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.195) 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.198 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.199 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.200 #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(X  25) = f(25) + f(20) = =.35 d. P(X>30) = f(35) =.40 Homework: p.201 #10, p.201 #14

a. Find valid f(200): f(200) = 1, so f(200) =.05 b. P(?) P(X>0) = f(50)+f(100)+f(150)+f(200) =.7 c. P(?) P(X  100) = f(100)+f(150)+f(200) =.4(“at least”) I. Basic Definitions Example: p.202 #14

II. Summary Measures for Discrete Random Variable Expected Value (mean): E(X) or  (p.203) E(X) =  xf(x) Variance: Var(X) or  2 (p.203) Var(X) =  (x-  ) 2 f(x) Homework: p.204 #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-  ) 2 f(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 p.997). Given n, p, x  f(x). Poisson Distribution: Table 7 (p p.1004). 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 Distributionp.207 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) = npp.214 Var(X) = np(1-p)p.214 Binomial Probability Distributionp.212 — f(x) = P(X=x) = — Table 5 (p.989-p.997): Given n, p and x, find f(x). Homework: p.216 #26, #27, #29, #30 c, d.

Example: p.216 #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(X  1) = ? f(2) =.16) P(X  1) = f(1) + f(2) =.64 e. E(X) = np =.8 Var(X) = np(1-p) =.48  = ?

Example: p.217 #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) = =.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)] = =.5886 d. E(X) = np = (20)(.2) = 4.

III. Two Popular Discrete Probability Distributions 2. Poisson Distributionp.218 Random variable X: x “occurrences” per unit (the number of “occurrences” in a unit - time, size,...). Example: X = the number of calls per hour. Does X have a Poisson distribution? Answer: Yes. Because “Occurrence”: a call. Unit: an hour. X: the number of “occurrences” (calls) per unit (hour). Reading: p.216 #29, #30, and p.220 #40, p.221 #42 Binomial or Poisson? If Binomial, “success”? p? n? If Poisson, “occurrence”?  ? unit?

III. Two Popular Discrete Probability Distributions Summary Measures for Poisson Distribution E(X) =  (  is given) Var(X) =  Poisson Probability Distributionp.219 — f(x) = P(X=x) = — Table 7 (p.999 through p.1004): Given  and x, find f(x). Note: Keep the unit of  consistent with the question. Homework: p.220 #38, p.221 #42, #43

Example: p.220 #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: a. f(x) = = b. What is the average number of occurrences in three time periods? (Different unit!)  3 = (3)(  ) = 6.c. f(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.221 #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(X  3) = f(0)+f(1)+f(2)+f(3) (Table 7.  =10, x = ?) = =.0104 c. P(X=0) = f(0) =.0821 (Table 7.  =(10)(15/60)=2.5 for unit of 15 seconds, x = 0) d. P(X  1) = ?(Table 7.  =2.5, x = ?) P(X  1) = f(1) + f(2) + f(3) + … ??? = 1 - P(X<1) = 1 - f(0) = =.9179.

Chapter 5 Summary Binomial distribution: X = # of “successes” out of n trials Poisson distribution: X = # of “occurrences” 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(X  1) = f(1) + f(2) + f(3) From Table 5, P(X  1) = =.271