of values. 0 1 1/2 1/4 1/16 Continuous random variable After the first value is defined the second value, and any value thereafter are known. Therefore, the number of values is countable After the first value is defined, any number can be the next one Discrete random variable Therefore, the number of values is uncountable 0123... Discrete and Continuous Random Variables

3. 3 If a random variable can assume values x i, then the following must be true: Requirements for a Discrete Distribution

4. 4 Population Mean (Expected Value) Given a discrete random variable X with values x i, that occur with probabilities p(x i ), the population mean of X is.

5. 5 Let X be a discrete random variable with possible values x i that occur with probabilities p(x i ), and let E(x i ) =  The variance of X is defined by Population Variance

6. 6 Laws of Expected Value  E(c) = c  E(X + c) = E(X) + c  E(c X ) = cE( X ) Laws of Variance  V(c) = 0  V(X + c) = V(X)  V(cX) = c 2 V(X) Laws of Expected Value and Variance

7. 7 Example 7.4 The monthly sales at a computer store have a mean of $25,000 and a standard deviation of $4,000. Profits are 30% of the sales less fixed costs of $6,000. Find the mean and standard deviation of the monthly profit. Laws of Expected Value Variance

8. 8 V(X + c) = V(X) V(cX) = c 2 V(X) V(Profit) = V(.30(Sales) – 6,000] = V[(.30)(Sales)] = (.30) 2 V(Sales) = 1,440,000 E(c X ) = cE( X ) E(X + c) = E(X) + c Profit =.30(Sales) – 6,000 E(Profit) = E[.30(Sales) – 6,000] = E[.30(Sales)] – 6,000 =.30E(Sales) – 6,000 =.(30)(25,000) – 6,000 = 1,500 Laws of Expected Value and Variance  = [1,440,000] 1/2 = 1,200 Solution Laws of Expected Value and Variance

9. 9 The bivariate (or joint) distribution is used when the relationship between two random variables is studied. The probability that X assumes the value x, and Y assumes the value y is denoted p(x,y) = P(X=x and Y = y) 7.4 Bivariate Distributions

10. 10 Example Xavier and Yvette are two real estate agents. Let X and Y denote the number of houses that Xavier and Yvette will sell next week, respectively. The bivariate probability distribution is presented next. Bivariate Distributions

11. 11 p(x,y) Bivariate Distributions X Y X=0X=2X=1 y=1 y=2 y=0 0.42 0.12 0.21 0.07 0.06 0.02 0.06 0.03 0.01 Example 7.5 – continued X Y 0 1 2 0.12.42.06 1.21.06.03 2.07.02.01

12. 12 Marginal Probabilities Example 7.5 – continued Sum across rows and down columns X Y 0 1 2p(y) 0.12.42.06.60 1.21.06.03.30 2.07.02.01.10 p(x).40.50.101.00 p(0,0) p(0,1) p(0,2) The marginal probability P(X=0) P(Y=1), the marginal probability.

13. 13 Describing the Bivariate Distribution The joint distribution can be described by the mean, variance, and standard deviation of each variable. This is done using the marginal distributions. x p(x) y p(y) 0.4 0.6 1.5 1.3 2.1 E(X) =.7 E(Y) =.5 V(X) =.41 V(Y) =.45

14. 14  = Describing the Bivariate Distribution To describe the relationship between the two variables we compute the covariance and the coefficient of correlation Covariance: COV(X,Y) =  (X –  x )(Y-  y )p(x,y) Coefficient of Correlation COV(X,Y)  x  y To describe the relationship between the two variables we compute the covariance and the coefficient of correlation Covariance: COV(X,Y) =  (X –  x )(Y-  y )p(x,y) Coefficient of Correlation COV(X,Y)  x  y

15. 15 Example 7.6 Calculate the covariance and coefficient of correlation between the number of houses sold by the two agents in Example 7.5 Solution COV(X,Y) =  (x-  x )(y-  y )p(x,y) = (0-.7)(0-.5)p(0,0)+…(2-.7)(2-.5)p(2,2) = -.15  =COV(X,Y)/  x  y = -.15/(.64)(.67) = -.35 Describing the Bivariate Distribution

