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The Geometric Distributions Section 8.2.1. Starter 8.2.1 Fred Funk hits his tee shots straight most of the time. In fact, last year he put 78% of his.

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Presentation on theme: "The Geometric Distributions Section 8.2.1. Starter 8.2.1 Fred Funk hits his tee shots straight most of the time. In fact, last year he put 78% of his."— Presentation transcript:

1 The Geometric Distributions Section 8.2.1

2 Starter 8.2.1 Fred Funk hits his tee shots straight most of the time. In fact, last year he put 78% of his tee shots in the fairway. In yesterday’s round, he hit only 7 of 14 shots in the fairway. What is the likelihood that he would have that bad a day (or worse) from the tee?

3 Objectives Identify whether a random variable is in a geometric setting Use a formula or a calculator command to find the PDF and CDF of a geometric random variable Express the PDF and CDF as either a table or a histogram California Standard 7.0 Students demonstrate an understanding of the standard distributions (normal, binomial, and exponential) and can use the distributions to solve for events in problems in which the distribution belongs to those families.

4 The Geometric Setting 1.There are only two outcomes: success and failure. 2.The variable on interest (X) is the number of trials needed to get the FIRST success. So there is NOT a pre-determined number of trials 3.Each trial is independent of all other trials. 4.The probability of success is the same for each trial.  Summary of binomial vs. geometric 1.BINOMIAL has a fixed number of trials and we count successes. 2.GEOMETRIC has a fixed number of successes (1) and we count trials.

5 Calculating the PDF Consider once again the dice game where I have a probability of 1/3 of succeeding. What is the probability that I win on my first try? P(X = 1) = (1/3) Probability I win on second try? –That means I fail and then win P(X = 2) = (2/3)(1/3) = 2/9 Probability I win on third try? P(X = 3) = (2/3)(2/3)(1/3) = 4/27

6 Generalize to a Formula If I win on the nth trial, I must have failed on the first (n – 1) trials and succeeded exactly ONCE. So there will be (n – 1) factors of q and one factor of p That means the formula must be: P(X = n) = q n-1 p Notice that this means the PDF never ends! –So to write an actual PDF, first decide how far you want X to go.

7 Write a PDF for the Dice Game Use your calculator and the formula to write the first 6 entries of the dice game PDF –Since the PDF in theory goes forever, we arbitrarily decide to end it at X = 6 –Round to.001 –Recall we already computed the first three entries! X123456 P(X)

8 Write a PDF for the Dice Game Use your calculator and the formula to write the first 6 entries of the dice game PDF –Since the PDF in theory goes forever, we arbitrarily decide to end it at X = 6 –Round to.001 –Recall we already computed the first three entries! X123456 P(X).333.222.148

9 Write a PDF for the Dice Game Use your calculator and the formula to write the first 6 entries of the dice game PDF –Since the PDF in theory goes forever, we arbitrarily decide to end it at X = 6 –Round to.001 –Recall we already computed the first three entries! X123456 P(X).333.222.148.099.066.044

10 Write a CDF for the Dice Game Add appropriate entries in the PDF to write the first 6 entries of the dice game CDF X123456 P(X).333.222.148.099.066.044 C(X)

11 Write a CDF for the Dice Game Add appropriate entries in the PDF to write the first 6 entries of the dice game CDF X123456 P(X).333.222.148.099.066.044 C(X).333.555

12 Write a CDF for the Dice Game Add appropriate entries in the PDF to write the first 6 entries of the dice game CDF X123456 P(X).333.222.148.099.066.044 C(X).333.555.704.802.868.912

13 Why Don’t Probabilities Add to 1? In the binomial CDF we saw the last entry was always 1 because it adds the probabilities of all possible outcomes. Why doesn’t that happen here? Actually, it would if we could write all of the infinite number of outcomes! In fact, we can. The CDF is an infinite geometric series whose sum we can find by a simple formula. When we do, we always get a sum of 1. –See the author’s demonstration on page 437

14 Calculator Commands To get the probability that X exactly equals one value, use Geometpdf(p,X) –Example: Geometpdf(1/3, 3) =.148 (or 4/27) –That is the same as we got from the formula. To get the probability that X equals at most one value, use Geometcdf(p,X) –Example: Geometcdf(1/3, 3) =.704 –That is the same as 1/3 + 2/9 + 4/27

15 Write PDF & CDF on the TI To construct the table on the calculator, decide how many outcomes you wish to show. –In most cases, 10 will be more than enough. Put the outcomes in L 1 with a sequence command. –What is the least possible value for X? Use the geometpdf and geometcdf commands to fill L 2 and L 3 –geometpdf(p, L 1 )→L 2 –geometcdf(p, L 1 )→L 3

16 Draw the PDF & CDF Histograms After creating the PDF and CDF tables, use Stat Plots to draw the histograms. Sketch and label each graph. Write a brief description of the shape of each graph.

17 Another Look at the Dice Game Suppose we change the game so that I only get three rolls to win. What is the probability that I succeed WITHIN three rolls? –From the CDF, P(X÷3) =.704 So what’s the probability that it takes more than 3 tries to succeed? –P(X>3) = 1 –.704 =.296 There’s another old formula (now somewhat obsolete because of the TI-83) that also works: –P(X>n) = q n –In this case, P(X>3) = (2/3) 3 =.296

18 Objectives Identify whether a random variable is in a geometric setting Use a formula or a calculator command to find the PDF and CDF of a geometric random variable Express the PDF and CDF as either a table or a histogram California Standard 7.0 Students demonstrate an understanding of the standard distributions (normal, binomial, and exponential) and can use the distributions to solve for events in problems in which the distribution belongs to those families.

19 Homework Read pages 434 – 441 Do problems 24 – 26


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