Jiaping Wang Department of Mathematical Science 03/20/2013, Monday

1. Chapter 5. Continuous Probability DistributionsSections 5.2, 5.3: Expected Value of Continuous Random Variables and Uniform Distribution Jiaping Wang Department of Mathematical Science 03/20/2013, Monday

2. Outline EV: Definitions and Theorem EV: Examples Uniform Distribution: Density and Distribution Functions Uniform Distribution: Mean and Variance More Examples Homework #8

3. Part1. EV: Definitions and Theorem

4. Definition and Theorem Definition 5.3: The expected value of a continuous random variable X that has density function f(x) is given by 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 X, then

5. Variance Definition 5.4: For a random variable X with probability density function f(x), the variance of X is given by V Where μ=E(X). For constants a and b, we have E(aX+b)=aE(X)+b V(aX+b)=a2V(X)

6. Part 2. EV: Examples

7. 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 Find the mean proportion of time during a 40-hour workweek the teller directly serve customers. Find the variance of the proportion of time during a 40-hour workweek the teller directly serves customers. Find an interval that, for 75% of the weeks, contains the proportion of time that the teller spends directly serving customers.

8. Answer: 1. Based on the definition, Thus, on average, the teller spends 75% of his time each week directly serving customers. 2. We need to compute the E(X2): Then, V(X)=E(X2)-E2(X)=0.60-(0.75)2=0.0375. 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=0.125a=0.5, P(X>b)=1-b3=0.125b=0.956. That is, for 75% of the weeks, the teller spends between 50% and 95.6% of his time directly serving customers.

9. Example 5.5 The weekly demand X, in hundreds of gallons, for propane at a certain supply station has a density function given by 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. Find the expected weekly demand. Find the expected weekly profit.

10. Answer: 1. Based on the definition, 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.

11. Tchebysheff’s Theorem and Example 5.6 The Tchebysheff’s theorem holds for the continuous random variable, X, ie., P(|X-μ|<kσ) ≥ 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  k=2. So the interval is given by [1565-2(428)1/2, 1565+2(428)1/2].

12. Part 3. Uniform Distribution: Density and Distribution Functions

13. 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

14. Cumulative Distribution Function The distribution function for a uniformly distributed X is given by For (c, c+d) contained within (a, b), we have P(c≤X≤c+d)=P(X≤c+d)-P(X≤c)=F(c+d)-F(c)=d/(b-a), which this probability only depends on the length d.

15. Mean and Variance -which depends only on the length of the interval [a, b].

16. 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

17. Part 3. More Examples

18. Additional Example 1 Let X have the density function given by Find the value c. Find F(x). P(0≤X≤0.5). E(X). Answer: 3. P(0≤X≤0.5)=0.25, 4. E(X)=0.4

19. Additional Example 2 Let X have the density function Find E(lnX). Answer:

20. 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).