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Chapter 5 Discrete Probability Distribution I. Basic Definitions II. Summary Measures for Discrete Random Variable • Expected Value (Mean) • Variance and Standard Deviation III. Two Popular Discrete Probability Distributions • Binomial Distribution • Poisson Distribution I. Basic Definitions • Random Variable: (p.188) a numerical description of the outcomes of an experiment. Assign values to outcomes of an experiment so the experiment can be represented as a random variable. Example: Test scores: 0 X 100 Toss coin: X = {0, 1} Roll a die: X = {1, 2, 3, 4, 5, 6} I. Basic Definitions • Discrete Random Variable: It takes a set of discrete values. Between two possible values some values are impossible. • Continuous Random Variable: Between any two possible values for this variable, another value always exists. Example: Roll a die: X = {1, 2, 3, 4, 5, 6}. X is a discrete random variable. Weight of a person selected at random: X is a continuous random variable. I. Basic Definitions Two ways to present a discrete random variable • Probability Distribution: (p.191 Table 5.3) a list of all possible values (x) for a random variable and probabilities (f(x)) associated with individual values. • Probability Function: probability may be represented as a function of values of the random variable. Example: p.192 Consider the experiment of rolling a die and define the random variable X to be the number coming up. 1. Probability distribution 2. Probability function: f(x) = 1/6. I. Basic Definition • Valid Discrete Probability Function 0 f(x) 1 for all f(x) AND f(x) = 1 Example: p.193 #7 a. The probability distribution is proper because all f(x) meet requirements 0 f(x) 1 and f(x) = 1. b. P(X=30) = f(30) = .25 c. P(X25) = f(25) + f(20) = .15 + .20 = .35 d. P(X>30) = f(35) = .40 Homework: p.194 #10, p.195 #14 I. Basic Definitions Example: p.195 #14 a. Find valid f(200): .1+.2+.3+.25+.1+f(200) = 1, so f(200) = .05 b. P(?) P(X>0) = f(50)+f(100)+f(150)+f(200) = .7 c. P(?) P(X100) = f(100)+f(150)+f(200) = .4 (―at least‖) II. Summary Measures for Discrete Random Variable • Expected Value (mean): E(X) or (p.196) E(X) = xf(x) • Variance: Var(X) or 2 (p.196) Var(X) = (x- )2f(x) Homework: p.198 #16 Example: Consider the experiment of tossing coin and define X to be 0 if head and 1 if tail. Find the expected value and variance. E(X) = xf(x) = (0)(.5)+(1)(.5) = .5 Var(X) = (x- )2f(x) = (0 - .5)2(.5)+(1 - .5)2(.5) = .25 III. Two Popular Discrete Probability Distributions — Binomial and Poisson Outlines: 1. Probability Distribution: Binomial Distribution: Table 5 (p.935 - p.937). Given n, p, x f(x). Poisson Distribution: Table 7 (p.939 - p.944). Given , x f(x). 2. Applications: Difference between Binomial and Poisson. 3. Applications: criterion to define ―success‖. III. Two Popular Discrete Probability Distributions 1. Binomial Distribution p.201 • Random variable X: x ―successes‖ out of n trials (the number of ―successes‖ in n trials). • Three conditions for Binomial distribution: — n independent trials — Two outcomes for each trial: ―success‖ and ―failure‖. — p: probability of a ―success‖ is a constant from trial to trial. III. Two Popular Discrete Probability Distributions Example: Toss a coin 10 times. We define the ―success‖ as a head and X is the number of heads from 10 trials. Does X have a binomial distribution? Answer: Yes. Follow-up: Why? n? p? Example: Roll a die 10 times. We define the ―success‖ as 5 or more points coming up and X is the number of successes from 10 trials. Does X have a binomial distribution? Answer: Yes. Why? n? p? III. Two Popular Discrete Probability Distributions • Summary Measures for Binomial Distribution E(X) = np p.207 Var(X) = np(1-p) p.207 • Binomial Probability Distribution p.205 n x — f(x) = P(X=x) = p (1 p) n x x — Table 5 (p.935-p.937): Given n, p and x, find f(x). Homework: p.209 #26, #27, #28, #29, #30 c, d. Example: p.209 #25 Given: Binomial, n = 2, p = .4 ―Success‖ X: # of successes in 2 trials Answer: b. f(1) = ? (Table 5) f(1) = .48 c. f(0) = ? f(0) = .36 d. f(2) = ? e. P(X1) = ? f(2) = .16) P(X1) = f(1) + f(2) = .64 e. E(X) = np = .8 Var(X) = np(1-p) = .48 = ? Example: p.210 #35 (Application) Binomial distribution? 