1 / 14

Chapter 3: Random Variables and Probability Distributions

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.

nailah
Download Presentation

Chapter 3: Random Variables and Probability Distributions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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. EGR 252.001 Spring 2010

  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) EGR 252.001 Spring 2010

  3. Discrete Random Variables • Problem 2.53 Page 55 Modified • 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} EGR 252.001 Spring 2010

  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 EGR 252.001 Spring 2010

  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: EGR 252.001 Spring 2010

  6. Example 3.8, pg 80 • Shipment: 8 computers of which 3 are defective • Random purchase of 2 computers • What is the probability distribution for the random variable X = defective computers purchased? Possibilities: X = 0 X =1 X = 2 Let’s start with P(X=0) P = specified target / all possible (0 defectives and 2 nondefectives are selected) (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) EGR 252.001 Spring 2010

  7. Discrete Probability Distributions • The discrete probability distribution function (pdf) • f(x) = P(X = x) ≥ 0 • Σxf(x) = 1 • The cumulative distribution,F(x) • F(x) = P(X ≤ x) = Σt ≤ xf(t) • Note the importance of case: F not same as f EGR 252.001 Spring 2010

  8. 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 •  (OR 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 •  (OR f(2) + f(3)) EGR 252.001 Spring 2010

  9. Another View • The probability histogram EGR 252.001 Spring 2010

  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 associated with the selected boards being from line A. EGR 252.001 Spring 2010

  11. 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, EGR 252.001 Spring 2010

  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 EGR 252.001 Spring 2010

  13. Continuous Probability Calculations • The continuous probability density function (pdf) f(x) ≥ 0, for all x ∈R • The cumulative distribution,F(x) EGR 252.001 Spring 2010

  14. Example: Problem 3.7, pg. 88 The total number of hours, measured in units of 100 hours x, 0 < x < 1 f(x) = 2-x, 1 ≤ x < 2 0, elsewhere • 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? { EGR 252.001 Spring 2010

More Related