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Chapter 8: Binomial and Geometric Distributions

Chapter 8: Binomial and Geometric Distributions. Binomial vs. Geometric. The Binomial Setting The Geometric Setting. Each observation falls into one of two categories. 1. Each observation falls into one of two categories. The probability of success

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Chapter 8: Binomial and Geometric Distributions

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  1. Chapter 8: Binomial and Geometric Distributions

  2. Binomial vs. Geometric The Binomial Setting The Geometric Setting • Each observation falls into • one of two categories. 1. Each observation falls into one of two categories. • The probability of success • is the same for each • observation. 2. The probability of success is the same for each observation. • The observations are all • independent. • The observations are all • independent. • There is a fixed number n • of observations. 4. The variable of interest is the number of trials required to obtain the 1st success.

  3. Are Random Variables and Binomial Distributions Linked? RECALL: Suppose that each of three randomly selected customers purchasing a hot tub at a certain store chooses either an electric (E) or a gas (G) model. Assume that these customers make their choices independently of one another and that 40% of all customers select an electric model. The number among the three customers who purchase an electric hot tub is a random variable. What is the probability distribution?

  4. Are Random Variables and Binomial Distributions Linked? X = number of people who purchase electric hot tub (# of E’s) X 0 1 2 3 .288 P(X) .216 .432 .064 (.6)(.6)(.6) GGG EEG GEE EGE (.4)(.4)(.6) (.6)(.4)(.4) (.4)(.6)(.4) EGG GEG GGE (.4)(.6)(.6) (.6)(.4)(.6) (.6)(.6)(.4) EEE (.4)(.4)(.4)

  5. Combinations Formula: Practice:

  6. Developing the Formula Blood type is inherited. If both parents carry genes for the O and A blood types, each child has a probability 0.25 of getting two O genes, thus having blood type O. Let X = the number of O blood types among 3 children. Different children inherit genes independently of each other. Since X has the binomial distribution with n = 3 and p = 0.25. We say that X is B(3, 0.25)

  7. Developing the Formula Outcomes Probability Rewritten OcOcOc OOcOc OcOOc OcOcO OOOc OOcO OcOO OOO

  8. Developing the Formula Be able to recognize the formula even though you can set up a Prob. Distribution without it. Rewritten n = # of observations p = probablity of success k = given value of variable

  9. Working with probability distributions State the distribution to be used Binomial vs. Geometric State important numbers Binomial: n & p Geometric: p Define the variable Write proper probability notation

  10. Twenty-five percent of the customers entering a grocery store between 5 p.m. and 7 p.m. use an express checkout.Consider five randomly selected customers, and let X denote the number among the five who use the express checkout. binomial n = 5 p = .25 X = # of people use express

  11. What is the probability that two used the express checkout? Get started with the required components: • Binomial, n = 5, p = .25 • X = # of people use express • P(X = 2)

  12. Calculator option #1:nCr = n choose r = 5 -> Math -> PRB -> 3:nCr -> 2 -> enter REQUIRED COMPONENTS 1, 2, 3 • Define the situation: Binomial, with n = 5 and p = 0.25 • Identify X: let X = # of people use express • Write proper probability notation that the • question is asking: P( X = 2) Calculator option #2: 2nd Vars 0:binompdf(5, 0.25, 2) = .2637

  13. What is the probability that no more than three used express checkout? • Binomial, n = 5, p = .25 • X = # of people use express • P(X ≤3) P(X ≤3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) Calculator option: P(X ≤3) = binomcdf(5, 0.25, 3) = .9844

  14. What is the probability that at least four used express checkout? • Binomial, n = 5, p = .25 • X = # of people use express • P(X ≥ 4) Calculator option: P(X 4) = 1 - P(X<4) = 1 - binomcdf(5, 0.25, 3) = .0156

  15. “Do you believe your children will have a higher standard of living than you have?” This question was asked to a national sample of American adults with children in a Time/CNN poll (1/29,96). Assume that the true percentage of all American adults who believe their children will have a higher standard of living is 0.60. Let X represent the number who believe their children will have a higher standard of living from a random sample of 8 American adults. binomial n = 8 p = .60 X = # of people who believe…

  16. Interpret P(X = 3) and find the numerical answer. binomial n = 8 p = .60 X = # of people who believe The probability that 3 of the people from the random sample of 8 believe their children will have a higher standard of living.

  17. Interpret P(X = 3) and find the numerical answer. binomial n = 8 p = .60 X = # of people who believe P(X=3) = binomialpdf(8, .6, 3) = .1239 or

  18. Find the probability that none of the parents believe their children will have a higher standard. • binomial, n = 8, p = 0.60 • X = # of people who believe • P(X = 0 )

  19. Find the probability that none of the parents believe their children will have a higher standard. binomial n = 8 p = .60 X = # of people who believe or P(X=0) = binomialpdf(8, 0.6, 0) = 0.00066

  20. Binomial vs. Geometric The Binomial Setting The Geometric Setting • Each observation falls into • one of two categories. 1. Each observation falls into one of two categories. • The probability of success • is the same for each • observation. 2. The probability of success is the same for each observation. • The observations are all • independent. • The observations are all • independent. • There is a fixed number n • of observations. 4. The variable of interest is the number of trials required to obtain the 1st success.

  21. Developing the Geometric Formula Probability X

  22. The Mean and Standard Deviation of a Geometric Random Variable If X is a geometric random variable with probability of success p on each trial, the expected value of the r.v. (the expected number of trials to get the first success) is X

  23. Suppose we have data that suggest that 3% of a company’s hard disc drives are defective. You have been asked to determine the probability that the first defective hard drive is the fifth unit tested. • geometric p = .03 • X = # of disc drives till defective • Find P(X = 5)

  24. A basketball player makes 80% of her free throws. We put her on the free throw line and ask her to shoot free throws until she misses one. Let X = the number of free throws the player takes until she misses. • Geometric, p = .20 • X = # of free throws till miss

  25. What is the probability that she will make 5 shots before she misses? (When is the 1stsuccess?) geometric p = .20 X = # of free throws till miss What is the probability that she will miss 5 shots before she makes one? geometric p = .80 Y = # of free throws till make

  26. What is the probability that she will make at most 5 shots before she misses? geometric p = .20 X = # of free throws till miss Or1 – P(makes all 6 shots) = 1 – (.806) = .7379

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