Survey

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Survey

Document related concepts

Transcript

Discrete Probability Distributions 6- 1 Chapter Six McGraw-Hill/Irwin © 2006 The McGraw-Hill Companies, Inc., All Rights Reserved.. Random variable A result from an experiment that, by chance, can take different values. Egs. 1. No. of heads you would get in 3 tosses of a coin 2. No. of employees absent on a shift 3. No. of students at CSUN in a semester 4. No. of minutes to drive home from CSUN 5. Inches of rainfall in LA during a year 6. Tire pressure in PSI of a car tire Examples 1-3 are ‘discrete’ 4-6 are ‘continuous’ random variables In this chapter, we focus on the ‘discrete’. Probability Distribution A listing of all possible outcomes of an experiment and the corresponding probability. c P F T Possible Outcomes N = 1/8 = 3/8 = 3/8 = 1/8 Probability Distribution Note the similarity to histogram Mean of Discrete Probability Distribution [ xP( x)] where represents the mean o x is each outcome o P(x) is the probability of the various outcomes x. Calculating Mean/Expected Value of Heads in 3 coin-toss experiment [ xP( x )] μ = 0 * .125 + 1 * .375 + 2 * .375 + 3 * .125 = 1.5 (which by intuition makes sense because for each toss you have 50% chance of getting a head). is a weighted average. It is also referred to as Expected Value, E(X). Binomial Probability Distribution •An outcome of an experiment is classified into one of two mutually exclusive categories, such as a success or failure (bi means two). •The data collected are the results of counts (hence, a discrete probability distribution). •The probability of success stays the same for each trial (independence). Binomial Probability Distribution P( x)n Cx (1 ) x n x Can you logically explain this formula? n is the number of trials x is the number of observed successes π is the probability of success on each trial Cx n n! x!(n-x)! Let us re-visit the 3-toss coin experiment n=3 π = .5 X = 0,1,2,3 (number of heads) P(0) = 3C0 * (.5)0 * (1 - .5)3-0 P(1) = 3C1 * (.5)1 * (1 - .5)3-1 P(2) = 3C2 * (.5)2 * (1 - .5)3-2 P(3) = 3C3 * (.5)3 * (1 - .5)3-3 = .125 = .375 = .375 = .125 You can also use Appendix A (page 489) Practice time! Do Self-Review 6-3, Page 168 Use Appendix A, Page 490 Cumulative Binomial Probability Distribution Problem 21, Page 173 Use table in Page 491 a. 0.387 ( n=9; x=9; straight from table ) b. 0.001 ( P(x<5) = P(x ≤ 4) ) c. 0.992 ( 1 – P(x ≤ 5) = 1 – 0.008 ) d. 0.946 [ (P(x=7) + (P(x=8) + (P(x=9) ] Mean of the Binomial Distribution n Logic: If you throw a coin say 100 times, how many times you would expect to get a head? For each throw, the probability of getting a head is .5. So, in 100 trials, you would expect 50 heads (which is 100*.5; ie. nπ ) Variance of the Binomial Distribution n (1 ) 2 For the previous example, = 100*.5*(1-.5) σ 2 = 25 S.D. σ = 5