1 / 32

Chapter 5

Chapter 5. Discrete Probability Distributions. Random Variable. A numerical description of the result of an experiment. Discrete Random Variable. A random variable that can only assume some finite number of values. Continuous Random Variable.

stacey-morris
Download Presentation

Chapter 5

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 5 Discrete Probability Distributions

  2. Random Variable A numerical description of the result of an experiment.

  3. Discrete Random Variable A random variable that can only assume some finite number of values.

  4. Continuous Random Variable A random variable that can assume any numerical value in an interval.

  5. Discrete Probability Distribution A list of the possible values of a discrete random variable and the associated probabilities.

  6. Discrete Probability Function, f(x) A function that provides the probability for each value of the discrete random variable

  7. Required Conditions for a Discrete Probability Function f(x) > 0 for all x Sf(x) = 1

  8. Discrete Uniform Probability Function f(x) = 1/n Where n is the number of values the random variable can assume.

  9. Expected Value of a Random Variable The average value of the variable over an infinite number of experiments. E(x) = m = Sxf(x)

  10. Example Suppose your population consisted of 100 families with children where:

  11. Example, cont. • Using the formula for the weighted mean, we could write: • = (Swixi)/(Swi) • = [(1)(30)+(2)(50)+(3)(20)]/[30+50+20] • = (1)(30/100)+(2)(50/100)+(3)(20/100) • = (1)(.3)+(2)(.5)+(3)(.2) • More generally: m = S[xi(wi/Swi)] = Sxf(x)

  12. Variance of a Random Variable The sum of the squared deviations from the mean weighted by the probabilities of a value occurring. Var(x) = s2 = S(x-m)2f(x)

  13. Standard Deviation of a Random Variable

  14. Random Variable, Example Assume a random variable can take on the following three values with the corresponding probabilities: xf(x) 1 0.1 2 0.4 3 0.5 Find the expected value and variance of the random variable

  15. Random Variable, Example E(x) = (1)(.1) + (2)(.4) + (3)(.5) = 2.4 s2 = (1 - 2.4)2(.1) + (2 - 2.4) 2(.4) + (3 - 2.4) 2(.5) = (1.96)(.1) + (.16)(.4) + (.36)(.5) = .196 + .064 + .18 = .44

  16. Random Variable, Example Random variables.xlsx

  17. Binomial Experiment The experiment consists of a sequence of n identical trials. Two outcomes are possible on each trial, one outcome is labeled success the other failure. The probability of success, denoted by p, does not change from trial to trial. The trials are independent.

  18. Binomial Probability Function

  19. Binomial Example A die will be rolled three times. Success is defined as rolling a 1 or a 2. Failure is defined as rolling a 3 or higher. Assume we want to find the probability of having one success. Given that definition: x = 1 n = 3 p = 1/3 (1 - p) = 2/3

  20. Binomial Example, Cont.

  21. Binomial Example, Cont.

  22. Binomial Example, Cont. The portion of the binomial formula: px(1-p)(n-x) represents the probability of going down one branch where there are x successes In this case: (1/3)1(2/3)2

  23. Binomial Example, Cont. The other portion of the formula calculates how many ways we can have a given number of successes and failures. n! x!(n-x)! Getting x successes in n trials can be thought of as how many ways can x items be drawn from a group of n; a questions answered by the combinations formula.

  24. Binomial Example, Cont. In this case: S F F Position 1 is drawn F S F Position 2 is drawn F F S Position 3 is drawn n! = 3! = 3! = 3 x!(n-x)! 1!(3-1)! 2!

  25. Expected Value and Variance for the Binomial Distribution E(x) = m = np Var(x) = s2 = np(1 – p)

  26. Practice Assume the probability of getting a bad part (success) is 20 percent and the probability of getting a good part (failure) is 80 percent. If we draw 6 parts and test them, what is probability of drawing 3bad parts? If we drawing at least 2 bad parts (given we draw 5)? What is the expected value of the distribution? The variance?

  27. Practice, cont.

  28. Practice, cont.

  29. Practice, cont.

  30. Binomial Applet http://stattrek.com/Tables/Binomial.aspx

  31. Graded Homework P. 199-200, #21, 23 P. 209-210, #29,31

  32. Binomial Applet http://onlinestatbook.com/stat_sim/index.html

More Related