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Chapter 5. Continuous Probability Distributions Sections 5.2, 5.3: Expected Value of Continuous Random Variables and Uniform Distribution Jiaping Wang Department of Mathematical Science 03/20/2013, Monday The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Outline EV: Definitions and Theorem EV: Examples Uniform Distribution: Density and Distribution Functions Uniform Distribution: Mean and Variance More Examples Homework #8 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Part 1. EV: Definitions and Theorem The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Definition and Theorem Definition 5.3: The expected value of a continuous random variable X that has density function f(x) is given by β πΈ π = οΏ½ π₯π π₯ ππ . ββ Note: we assume the absolute convergence of all integrals so that the expectations exist. Theorem 5.1: If X is a continuous random variable with probability density f(x), and if g(X) is any real-valued function of β X, then πΈ π π = β«ββ π π₯ π π₯ ππ The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Variance Definition 5.4: For a random variable X with probability density function f(x), the variance of X is given by β V π = πΈ π β π 2 = β«ββ π₯ β π 2π π₯ ππ = πΈ π2 β π2 . Where ΞΌ=E(X). For constants a and b, we have E(aX+b)=aE(X)+b V(aX+b)=a2V(X) The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Part 2. EV: Examples The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Example 5.4 For a given teller in a bank, let X denote the proportion of time, out of a 40-hour workweek, that he is directly serving customers. Suppose that X has a probability density function given by 3π₯2, 0 β€ π₯ β€ 1 f x =οΏ½ 0, πππππππππ 1. Find the mean proportion of time during a 40-hour workweek the teller directly serve customers. 2. Find the variance of the proportion of time during a 40-hour workweek the teller directly serves customers. 3. Find an interval that, for 75% of the weeks, contains the proportion of time that the teller spends directly serving customers. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Answer: 1. Based on the definition, β 1 πΈ π = οΏ½ π₯π₯ π₯ ππ = οΏ½ π₯ 3π₯2 ππ = 3/4. ββ 0 Thus, on average, the teller spends 75% of his time each week directly serving customers. 2. We need to compute the E(X2): β 1 πΈ π = οΏ½ π₯2π π₯ ππ = οΏ½ π₯2 3π₯2 ππ = 0.60. ββ 0 V(X)=E(X2)-E2(X)=0.60-(0.75)2=0.0375. Then, 3. There are lots of ways to construct the interval such that the proportion of time that the teller spends directly serving customers for 75% of the weeks, for example, P(X<a)=0.12, P(X>b)=0.13, or P(X<a)=0.10, P(X>b)=0.15, for the other 25% of the weeks. We choose the half of 25% for the two sided tails, ie., P(X<a)=0.125 and P(X>b)=0.125 for some a and b. So we have P(X<a)=a3=0.125ο a=0.5, P(X>b)=1-b3=0.125ο b=0.956. That is, for 75% of the weeks, the teller spends between 50% and 95.6% of his time directly serving customers. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Example 5.5 The weekly demand X, in hundreds of gallons, for propane at a certain supply station has a density function given by π₯ ,0 β€ π₯ β€ 2 4 π π₯ = 1 ,2 < π₯ β€ 3 2 0, πππππππππ It takes $50 per week to maintain the supply station. Propane is purchased for $270 per hundred gallons and redistributed by the supply station for $1.75 per gallon. 1. Find the expected weekly demand. 2. Find the expected weekly profit. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Answer: 1. Based on the definition, β 2 πΈ π = οΏ½ π₯π₯ π₯ ππ = οΏ½ π₯ ββ 0 3 4 1 ππ + οΏ½ π₯ ππ = 1.92. π₯ 2 2 Thus, on average, the weekly demand for propane will be 192 gallons at this supply station. 2. The propane is purchased for $270 per hundred gallons and sold for $175 per hundred gallons, yielding a profit of $95 per hundred gallons sold. The weekly profit P is given as P=95X-50, so E(P)=95E(X)-50=95(1.92)-50=132.40. