Chapter 07 Textbook Study Guide

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Department
Operations Management and Information System
Course
OMIS 2010
Professor
Alan Marshall
Semester
Fall

Description
Chapter 7: Random Variables and Discrete Probability Distributions 7.1 Random Variables and Probability Distributions This section introduced the concept of a random variable, which assigns a num erical value to each simple event in a sample space, thereby enabling us to work with numerical-valued outcomes. A random variable is discrete if the number of possi ble values that it can assume is finite or countably infinite; otherwise, the random variable is continuous. The second i mportant concept introduced was t hat of a discrete probability distribution, whic h lists all possible values that a discrete random variable can take on, together with the probability that each value will be assumed. When constructing a discrete probabil- ity distribution, bear in mind that the probabilities must sum to 1. Example 7.1 The local taxation office claims that 10% of the income tax forms processed this year contain errors. Suppose that two of these form s are selected at random. Find the probability distribution of the random variable X, defined as the number of forms selected that contain an error. Solution A sample space for this random experiment is S = F F ,F F ,F F ,F F } 1 2 1 2 1 2 1 2 where, for example, F1F2 indicates that the first form selected contains an error and t he second form does not. The random variable X assigns a value of 0, 1, or 2 to each of these simple events. To find the probability that each of these values will be assumed, we must determine the simple events that result in each of these values being assumed and the probabilities that the simple events will occur. For example, X = 2 if and only if the event1F 2 occurs. Since the occurrence of an error on one form is independent of the occurrence of an error on the other form, P(X = 2) = p(2) = P(FF ) = P(F )P(F ) = (.1)(.1) = .01 1 2 1 2 Similarly, P(X =1) = p(1) = P(FF ) 1 2(FF ) 1 2 = P(F 1P(F )2+ P(F )P1F ) 2 = (.1)(.9) + (.9)(.1) = .18 P ( = 0) = p (0) = ( FF1 2) = ( ) 1 F 2) = (.9)(.9) = .81 Summarizing the probability distribution of X in tabular form, we obtain 72 x p(x) 0 .81 1 .18 1 .01 Note that we could have used a p robability tree to identify the simple events and their associated probabilities. EXERCISES 7.1 For each of the following random variables, indicate whether the variable is discrete or continu- ous and specify the possible values that it can assume. a) X = the number of customers served by a restaurant on a given day b) X = the time in minutes required to complete a particular assembly c) X = the number of accidents at a particular intersection in a given week d) X = the number of questions answered correctly by a randomly selected student who wrote a quiz consisting of 15 multiple choice questions e) X = the weight in ounces of a newborn baby chosen at random 7.2 The following table gives the probability distribution of X, the number of telephones in a ran- domly selected home in a certain community. x p(x) 0 .021 1 .412 2 .283 3 .188 73 4 .096 If one home is selected at random, find the probability that it will have: a) no telephone; b) fewer than two telephones; c) at least three telephones; d) between one and three telephones (inclusive). 7.3 The security force on a particular campus estimates that 80% of the cars that are parked illegally on campus are detected and issu ed a $10 parking ticket. Suppose your car will be parked ille- gally on two occasions during the com ing year. Determine the probability distribution of the random variable X = the total dollar amount of parking tickets that you will receive. 7.2 Expected Value and Variance This section dealt with the two main descriptive measures of a discrete random variable: its expected value and its variance. As well as being able to calculate the numerical values of these measures, you are expected to have a general understanding of their meanings and to be familiar with the laws of expected value and of variance. 74 The expected value (or mean value) of a di screte random variable X is a weighted average of the values that X can assume: n E(X)= = x i(x i i=1 The variance of a di screte random variable X is a wei ghted average of t he squared devi ations of t he values of X from their mean : n V(X) = = 2 (xi p(x ) i i=1 When computing the variance of X by hand, it is usually easier to use the shortcut formula: 2 2 2 = E(X ) A related measure of the dispersion of the values of X about their mean value is the standard deviation of X, defined to be the positive square root of the variance of X. The standard deviation of X, denoted has the advantage of being expressed in the same units as both the values of X and their mean value . Example 7.2 Let X be a random variable with the following probability distribution: x p(x) 2 .1 4 .3 6 .4 8 .2 a) Find the mean and standard deviation of X. b) Find the mean and standard deviation of Y = 2X + 4. Solution a) From the definitions of E(X) and V(X): 75
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