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## Chapter 5

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**Chapter 5**Discrete Random Variables • Two Types of Random Variables • Discrete Probability Distributions • The Binomial Distribution Dr. Constance Lightner- Fayetteville State University**Random Variables**• A random variableis a numerical description of the outcome of an experiment. • A discrete random variablemay assume either a finite number of values or an infinite sequence of values. • A continuous random variablemay assume any numerical value in an interval or collection of intervals. Dr. Constance Lightner- Fayetteville State University**Discrete random variable with a finite number of values:**Let x = number of raffle tickets sold in one day where the seller has 20 tickets to sell. Clearly, x could equal 0, 1, 2, 3, ….., 20. • Discrete random variable with an infinite sequence of values: Let x = number of customers that enter a McDonald’s restaurant in one day. Here x can take on the values 0, 1, 2, . . . We can count the customers arriving, but there is no finite upper limit on the number that might arrive. • Continuous Random Variables Let x = the time that it takes the server at McDonalds to fill your order. In my experiences x could equal any value between 0 and 5 minutes. Dr. Constance Lightner- Fayetteville State University**Probability Distribution**• The probability distributionfor a discrete random variable provides the associated probabilities that the random variable will take on each possible value that it may assume. • We can describe a discrete probability distribution with a table, graph, or equation. • The discrete probability distribution is defined by a probability function p(x). In general probability functions are defined by a probability function f(x). For discrete probability functions p(x) is the same as f(x). • The function p(x) gives the probability that the random variable takes on the value x. Dr. Constance Lightner- Fayetteville State University**Example 5.1 (Probability distribution –Tabular)**Using past data on digital camera sales at a retail store, a tabular representation of the probability distribution for digital camera sales was developed. Number Units Soldof Daysx p(x) 0 70 0 .35 1 60 1 .30 2 38 2 .19 3 12 3 .06 4 20 4 .10 200 1.00 Dr. Constance Lightner- Fayetteville State University**Example 5.1 (Probability distribution –Graphical)**Dr. Constance Lightner- Fayetteville State University**Example 5.2**A Casino has a slot machine that costs $1 to play. The machine is set up so that a person could lose $1 if the play, win $0 (i.e. they put a dollar in the machine and win $1), or gain $1, $2, or $3. The following equation describes the probabilities associated with potential “winnings”. p(x)= x = -1, 0, 1, 2, 3 (i.e.. x=The amount of money “won” playing the slot machine) Dr. Constance Lightner- Fayetteville State University**Required Conditions for a Discrete Probability Function**• All probabilities must be greater than or equal to 0, i.e., p(x) 0. • The sum of the probabilities of all potential discrete random variable values occurring must equal one, i.e. p(x)=1. Dr. Constance Lightner- Fayetteville State University**Example 5.1 Revisited (Required Conditions)**• Refer to slide 5, Note that p(0), p(1), p(2), p(3), and p(4) are all greater than zero. • Also if you sum these probabilities 0.35 + 0.3+ 0.19 + 0.06 + 0.10=1 Dr. Constance Lightner- Fayetteville State University**Example 5.2 Revisited (Required Conditions)**Recall that p(x)= x = -1, 0, 1, 2, 3 Thus, Clearly all probabilities are greater than 0. Also all probabilities sum up to 1. Dr. Constance Lightner- Fayetteville State University**The Expected Value of a Discrete Random Variable**• The expected value of a discreterandom variable is the same as the population mean. The mean of a discrete random is denoted x. • The expected value of the random variable x is denoted E(x). • The expected value gives the long run average value of the random variable. For example refer to Example 4.2. If we were to watch thousands of people play the slot machine in the example, the expected value would be the average winnings that were observed. E(x)= x =xp(x) Dr. Constance Lightner- Fayetteville State University**Example 5.1 Revisited (Expected value)**How many digital cameras should the retail store expect to sale on a given day? x p(x)xp(x) 0 .35 0 1 .30 .3 2 .19 .38 3 .06 .18 4 .10 .4 E(x)= x =1.26 On average, the retail store will sell 1.26 digital cameras per day. Dr. Constance Lightner- Fayetteville State University**Example 5.2 Revisited (Expected value)**How much money should a person expect to win at this casino slot machine? x p(x)xp(x) -1 5/15 -5/15 0 4/15 0 1 3/15 3/15 2 2/15 4/15 3 1/15 3/15 E(x)= x =5/15=.33 On average, a person will win .33 dollars or 33¢.