Chapter 3 Random Variables and Probability Distributions Definition
Chapter 3: Random Variables and Probability Distributions § Definition and nomenclature § A random variable is a function that associates a real number with each element in the sample space. § We use a capital letter such as X to denote the random variable. § We use the small letter such as x for one of its values. § Example: Consider a random variable Y which takes on all values y for which y > 5. JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 1
Defining Probabilities: Random Variables § Examples: § Out of 100 heart catheterization procedures performed at a local hospital each year, the probability that more than five of them will result in complications is P(X > 5) § Drywall anchors are sold in packs of 50 at the local hardware store. The probability that no more than 3 will be defective is P(Y < 3) JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 2
Discrete Random Variables § Pr. 2. 51 P. 59 (Modified) A box contains 500 envelopes (75 have $100, 150 have $25, 275 have $10) § Assume someone spends $75 to buy 3 envelopes. The sample space describing the presence of $10 bills (H) vs. bills that are not $10 (N) is: § S = {NNN, NNH, NHN, HNN, NHH, HNH, HHN, HHH} § The random variable associated with this situation, X, reflects the outcome of the experiment § X is the number of envelopes that contain $10 § X = {0, 1, 2, 3} § Why no more than 3? Why 0? JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 3
Discrete Probability Distributions 1 § The probability that the envelope contains a $10 bill is 275/500 or. 55 § What is the probability that there are no $10 bills in the group? P(X = 0) =(1 -0. 55) *(1 -0. 55) = 0. 091125 P(X = 1) = 3 * (0. 55)*(1 -0. 55)* (1 -0. 55) = 0. 334125 § Why 3 for the X = 1 case? § Three items in the sample space for X = 1 § NNH NHN HNN JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 4
Discrete Probability Distributions 2 P(X = 0) =(1 -0. 55) *(1 -0. 55) = 0. 091125 P(X = 1) = 3*(0. 55)*(1 -0. 55)* (1 -0. 55) = 0. 334125 P(X = 2) = 3*(0. 55^2*(1 -0. 55)) = 0. 408375 P(X = 3) = 0. 55^3 = 0. 166375 § The probability distribution associated with the number of $10 bills is given by: x 0 1 2 3 P(X = x) JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 5
Another View § The probability histogram JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 6
Another Discrete Probability Example § Given: § A shipment consists of 8 computers § 3 of the 8 are defective § Experiment: Randomly select 2 computers § Definition: random variable X = # of defective computers selected § What is the probability distribution for X? § Possible values for X: X = 0 X =1 X = 2 § Let’s start with P(X=0) [0 defectives and 2 nondefectives are selected] Recall that P = specified target / all possible (all ways to get 0 out of 3 defectives) ∩ (all ways to get 2 out of 5 nondefectives) (all ways to choose 2 out of 8 computers) JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 7
Discrete Probability Example § What is the probability distribution for X? § Possible values for X: X = 0 X =1 X = 2 § Let’s calculate P(X=1) [1 defective and 1 nondefective are selected] (all ways to get 1 out of 3 defectives) ∩ (all ways to get 1 out of 5 nondefectives) (all ways to choose 2 out of 8 computers) x P(X = x) JMB Chapter 3 Lecture 1 9 th ed 0 1 EGR 252 2012 2 Slide 8
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 JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 9
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) JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 10
Your Turn … § The output of the same type of circuit board from two assembly lines is mixed into one storage tray. In a tray of 10 circuit boards, 6 are from line A and 4 from line B. If the inspector chooses 2 boards from the tray, show the probability distribution function for the number of selected boards coming from line A. x 0 P(x) 1 2 JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 11
Continuous Probability Distributions In general, § 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 JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 12
Visualizing Continuous Distributions § 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 JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 13
Continuous Probability Calculations § The continuous probability density function (pdf) f(x) ≥ 0, for all x ∈ R § The cumulative distribution, F(x) JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 14
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? JMB Chapter 3 Lecture 1 9 th ed EGR 252 2012 Slide 15
- Slides: 15