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Scott N Gillespie Engineering 323 Beautiful Homework #8 Chapter 4 Problem #1 Fall 2000 Page 1 Let X denote the amount of time for which a book on 2-hour reserve at a college library is checked out by a randomly selected student and suppose that X has a density function as follows. .5x 0≤x≤2 0 Otherwise f(x) = Calculate the following probabilities: A) P(X ≤ 1) B) P(.5 ≤ X ≤ 1.5) C) P(1.5 < X) CONTINUOUS RANDOM VARIABLES A random variable is said to be continuous if its set of possible values is an entire interval of numbers. That is, for some set A to B, any number x between A and B is possible. Definition: Let X be a continuous random variable. Then X has a Probability Density Function f(x) for any two number A and B (where A is less that B). This is denoted in the function below (Devore, p 140). b P(a ≤ X ≤ b) = ∫ f(x) dx a This is the probability that the random variable X takes on a value within the interval from a to b. Figure 1 is a representation of this probability. Figure 1 Graphical representation of P(a < X <b) (Devore, p140). Scott N Gillespie Engineering 323 Beautiful Homework #8 Chapter 4 Problem #1 Fall 2000 Page 2 For a continuous random variable the probability that X is equal to a certain single value is always 0. These functions have to be evaluated over an interval to determine a positive probability. As shown in Figure1, there has to be an “area” underneath the graph. A thin vertical line would represent a single value. For example, let’s let a and b both be equal to five. This statement is valid for all X. b ∫ f(x) dx = F(b) – F(a) = F(5) – F(5) = 0 a The graph of the PDF is shown in Figure 2 1.2 1 f(x) 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 X = Hours that a book is checked out 2.5 Figure 2 Probability Density Function for problem 1. Expected Value [E(X)] ∝ ∝ ∝ E(X) = ∫ x f(x) dx = ∫ x .5x dx = ∫ .5 x 2 -∝ -∝ 2 2 dx = (.5/3)x = 4/3 Hours 0 0 Variance [V(X)] V(X) = E(X2 ) - [E(X)]2 ∝ ∝ 2 E(X2 ) = ∫ x2 f(x) dx = ∫ x2 .5x dx = ∫ .5x3 dx = (.5/4) x4 -∝ -∝ 0 V(X) = 2 – (4/3)2 = 2/9 (Hours)2 2 = 2 Hours 0 Scott N Gillespie Engineering 323 Beautiful Homework #8 Chapter 4 Problem #1 Fall 2000 Page 3 Cumulative Distribution Function X X F(x) = P (X ≤ x) = ∫ f(y) dy = ∫ .5y dy = .25y2 -∞ 0 X 0 Mathematical Representation: F(x) = 0 .25 x2 1 x<0 0≤x≤2 x>2 This probability is shown in Figure 3 1.2 1 F(x) 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 x = hours that a book is checked out Figure 3 Cumulative Distribution Function for problem 1. A ) P(X ≤ 1) What is the probability that the continuous random variable X is less than or equal to 1? Let X equal the amount of time that a book on 2 hour reserve is checked out. B 1 P(a ≤ X ≤ b) = ∫ f(x) dx P(0 ≤ X ≤ 1) = ∫ .5x dx A 0 P(0 ≤ X ≤ 1) = .25x 1 2 = .25 Hours 0 Scott N Gillespie Engineering 323 Beautiful Homework #8 Chapter 4 Problem #1 Fall 2000 Page 4 This probability is show in Figure 2 1.2 1 f(x) 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 X = Hours that a book is checked out 2.5 Figure 2 Probability that (0<x<.5). B ) P(.5 ≤ X ≤ 1.5) What is the probability that the continuous random variable X is greater than or equal to .5 and less than or equal to 1.5? Let X equal the amount of time that a book on 2 hour reserve is checked out. B P(a ≤ X ≤ b) = ∫ f(x) dx 1.5 P(.5 ≤ X ≤ 1.5) = ∫ .5x dx A .5 P(.5 ≤ X ≤ 1.5) = .25x2 1 = .5 Hours 0 This probability is show in Figure 3 1.2 1 f(x) 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 X = Hours that a book is checked out Figure 3 Probability that (.5<x<1.5). 2.5 Scott N Gillespie Engineering 323 Beautiful Homework #8 Chapter 4 Problem #1 Fall 2000 Page 5 C ) P(X > 1.5) What is the probability that the continuous random variable X is greater that 1.5? Let X equal the amount of time that a book on 2 hour reserve is checked out. B 2 P(a ≤ X ≤ b) = ∫ f(x) dx P(1.5 ≤ X ≤ 2) = ∫ .5x dx A 1.5 P(1.5 ≤ X ≤ 2) = 1 .25x2 = .4375 Hours 0 This probability is shown in Figure 4 1.2 1 f(x) 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 X = Hours that a book is checked out Figure 4 Probability that (1.5<x<2). 2.5