1. Is there a criterion for ―success‖ and ―failure‖? 2. n trials? n=? 3. p=? Given: ―Success‖: withdraw; n = 20; p = .20 Answer: a. P(X 2) = f(0) + f(1) + f(2) = .0115 + .0576 + .1369 = .2060 b. P(X=4) = f(4) = .2182 c. P(X>3) = 1 - P(X 3) = 1 – [f(0) + f(1) + f(2) + f(3)] = 1 - .0115 - .0576 - .1369 - .2054 = .5886 d. E(X) = np = (20)(.2) = 4. III. Two Popular Discrete Probability Distributions 2. Poisson Distribution p.210 • Random variable X: x ―successes‖ per unit (the number of ―successes‖ in a unit - time, size, ...). Example: X = the number of calls per hour. Does X have a Poisson distribution? Answer: Yes. Because ―Success‖: a call. Unit: an hour. X: the number of ―successes‖ (calls) per unit (hour). Reading: p.209 #29, #30, and p.213 #40, p.214 #42 Binomial or Poisson? If Binomial, ―success‖? p? n? If Poisson, ―success‖? ? unit? III. Two Popular Discrete Probability Distributions • Summary Measures for Poisson Distribution E(X) = ( is given) Var(X) = • Poisson Probability Distribution p.211 xe — f(x) = P(X=x) = x! — Table 7 (A-19 through p.A-24): Given and x, find f(x). Note: Keep the unit of consistent with the question. Homework: p.213 #38, p.214 #42, #43 Example: p.213 #39 Given: •Poisson distribution, because X = the number of occurrences per time period. • = 2 and unit is ―a time period‖. Note: Unit in questions may be different. Answer: xe 2 x e 2 a. f(x) = = x! x! b. What is the average number of occurrences in three time periods? (Different unit!) 6 x e 6 3 = (3)( ) = 6. c. f(x) = x! d. f(2) = P(X=2) = .2707 (Table 7. =2, x = 2) e. f(6) = P(X=6) = .1606 (Table 7. =6, x = 6) Example: p.214 #43 Given: • Poisson distribution, because X = the number of arrivals per time period. • = 10 and unit is ―per minute‖. Note: Unit in questions may be different. Answer: a. f(0) = 0 (Table 7. =10, x = 0) b. P(X3) = f(0)+f(1)+f(2)+f(3) (Table 7. =10, x = ?) = 0 + .0005 + .0023 + .0076 = .0104 c. P(X=0) = f(0) = .0821 (Table 7. =(10)(15/60)=2.5 for unit of 15 seconds, x = 0) d. P(X1) = ? (Table 7. =2.5, x = ?) P(X1) = f(1) + f(2) + f(3) + … ??? = 1 - P(X<1) = 1 - f(0) = 1 - .0821 = .9179. Chapter 5 Summary • Binomial distribution: X = # of ―successes‖ out of n trials • Poisson distribution: X = # of ―successes‖ per unit • For Binomial distribution, E(X)=np, Var(X)=np(1-p) • For Poisson distribution, E(X) = = Var(X) • Binomial distribution table: Table 5 • Poisson distribution table: Table 7 • Poisson Approximation of Binomial distribution If p .05 AND n 20, Poisson distribution (Table 7) can be used to find probability for Binomial distribution. Example: A Binomial distribution with n = 250 and p = .01, f(3) = ? Answer: = np =(250)(.01)=2.5. From Table 7, f(x)=.2138 • Sampling with replacement and without replacement. Example: A bag of 100 marbles contains 10% red ones. Assume that samples are drawn randomly with replacement. a. If a sample of two is drawn, what is the probability that both will be red? b. If a sample of three is drawn, what is the probability that at least one will be red? (―without replacement‖: Hypergeometric distribution p.214) Answer: a. n = 2, p = .1, f(2) = ? From Table 5, f(2) = .01 b. n = 3, p = .1, P(X1) = f(1) + f(2) + f(3) From Table 5, P(X1) = .2430 + .0270 + .0010 = .271 Chapter 5 Homework Review Basics for Discrete Distribution: Slide 5. p. 194 #10, p.195 #14 Slide 7. p. 198 #16 Binomial Distribution Slide 11. p.209#26, #28 Poisson Distribution Slide 15. p.214 #42, #43

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