Loading in 5 sec....

8.2 THE GEOMETRIC DISTRIBUTIONS (Pages 434 -444)PowerPoint Presentation

8.2 THE GEOMETRIC DISTRIBUTIONS (Pages 434 -444)

- 88 Views
- Uploaded on
- Presentation posted in: General

8.2 THE GEOMETRIC DISTRIBUTIONS (Pages 434 -444)

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

8.2 THE GEOMETRIC DISTRIBUTIONS (Pages 434 -444)

“Blessed are those that nought expect,for they shall not be disappointed.”

John Wolcot

- The Advanced Placement Statistics Syllabus states that students need only know how to obtain geometric probabilities through simulation. The geometric setting is somewhat similar to the binomial setting, the basic difference being that the geometric setting does not have a fixed number of observations.

- Each observation is in one of two categories: success or failure.
- The probability of success is the same for each observation.
- Observations are independent.
- Knowing the result of one observation tells you nothing about the other observations.

- The variable of interest is the number of trials required to obtain the first success.

- Question: How many times would you expect to have to roll a single die in order to get a "6"?
- Here is a simulation approach (ten trials) using randInt(1,6,15) on the TI-83.

- Trial 1: 3 5 2 2 3 6 ... (6 rolls)
- Trail 2: 1 3 3 2 1 4 2 1 1 6 ... (10 rolls)
- Trial 3: 5 1 5 6 ... (4 rolls)
- Trial 4: 4 1 4 3 6 ... (5 rolls)
- Trial 5: 4 5 3 4 6 ... (5 rolls)
- Trial 6: 5 5 1 3 3 2 2 1 3 6 ... (10 rolls)
- Trial 7: 5 3 4 1 6 ... (5 rolls)
- Trial 8: 5 2 4 2 1 6 ... (6 rolls)
- Trial 9: 4 5 1 4 6 ... (5 rolls)
- Trial 10: 2 5 4 2 1 2 3 4 6 (9 rolls)

- Now, if p is the probability of success, and q = 1-p is the probability of failure, then the probability of success on the first trial is p, on the second trial it is qp, on the third trial it is q2p, etc. Hence, If X is a variable representing the number of trials until the first success, the expected value of X is
- mx = 1p + 2qp + 3q2p + 4q3p + .... = p(1 + 2q + 3q2 + 4q3 + ...).

- Observe that 1 + 2q + 3q2 + 4q3 + ... = (1+q+q2+q3+ ...)(1+q+q2+q3+ ...)
- Also, for 0<q<1, the sum of the infinite series 1 + q + q2 + q3 = 1/(1-q).
- Hence,
- mx = p (1+q+q2+q3+ ...)(1+q+q2+q3+ ...)
- = p[1/(1-q)][1/(1-q)]
- = p/(1-q)2
- = p/p2
- = 1/p.

- In the die-rolling example, the probability of rolling a 3 is 1/6. Hence, the expected number of rolls before the first success is 1/(1/6) = 6.

- In the present California Lottery, one chooses 6 numbers from the set 1,2,3,...,49,50,51. The State of California then randomly selects 6 numbers from the set. If you happen to match the six number chosen by the State, you win millions of dollars. Your probability of matching six is 1/51C6 = 1/18,009,460. That is, you would expect to have to play 18,009,460 times to get your first success. If you played once a week, you would expect your first success after 346,336 weeks, or 6,660 years. Good luck!

Read

Pages 464-475

Work

8.37 – 8.45 odd