1. (c) 2007 IUPUI SPEA K300 (4392) Outline Random Variables Expected Values Data Generation Process (DGP) Uniform Probability Distribution Binomial Probability Distribution

2. (c) 2007 IUPUI SPEA K300 (4392) Random Variables Variables whose values change randomly Neither predictable, nor constant Not arbitrary, but stochastic (followed by a data generation process) Statistically independent P(x=v): probability that X takes a particular value of v. Examples The number of heads when tossing a coin The number you get when rolling a die

3. (c) 2007 IUPUI SPEA K300 (4392) Discrete versus Continuous Discrete if the set of outcomes is either finite in number or countably infinite Example: tossing a coin, rolling a die, the number of customers arrived per hour Continuous if the set of outcomes is infinitively divisible and not countable. Example: GPA, height, gas mileage

4. (c) 2007 IUPUI SPEA K300 (4392) Level of Measurement RankDifferenceRatioContinuousOperator NominalN/A Discrete= OrdinalYesN/A Discrete=, >, < IntervalYes N/AYes=, >, <, +, -, x, ÷ RatioYes =, >, <, +, -, x, ÷ CountYes Discrete=, >, <, +, -, x, ÷

5. (c) 2007 IUPUI SPEA K300 (4392) Rules of Probability 0<=P(X=v)<=1 ∑P(X=v)=1 P(X=v)=0 when never happening P(X=v)=1 when always happening

6. (c) 2007 IUPUI SPEA K300 (4392) Uniform Distribution Each event has the same probability Examples Tossing a coil: 1/2 rolling a die: 1/6 X123456 P(X)1/6

7. (c) 2007 IUPUI SPEA K300 (4392) Expected Value 1 Expected value of a probability distribution is equivalent to mean ∑xP(x)=x 1 *P(x 1 )+ x 2 *P(x 2 )+… Expected value of rolling a die: 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 +6*1/6 = 21/6=3.5

8. (c) 2007 IUPUI SPEA K300 (4392) Expected Value 2 Let us play a game of tossing a coin. You have 50% chance to get a head. If you get a head, you will get $500; otherwise (getting a tail) you will lose $400. Expected value:+$500*1/2 + (-$400)*1/2 =$50 Do you want to play this game?

9. (c) 2007 IUPUI SPEA K300 (4392) Variance of Probability Distribution σ 2 = ∑[(x-µ) 2 P(x)] σ 2 = ∑x 2 P(x)-µ 2 Example 5-9. µ=3.5 σ 2 = (1-3.5) 2 *1/6 +(2-3.5) 2 *1/6 …=2.9 σ 2 = 1 2 *1/6 +2 2 *1/6 … -(3.5) 2 =2.9

10. (c) 2007 IUPUI SPEA K300 (4392) Data Generation Process (DGP) How are data (random variables) generated? “Tossing a coin” is a DGP that has only two outcomes (head/tail) with equal P “Rolling a die” is a DGP that has six outcomes with equal P (1/6) How about GPA of SPEA students? How do we formulate these DGPs

11. (c) 2007 IUPUI SPEA K300 (4392) Binomial Experiment Each trial: Bernoulli process (or trial) There must be a fixed number of trials, n Each trial can have only two outcomes, x (yes/no, true/false, success/failure) The outcome of each trial must be (statistically) independent of each other The probability of a success, p, must remain the same for each trial

12. (c) 2007 IUPUI SPEA K300 (4392) Binomial Distribution 1 Consists of N times of the Bernoulli trial with two outcomes Probabilities of these outcomes, which are not necessarily equal Examples: What is the probability that you will get 10 heads when tossing a coin 20 times? Given.01 of getting a certain disease, what is the probability that 5 out of 10 randomly selected subjects get the disease?

13. (c) 2007 IUPUI SPEA K300 (4392) Binomial Distribution 2 Tossing a coin four times (16 outcomes=2^4) X is the number of times to get the head (0 through four) P is the probability of getting the head 4 C 0 = 4 C 4 =1, 4 C 1 = 4 C 3 =4 4 C 2 =4!/[2!(4-2)!]=4!/(2!2!)=6 X01234Total P(X)1/164/166/164/161/161

14. (c) 2007 IUPUI SPEA K300 (4392) Binomial Distribution 3 P(p) probability of success P(q) probability of failure, P(q)=1-P(p) N the number of trials X the number of successes in n trials

15. (c) 2007 IUPUI SPEA K300 (4392) Binomial Distribution 4 What is the probability that you will get 10 heads when tossing a coin 20 times? Given.01 of getting a certain disease, what is the probability that 5 out of 10 randomly selected subjects get the disease?

16. (c) 2007 IUPUI SPEA K300 (4392) Binomial Distribution 5 Sick and tired of the formula? Good news. Someone computed the probability of binomial distribution for you. See example 5-18, p 265. What you have to know is N, X, and P Go to page 626 What is the probability that 5 out of 10 get the disease? P=.01

17. (c) 2007 IUPUI SPEA K300 (4392) Binomial Distribution 6 Expected value: µ=np Variance: npq Standard deviation: sqrt(npq) Example 5.21, p. 267. µ=np=4 X ½= 2 σ 2 =npq=4X½ X½ =1 σ=sqrt(1)=1

18. (c) 2007 IUPUI SPEA K300 (4392) Illustration 0 Tossing a coin Two outcomes: head and tail P(head)=1/2 http://www.uwsp.edu/psych/stat/9/hyptes td.htm

19. (c) 2007 IUPUI SPEA K300 (4392) Illustration 1, N=1 and 2

20. (c) 2007 IUPUI SPEA K300 (4392) Illustration 2, N=3 and 4

21. (c) 2007 IUPUI SPEA K300 (4392) Illustration 3, N=5 and 10

22. (c) 2007 IUPUI SPEA K300 (4392) Illustration 4, N=50 and 100

23. (c) 2007 IUPUI SPEA K300 (4392) Illustration 3