**Variance and Standard Deviation**• The variance of a discrete random variable is: • The standard deviation is the square root of the variance. Dr. Constance Lightner- Fayetteville State University**Example 5.1 Revisited (Variance and Standard Deviation)**Recall x = 1.26 x p(x)(x- x ) (x- x )2(x- x )2p(x) 0 .35 -1.26 1.5876 0.55566 1 .30 -.26.0676 0.02028 2 .19 .74 .5476 0.104044 3 .06 1.74 3.0276 0.181656 4 .10 2.74 7.5076 0.75076 x2= 1.6124 Thus x = Dr. Constance Lightner- Fayetteville State University**The Binomial Distribution**The Binomial Experiment 1. Experiment consists of n identical trials (or a sample size of n) 2. Each trial results in exactly two outcomes. One which you call a “success” and the other you deem a “failure”. 3. Probability of success is denoted p, and is constant from trial to trial. Thus, the probability of a failure is 1-p. 4. Trials are independent. Dr. Constance Lightner- Fayetteville State University**Binomial Probability Distribution**• Our interest is in the number of successes occurring in the n trials. • We let x denote the number of successes occurring in the n trials. Dr. Constance Lightner- Fayetteville State University**Binomial Probability Distribution**If x = the total number of successes in n trials of a binomial experiment, then x a binomial random variable and the probability of x successes in n trials is: where q=1-p Dr. Constance Lightner- Fayetteville State University**Example 5.3**Evans is concerned about a low retention rate for employees. On the basis of past experience, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employees chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year? Anderson, Sweeney and Williams Dr. Constance Lightner- Fayetteville State University**Example 5.3 (continued)**First let us review the 4 characteristics of the binomial experiment to verify that this is a binomial experiment: • There are 3 identical employees (in terms of their positions) • There are only two possible outcomes for each employee (either they will leave the job or stay). We will call leaving the company a success and staying a failure. [Hint: Always name a success in terms of the immediate question being asked. Sine we are interested in the probability of a customer leaving, name leaving a success.] • The probability of a success (i.e. employee leaving), p, equals 0.1. 4. One employee leaving does not affect whether another will leave. Dr. Constance Lightner- Fayetteville State University**Example 5.3 (continued)**Using the Binomial Probability Function Let: p = .10, n = 3, x = 1 There is a .243 probability that exactly 1 out of the 3 employees will leave. Dr. Constance Lightner- Fayetteville State University**Example 5.4**Suppose we were interested in the number of female applicants to an MBA program. Traditionally, 40% of applicants to the program are women. We reviewed 20 applications, what is the probability that 5 applications are from women? First let us review the 4 characteristics of the binomial experiment to verify that this is a binomial experiment: • There are 20 identical applications (in terms of the questions asked) • There are only two possible outcomes for each application (either the applicant is male or female). We will call a female applicant a success and a male applicant a failure. [Hint: Always name a success in terms of the immediate question being asked. Sine we are interested in the probability that 5 applicants are women, name a female applicant a success.] Dr. Constance Lightner- Fayetteville State University**Example 5.4 (Continued)**• The probability of a success (i.e. getting a female applicant), p, equals 0.4. [Since traditionally 40% of applicants are women.] • The gender of one applicant does not affect the gender of another. Thus n=20 and p=0.4. So the probability the we get 5 applicants out of the 20 applications received Equals 0.0746 Conclusion: 7.46 % of the time 5 out of 20 applications will be from women. Dr. Constance Lightner- Fayetteville State University**Using the Binomial Table in the Appendix**Dr. Constance Lightner- Fayetteville State University**Mean and Variance of a Binomial Random Variable**If x is a binomial random variable with parameters n and p, Then the mean random variable value is x =np The variance equals x 2 = npq (q=1-p) The standard deviation equals**Example 5.3 Revisited(Expected value, variance, and standard**deviation) The expected number of female applicants is E(x)= np=20*.4= 8 The variance is x2= npq= 20(.4)(.6)= 4.8 The standard deviation is x = Dr. Constance Lightner- Fayetteville State University**Example 5.4 Revisited(Expected value, variance, and standard**deviation) The expected number of employees that will leave is E(x)= np=3*.1= .3 The variance is x2= npq= 3(.1)(.9)=.27 The standard deviation is x = Dr. Constance Lightner- Fayetteville State University**The End**Dr. Constance Lightner- Fayetteville State University