16. 16 The following relationship can assist in calculating E(X+Y) and V(X+Y) E(X+Y) =E(X) + E(Y); V(X+Y) = V(X) +V(Y) +2COV(X,Y) When X and Y are independent COV(X,Y) = 0, and V(X+Y) = V(X)+V(Y). The Expected Value and Variance of X+Y

17. 17 7.6 The Binomial Distribution The binomial experiment can result in only one of two possible outcomes. Typical cases where the binomial experiment applies: A coin flipped results in heads or tails An election candidate wins or loses An employee is male or female A car uses 87octane gasoline, or another gasoline.

18. 18 Calculating the Binomial Probability In general, The binomial probability is calculated by:

19. 19 Example Pat Statsdud is registered in a statistics course and intends to rely on luck to pass the next quiz. The quiz consists on 10 multiple choice questions with 5 possible choices for each question, only one of which is the correct answer. Pat will guess the answer to each question Find the following probabilities Pat gets no answer correct Pat gets two answer correct? Pat fails the quiz Calculating the Binomial Probability

20. 20 Solution Checking the conditions An answer can be either correct or incorrect. There is a fixed finite number of trials (n=10) Each answer is independent of the others. The probability p of a correct answer (.20) does not change from question to question. Calculating the Binomial Probability

21. 21 Solution – Continued Determining the binomial probabilities: Let X = the number of correct answers Calculating the Binomial Probability

22. 22 = p(0) + p(1) + p(2) + p(3) + p(4) =.1074 +.2684 +.3020 +.2013 +.0881 =.9672 Solution – Continued Determining the binomial probabilities: Pat fails the test if the number of correct answers is less than 5, which means less than or equal to 4. Calculating the Binomial Probability This is called cumulative probability P(X  4 

23. 23 E(X) =  = np V(X) =  2 = np(1-p) Example 7.11 If all the students in Pat’s class intend to guess the answers to the quiz, what is the mean and the standard deviation of the quiz mark? Solution  = np = 10(.2) = 2.  = [np(1-p)] 1/2 = [10(.2)(.8)] 1/2 = 1.26. Mean and Variance of Binomial Variable Binomial Distribution- summary

24. 24 The Poisson experiment typically fits cases of rare events that occur over a fixed amount of time or within a specified region Typical cases The number of errors a typist makes per page The number of customers entering a service station per hour The number of telephone calls received by a switchboard per hour. 7.7 Poisson Distribution

25. 25 The number of successes (events) that occur in a certain time interval is independent of the number of successes that occur in another time interval. The probability of a success in a certain time interval is the same for all time intervals of the same size, proportional to the length of the interval. The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller. Properties of the Poisson Experiment

26. 26 The Poisson Random Variable The Poisson variable indicates the number of successes that occur during a given time interval or in a specific region in a Poisson experiment Probability Distribution of the Poisson Random Variable. The Poisson Variable and Distribution

27. 27 Poisson Distributions (Graphs) 0 1 2 3 4 5

28. 28 Poisson Distributions (Graphs) Poisson probability distribution with  =2 Poisson probability distribution with  =5 Poisson probability distribution with  =7 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

29. 29 Poisson Distribution Example The number of Typographical errors in new editions of textbooks is Poisson distributed with a mean of 1.5 per 100 pages. 100 pages of a new book are randomly selected. What is the probability that there are no typos? Solution P(X=0)= e -   x x! e -1.5 1.5 0 0!.2231==

30. 30 Poisson Distribution Example For a 400 page book calculate the following probabilities T here are no typos There are five or fewer typos Solution P(X=0)= P(X  5)=<use the formula to find p(0), p(1),…,p(5), then calculate p(0)+p(1)+…+p(5) =.4457 e -   x x! e -6 6 0 0!.002479== Important! A mean of 1.5 typos per100 pages, is equivalent to 6 typos per 400 pages. Finding Poisson Probabilities