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Chapter 3: Random Variables and Probability Distributions

Learn about random variables, discrete probability distributions, and examples to understand how to calculate probabilities in various scenarios. Explore functions and nomenclature to enhance your knowledge in probability theory.

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Chapter 3: Random Variables and Probability Distributions

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  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 an uppercasel letter such as X to denote the random variable. • We use a lowercase 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 JMB 2019

  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 JMB 2019

  3. Discrete Random Variables • Exercise 2.51 P.59 (Modified) A box contains 500 envelopes (75 have $100, 150 have $25, 275 have $10). • Someone spends $75 to buy 3 envelopes. The individual opens each envelope to see if the envelope contains a $10 bill. 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? EGR 252 JMB 2019

  4. Discrete Probability Distributions 1 • The probability that an envelope contains a $10 bill is 275/500 or .55 • What is the probability that there are no $10 bills in the group of three envelopes? P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) = 0.091125 • What is the probability that only one envelope contains a $10 bill? 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 JMB 2019

  5. Discrete Probability Distributions 2 • Summary of probabilities for all possible outcomes 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 distributionassociated with the number of $10 bills is given by: EGR 252 JMB 2019

  6. Another View • The probability histogram EGR 252 JMB 2019

  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) EGR 252 JMB 2019

  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) EGR 252 JMB 2019

  9. 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 JMB 2019

  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) EGR 252 JMB 2019

  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. EGR 252 JMB 2019

  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, EGR 252 JMB 2019

  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 EGR 252 JMB 2019

  14. Continuous Probability Calculations • The continuous probability density function (pdf) f(x) ≥ 0, for all x ∈R • The cumulative distribution,F(x) EGR 252 JMB 2019

  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 • 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 JMB 2019

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