1. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 1 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.

2. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 2 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)

3. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 3 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?

4. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 4 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) *(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

5. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 5 Discrete Probability Distributions 2 P(X = 0) =(1-0.55) * (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: x0123 P(X = x)

6. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 6 Another View  The probability histogram

7. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 7 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) (all ways to choose 2 out of 8 computers)

8. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 8 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) (all ways to choose 2 out of 8 computers) x012 P(X = x)

9. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 9 Discrete Probability Distributions  The discrete probability distribution function (pdf)  f(x) = P(X = x) ≥ 0  Σ x f(x) = 1  The cumulative distribution, F(x)  F(x) = P(X ≤ x) = Σ t ≤ x f(t)  Note the importance of case: F not same as f

10. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 10 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)

11. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 11 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. xP(x)P(x) 012012

12. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 12 Continuous Probability 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 In general,

13. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 13 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

14. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 14 Continuous Probability Calculations  The continuous probability density function (pdf) f(x) ≥ 0, for all x ∈ R  The cumulative distribution, F(x)

15. JMB Chapter 3 Lecture 1 9th edEGR 252 2013 Slide 15 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? {