ST371-4-OtherDiscrete

# ST371-4-OtherDiscrete - Other Discrete Distributions...

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Other Discrete Distributions Textbook 3.5, 3.6

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Motivating Example Interested in learning about a student-run club on campus, which has 20 members. 10 women and 10 men Sample 4 members at random without replacement* and record X = {number of females} (along with other variables) . What distribution does X follow?
Motivating Example *Note: Sampling with replacement means that we select a person from the population, measure them, then put them back into the population before we select again. Sampling without replacement means that we do not put the person back into the population before we select again.

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Motivating Example X not Bin(4,0.5) since selection of 2 nd person depends on that of the 1 st : 1 st person: P(Female) = 10/20 = 0.50 2 nd person: P(Female|1 st female) = 9/19 = 0.47; P(Female|1 st male) = 10/19 = 0.53 Probability of picking a female depends on who has already been picked; this becomes more pronounced as we continued to select people. Both ‘independent trials’ and ‘equal prob. of success’ conditions for the Bin dist. are violated
Motivating Example What if the club had 20,000 members, half of which were females? 1 st person: P(Female) = 10000/20000 = 0.50 2 nd person: P(Female|1 st female) = 9999/19999 = 0.49997; P(Female|1 st male) = 10000/19999 = 0.50002 P(Female) stays around 0.5 for the 3 rd and 4 th people selected, regardless of who was selected before => ok to use Bin dist.

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Note: We can use the Binomial Distribution… If the sample is a relatively small fraction of the population (Rule of thumb: Population at least 10 times larger than the sample) Or if we are sampling with replacement When sampling without replacement from a small population, we cannot use the Binomial Distribution, instead use the Hypergeometric distribution
Conditions for using the Hypergeometric Distribution 1. The population is finite and consists of N individuals or objects. 2. There are 2 possible outcomes for each individual/object: success and failure; and there are M successes in the population 3. A sample of n individuals are randomly selected without replacement If these conditions are met, then X = {number of successes} is a hypergeometric rv. The distribution of X depends on 3 parameters: n, M, and N

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Hypergeometric PMF 𝒑 𝒙 = 𝑴 𝒙 𝑵 − 𝑴 𝒏 − 𝒙 𝑵 𝒏
Hypergeometric PMF 𝒑 𝒙 = 𝑴 𝒙 𝑵 − 𝑴 𝒏 − 𝒙 𝑵 𝒏 the number of ways to have x successes & ( n-x ) failures

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Hypergeometric PMF 𝒑 𝒙 = 𝑴 𝒙 𝑵 − 𝑴 𝒏 − 𝒙 𝑵 𝒏 the number of ways to have x successes & ( n-x ) failures the number of ways to select a subset of n individuals out of group of N Thus, this formula is really just our basic definition of P(A) that was introduced in Chapter 2, presented in a new context.
Hypergeometric PMF 𝒑 𝒙 = 𝑴 𝒙 𝑵 − 𝑴 𝒏 − 𝒙 𝑵 𝒏 Notation: X ~ Hyper( n, M, N ) or h(x; n, M, N)

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Mean & Variance of X ~ Hyper( n,M,N ) 𝑬 𝑿 = 𝒏 ∙ 𝑴 𝑵 𝑽 𝑿 = 𝑵 − 𝒏 𝑵 − 𝟏 ∙ 𝒏 ∙ 𝑴 𝑵 𝟏 − 𝑴 𝑵