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OMIS 2010 (32)
Chapter 08

# Chapter 08 Textbook Study Guide

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

Description
Chapter 8: Continuous Probability Distributions 8.1 Introduction This chapter continued our discussion of probability distributions. It began by describing continuous probability distributions in general, and then took a detailed look at the normal distribution, which is the most important specific continuous distribution. At the completion of this chapter, you are expected to understand the following: 1. The basic differences between discrete and continuous random variables. 2. How to convert a normal random variable into the standard normal random variable, and how to use the table of standard normal probabilities. 3. How to recognize when it is appropriate to use an exponential distribution, and how to compute exponential probabilities. 8.2 Continuous Probability Distributions This section introduced the notion of a cont inuous random variable, which differs from a discrete random variable both in the type of numerical events of interest and the type of function used to find probabilities. A continuous random variable X can assume any value in some interval, such as 5 < x < 20. Since a continuous random variable can assum e an uncount ably infinite number of val ues, the probability that any particular value will be assumed is zero. Hence, for a continuous random variable X, it is only meaningful to talk about the probability that X will assume a value within a particular interval. Such a probability is found using the probability density function, f(x), associated with X. The probability that X will take a value in the interval a < x < b is given by the area under the graph of f(x) between the values a and b. For most of the specific continuous distributions that you will encounter (such as the normal distribution), you can easily compute probabilities such as this by using probability tables appearing in the appendices. Notice that because P(X = a) = P(X = b) = 0 the following equality holds for any continuous random variable X: P(a < X < b) = P(a ≤ X ≤ b) A specific cont inuous di stribution t hat i s especi ally si mple t o work wi th i s t he uniform distribution. A random variable X defined over an i nterval a ≤ x ≤ b is u niformly distributed if its probability density function is given by 1 f(x) = , a ≤ x ≤ b b − a 93 Because the graph of the density function is a horiz ontal line, the area under the graph is a rectangle, giving rise to an alternative name for the uniform distribution—the rectangular distribution. Notice that the values of a uniform random variable X are distributed evenly, or uniformly, across the domain of X. Example 8.1 A continuous random variable X has the following probability density function: ⎧ .01x + .1 for − 10 ≤ x ≤ 0 f(x) = ⎨⎩−.01x + .1 for 0 ≤ x ≤ 10 a) Graph the density function f(x). b) Verify that f(x) satisfies the requirements of a probability density function. c) Find P(X ≥ 5). Solution a) b) To verify that f(x) is indeed a probability density function, we observe that: 1. f(x) ≥ 0 for all values of x. 2. The total area under the graph of f(x) is 1. To see this, recall that the area of a triangle is bh/2, where b is the base and h is the height of the triangle. Therefore, the area under f(x) to the right of x = 0 is (10)(.1)/2 = .5. Due to symmetry, the area under f(x) to the left of x = 0 is also .5, so the total area under f(x) is .5 + .5 = 1. c) To find P(X ≥ 5), we must find the area under the graph of f(x) to the right of x = 5. This area is the shaded triangle shown on the graph in part a). The base of t his triangle is (10 – 5) = 5, and the height is f(5) = –.01(5) + .1 = .05. Hence, P(X ≥ 5) = (5)(.05)/2 = .125 94 EXERCISES 8.1 Consider a uniform random variable X with the following probability density function: 1 f(x) = 50 , 20 ≤ x ≤ 70 a) Graph the density function f(x). b) Verify that f(x) satisfies the requirements of a probability density function. c) Find P(X ≥ 40). d) Find P(X ≤ 39). 8.2 A continuous random variable X has the following probability density function: f(x) = –.08x + .4, 0 ≤ x ≤ 5 a) Graph the density function f(x). 95 b) Verify that f(x) satisfies the requirements of a probability density function. c) Find P(X ≥ 3). d) Find P(X ≤ 2). e) Find P(X = 2). 8.3 Normal Distribution The normal distribution is the most important specific continuous distribution. Given a problem in- volving a normal distribution, you should begin by clearly defining the relevant normal variable X. You should next sketch a graph of the normal distribution and label it with the given information concerning the mean, standard deviation, and probabilities. Table 3 in Appendix B tab ulates probabilities for the standard normal random variable Z. Hence, before using this table, you must convert values of the nor- mal random variable X into values of Z using the transformation. X −µ x Z = σ x . Example 8.2 Use Table 3 in Appendix B to find the following probabilities, where Z is the standard normal ran- dom variable: a) P(Z ≥ 1.25) b) P(–.82 ≤ Z ≤ 1.36) c) P(.47 ≤ Z ≤ 2.12) 96 Solution a) We begin by sketching a graph of the normal curve and shading the area under the curve to the right of Z = 1.25, which corresponds to the required probability. Recall that the areas tabulated in Table 3 in Appendix B are of t he form P(0 ≤ Z ≤ z )0for selected values of z 0 so required areas must be expressed in term s of areas of th is form before Table 3 can be used. Since the total area under the curve to the right of z = 0 is .5, P (Z ≥ 1.25) = .5 – P(0 ≤ Z ≤ 1.25) = .5 – .3944 = .1056 b) P(–.82 ≤ Z ≤ 1.36) = P(–.82 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 1.36) Because the normal distribution is symmetrical, the area between –.82 and 0 is equal to the area between 0 and .82. Hence, P(–.82 ≤ Z ≤ 0) is equal to the area between 0 and .82. Therefore, P(–.82 ≤ Z ≤ 1.36) = P(0 ≤ Z ≤ .82) + P(0 ≤ Z ≤ 1.36) = .2939 + .4131 = .7070 c) Expressing the required area i n terms of t he types of areas that are tabulated in Table 3, we obtain P(.47 ≤ Z ≤ 2.12) = P(0 ≤ Z ≤ 2.12) – P(0 ≤ Z ≤ .47) = .4830 – .1808 = .3022 97 Example 8.3 If Z is the standard normal variable, find the value0z for which: a) P(–z 0 Z ≤ z 0 = .95 b) P(Z ≥ z ) = .0025 0 Solution a) Since the normal distribution is symmetrical, we know that P(–z0≤ Z ≤ 0) = P(0 ≤ Z ≤ z 0 = .95/2 = .4750 Locating .4750 in the body of Table 3, we find that the corresponding z-value is z0= 1.96. b) Since the area to the right of z is .0025, which is less than .5, z must lie to the right of 0. The 0
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