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Tchebysheffβs Theorem and Example 5.6 The Tchebysheffβs theorem holds for the continuous random variable, X, ie., P(|X-ΞΌ|<kΟ) β₯ 1-1/k2 Example 5.6: The weekly amount X spent for chemicals by a certain firm has a mean of $1565 and a variance of $428. Within what interval should these weekly costs for chemicals be expected to lie in at least 75% of the time? Answer: To find the interval guaranteed to contain at least 75% of the probability mass for X, we need to have 1-1/k2=0.75 ο k=2. So the interval is given by [1565-2(428)1/2, 1565+2(428)1/2]. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Part 3. Uniform Distribution: Density and Distribution Functions The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Density Function Consider a simple model for the continuous random variable X, which is equally likely to lie in an interval, say [a, b], this leads to the uniform probability distribution, the density function is given as π π π = οΏ½π β π,π β€ π β€ π π, πππππππππ The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Cumulative Distribution Function The distribution function for a uniformly distributed X is given by 0, π₯ < 1 π₯βπ , πβ€π₯β€π πΉ π₯ =οΏ½ πβπ 1, π₯ > π For (c, c+d) contained within (a, b), we have P(cβ€Xβ€c+d)=P(Xβ€c+d)-P(Xβ€c)=F(c+d)-F(c)=d/(b-a), which this probability only depends on the length d. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Mean and Variance β π π₯ π+π πΈ π = οΏ½ π₯π₯ π₯ ππ = οΏ½ ππ = π β π 2 ββ π πΈ π2 β π 2 2 + ππ + π2 π₯ π = οΏ½ π₯2π π₯ ππ = οΏ½ ππ = 3 ββ π πβπ π π = πΈ ππ β πΈπΈ π = π2+ππ+π2 π+π 3 2 the length of the interval [a, b]. 2= 1 12 π β π 2 which depends only on The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Example 5.7 A farmer living in western Nebraska has an irrigation system to provide water for crops, primarily corn, on a large farm. Although he has thought about buying a backup pump, he has not done so. If the pump fails, delivery time X for a new pump to arrive is uniformly distributed over the interval from 1 to 4 days. The pump fails. It is a critical time in the growing season in that the yield will be greatly reduced if the crop is not watered within the next 3 days. Assuming that the pump is ordered immediately and the installation time is negligible, what is the probability that the farmer will suffer major yield loss? Answer: Let T be the time until the pump is delivered. T is uniformly distributed over The interval [1, 4]. The probability of major loss is the probability that the time until Delivery exceeds 3 days. So 4 1 1 π π > 3 = οΏ½ ππ = . 3 3 3 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Part 3. More Examples The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Additional Example 1 Let X have the density function given by 1. 2. 3. 4. Find the value c. Find F(x). P(0β€Xβ€0.5). E(X). Answer: β 0 1 π 1. β«ββ π π₯ ππ = 1 β β«β1 0.2ππ + β«0 0.2 + ππ ππ = 1 β 0.2 + 0.2 + 2 = 1 β π = 1.2 0, π₯ β€ β1 0.2π₯ + 0.2, β1 β€ π₯ β€ 0 2. πΉ π₯ = π π β€ π₯ = οΏ½ 3. P(0β€Xβ€0.5)=0.25, 0.2 + 0.2π₯ + 0.6π₯2, 0 < π₯ β€ 1 1, π₯ > 1 4. E(X)=0.4 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Additional Example 2 Let X have the density function Find E(lnX). β π ln π₯ π₯ Answer: πΈ πππ = β«ββ ln π₯ π π₯ ππ = β«1 1 2 ππ = . The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Homework #8 Page 199-200: 5.4, 5.7 Page 209: 5.22 Page 214-215: 5.28, 5.40. Additional Hw1: Let X have the density function Find the E(X). Additional Hw2: The density function of X is given by (a). Find a and b. (b). Determine the cumulative distribution function F(x). The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL