1 / 20

Defining Probabilities: Random Variables

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 __________

duyen
Download Presentation

Defining Probabilities: Random Variables

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. 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 __________ • Drywall anchors are sold in packs of 50 at the local hardware store. The probability that no more than 3 will be defective is __________ • In general, ___________

  2. Discrete Random Variables • Example: • Look back at problem 3, page 46. 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: _____________________________ • The random variable associated with this situation, X, reflects the outcome of the choice and can take on the values: _____________________________

  3. Discrete Probability Distributions • The probability that there are no $10 in the group is P(X = 0) = ___________________ (recall results from last time) • The probability distribution associated with the number of $10 bills is given by:

  4. Another Example • Example 3.3, pg 66 P(X = 0) = _____________________

  5. 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)

  6. Probability Distributions • From our example, the probability that no more than 2 of the envelopes contain $10 bills is P(X ≤ 2) = F(2) = _________________ • The probability that no fewer than 2 envelopes contain $10 bills is P(X ≥ 2) = 1 - P(X ≤ 1) = 1 - F(1) = ________________

  7. Another View • The probability histogram

  8. 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.

  9. Continuous Probability Distributions • Examples: • 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, __________

  10. Understanding 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

  11. Continuous Probability Distributions • The continuous probability density function (pdf) f(x) ≥ 0, for all x ∈R • The cumulative distribution,F(x)

  12. Probability Distributions • Example: Problem 7, pg. 73 x, 0 < x < 1 f(x) = 2-x, 1 ≤ x < 2 0, elsewhere 1st – what does the function look like? • P(X < 120) = ___________________ • P(50 < X < 100) = ___________________ {

  13. Your turn • Problem 14, pg. 73

  14. Additional useful information* … Joint probability distributions • Example 1 (discrete): the joint probability mass function (see defn. 3.8, pg. 75) In the development of a new receiver for a digital communication system, each received bit is rated as acceptable, suspect, or unacceptable, depending on the quality of the received signal, with the following probabilities: P(acceptable) = P(x) = 0.9 P(suspect) = P(y) = 0.08 P(unacceptable) = P(z) = 0.02 * Section 3.4 in your book (Optional)

  15. If we let X denote the number of acceptable bits and Y denote the number of suspect bits, then the joint probability associated with the number of acceptable and suspect bits in 4 transmitted bits is denoted by: fXY(x, y), where fXY(x, y)≥ 0 ∑x ∑yfXY(x, y) = 1 fXY(x, y) = P(X = x, Y = y) So, the probability of exactly two acceptable bits and exactly 1 suspect bit in the first 4 bits is fXY(2, 1) = P(X = 2, Y = 1) Assuming independence, we can determine the probability of a particular combination, say aasu, as: P(aasu) = P(a)*P(a)*P(s)*P(u) = 0.9*0.9*0.08*0.02 = 0.0013

  16. Recognizing that aasu is just one of several possible combinations of 4 bits, we next need to determine the number of possible permutations of 2 acceptable and 1 suspect bit in 4 tested bits. That is (from theorem 2.6, pg. 36), So, fXY(2, 1) = P(X = 2, Y = 1) = 12(0.0013) = 0.0156

  17. Joint probability distributions • Example 2 (continuous): the joint density function (see definition 3.9, pg. 76) If X denotes the time until a computer server connects to your machine (in milliseconds) and Y denotes the time until the server authorizes you as a valid user (in milliseconds.) Each measures the wait from a common starting time and X < Y. Assume that the joint probability density function for X and Y is given as: fXY(x, y) = 6 x 10-6exp(-0.001x – 0.002y) for x < y

  18. The probability that X < 1000 and Y < 2000 is:

  19. [Note: we can verify that this density function integrates to 1 as follows: ]

  20. For further study … • Read section 3.4 • Solve selected problems on pp. 84-86

More Related