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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. Outline. EV: Definitions and Theorem EV: Examples - PowerPoint PPT Presentation
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The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
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
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Part 1. EV: Definitions and Theorem
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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
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Variance
Definition 5.4: For a random variable X with probability density function f(x), the variance of X is given by VWhere μ=E(X).
For constants a and b, we have
E(aX+b)=aE(X)+bV(aX+b)=a2V(X)
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Part 2. EV: Examples
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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
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.
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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=0.125a=0.5, P(X>b)=1-b3=0.125b=0.956. That is, for 75% of the weeks, the teller spends between 50% and 95.6% of his time directly serving customers.
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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.
1. Find the expected weekly demand.2. Find the expected weekly profit.
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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.
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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].
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Part 3. Uniform Distribution: Density and Distribution
Functions
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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
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Cumulative Distribution Function
The distribution function for a uniformly distributed X is given by
For (c, c+d) contained within (a, b), we haveP(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.
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Mean and Variance
-which depends only on the length of the interval [a, b].
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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 overThe interval [1, 4]. The probability of major loss is the probability that the time until Delivery exceeds 3 days. So
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Part 3. More Examples
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Additional Example 1
Let X have the density function given by
1. Find the value c.2. Find F(x).3. P(0≤X≤0.5).4. E(X).
Answer:
1. 3. P(0≤X≤0.5)=0.25, 4. E(X)=0.4
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Additional Example 2
Let X have the density function
Find E(lnX).
Answer:
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Homework #8
Page 199-200: 5.4, 5.7Page 209: 5.22Page 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).