Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Discrete Probability Distributions The discrete probability distribution function (pdf) f(x) = P(X = x) ≥ 0 Σx f(x) = 1 The cumulative distribution, F(x) F(x) = P(X ≤ x) = Σt ≤ x f(t) Note the importance of case: F not same as f MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 1 Probability Distributions From our example, the probability that no more than 2 of the envelopes contain $10 bills is P(X ≤ 2) = F (2) = _________________ F(2) = f(0) + f(1) + f(2) = .833625 Another way to calculate F(2) (1 - f(3)) The probability that no fewer than 2 envelopes contain $10 bills is P(X ≥ 2) = 1 - P(X ≤ 1) = 1 – F (1) = ________ 1 – F(1) = 1 – (f(0) + f(1)) = 1 - .425 = .575 Another way to calculate P(X ≥ 2) is f(2) + f(3) MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 2 Continuous Probability Distributions b In general, P (a X b) f ( x )dx a The probability that the average daily temperature in Georgia during the month of August falls between 90 and 95 degrees is The probability that a given part will fail before 1000 hours of use is MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 3 Visualizing Continuous Distributions The probability that the average daily temperature in Georgia during the month of August falls between 90 and 95 degrees is -5 -3 -1 1 3 5 The probability that a given part will fail before 1000 hours of use is 0 MDH Chapter 3-4 Lecture 1 EGR 252 2015 5 10 15 20 25 30 Slide 4 Continuous Probability Calculations The continuous probability density function (pdf) f(x) ≥ 0, for all x ∈ R f ( x )dx 1 b P (a X b) f ( x )dx a The cumulative distribution, F(x) x F ( x ) P( X x ) f (t )dt MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 5 Example: Problem 3.7, pg. 92 The total number of hours, measured in units of 100 hours x, 0<x<1 f(x) = 2-x, 1≤x<2 0, elsewhere { a) P(X < 120 hours) = P(X < 1.2) = P(X < 1) + P (1 < X < 1.2) NOTE: You will need to integrate two different functions over two different ranges. b) P(50 hours < X < 100 hours) = Which function(s) will be used? MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 6 4.1 Mathematical Expectation Example: Repair costs for a particular machine are represented by the following probability distribution: x $50 $200 $350 P(X = x) 0.3 0.2 0.5 What is the expected value of the repairs? That is, over time what do we expect repairs to cost on average? MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 7 Expected Value – Repair Costs μ = E(X) μ = mean of the probability distribution For discrete variables, μ = E(X) = ∑ x f(x) So, for our example, E(X) = 50(0.3) + 200(0.2) + 350(0.5) = $230 MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 8 Another Example – Investment By investing in a particular stock, a person can take a profit in a given year of $4000 with a probability of 0.3 or take a loss of $1000 with a probability of 0.7. What is the investor’s expected gain on the stock? X $4000 P(X) 0.3 -$1000 0.7 E(X) = $4000 (0.3) -$1000(0.7) = $500 MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 9 Expected Value - Continuous Variables For continuous variables, μ = E(X) = E(X) = ∫ x f(x) dx Vacuum cleaner example: problem 7 pg. 92 f(x) = { x, 2-x, 0, 0<x<1 1≤x<2 elsewhere (in hundreds of hours.) 1 E(X) x dx 2 0 2 1 x3 1 2 x3 x 2 x dx x | 3 0 3 | 2 1 = 1 * 100 = 100.0 hours of operation annually, on average MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 10 Functions of Random Variables Ex 4.4. pg. 111: Probability of X, the number of cars passing through a car wash in one hour on a sunny Friday afternoon, is given by x P(X = x) 4 5 1/12 1/12 6 7 8 9 1/4 1/4 1/6 1/6 Let g(X) = 2X -1 represent the amount of money paid to the attendant by the manager. What can the attendant expect to earn during this hour on any given sunny Friday afternoon? E[g(X)] = Σ g(x) f(x) = Σ (2X-1) f(x) = (2*4-1)(1/12) +(2*5-1)(1/12) …+(2*9-1)(1/6) = $12.67 MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 11 4.2 Variance of a Random Variable Recall our example: Repair costs for a particular machine are represented by the following probability distribution: x $50 200 350 P(X = x) 0.3 0.2 0.5 What is the variance of the repair cost? – That is, how might we quantify the spread of costs? MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 12 Variance – Discrete Variables For discrete variables, σ2 = E [(X - μ)2] = ∑ (x - μ)2 f(x) = E (X2) - μ2 Recall, for our example, μ = E(X) = $230 Preferred method of calculation: σ2 = [E(X2)] – μ2 = 502 (0.3) + 2002 (0.2) + 3502 (0.5) – 2302 = $17,100 Alternate method of calculation: σ2 = E(X- μ)2 f(x) = (50-230)2 (0.3) + (200-230)2 (0.2) + (350-230)2 (0.5) = $17,100 MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 13 Variance - Investment Example By investing in a particular stock, a person can take a profit in a given year of $4000 with a probability of 0.3 or take a loss of $1000 with a probability of 0.7. What are the variance and standard deviation of the investor’s gain on the stock? E(X) = $4000 (0.3) -$1000 (0.7) = $500 σ2 = [∑(x2 f(x))] – μ2 = (4000)2(0.3) + (-1000)2(0.7) – 5002 = $5,250,000 σ = $2291.29 MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 14 Variance of Continuous Variables For continuous variables, σ2 = E [(X - μ)2] =[∫ x2 f(x) dx] – μ2 Recall our vacuum cleaner example pr. 7 pg. 88 { f(x) = x, 2-x, 0, 0<x<1 1≤x<2 elsewhere (in hundreds of hours of operation.) What is the variance of X? The variable is continuous, therefore we will need to evaluate the integral. MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 15 Variance Calculations for Continuous Variables b 2 2 (X) x f ( x ) dx a (Preferred calculation) 2 1 2 3 2 2 (X) x dx x 2 x dx 0 1 2 4 x 2 (X) 4 4 2 x3 x |0 3 4 1 2 | 1 12 0.1667 What is the standard deviation? σ = 0.4082 hours MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 16 Covariance/ Correlation A measure of the nature of the association between two variables Describes a potential linear relationship Positive relationship Large values of X result in large values of Y Negative relationship Large values of X result in small values of Y “Manual” calculations are based on the joint probability distributions Statistical software is often used to calculate the sample correlation coefficient (r) MDH Chapter 3-4 Lecture 1 EGR 252 2015 Slide 17