8 2 the geometric distributions pages 434 444
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8.2 THE GEOMETRIC DISTRIBUTIONS (Pages 434 -444). “Blessed are those that nought expect, for they shall not be disappointed.” John Wolcot. OVERVIEW:.

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8.2 THE GEOMETRIC DISTRIBUTIONS (Pages 434 -444)

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8 2 the geometric distributions pages 434 444

8.2 THE GEOMETRIC DISTRIBUTIONS (Pages 434 -444)

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

John Wolcot


Overview

OVERVIEW:

  • 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.


The geometric setting

The geometric setting:

  • 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.


Example

Example:

  • 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.


The mean number of rolls for the 10 trials is 6 5

The mean number of rolls for the 10 trials is 6.5.

  • 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)


Developing a geometric distribution

Developing a Geometric Distribution

  • 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 + ...).


M x p 1 2q 3q 2 4q 3

mx = 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.


California lottery

California Lottery

  • 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!


For section 8 2

Read

Pages 464-475

Work

8.37 – 8.45 odd

For Section 8.2:


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