Sampling without replacement: Suppose you are studying a defect in a trout speci

Question

Sampling without replacement: Suppose you are studying a defect in a trout species at the San Gregorio Reservoir . You know there are N 200 total trout in the a. (4 points) Suppose you take a sample of n = 10 trout from the reservoir without replacement. How many possible samples could you get? b. (4 points) Now suppose you know there are m = 20 trout in the reservoir with the defect you are looking for. Let X = the number of trout with the defect in your sample. How many different ways are there to sample X detected trout from m 20 possible defected trout, without replacement? c. (4 points) Now you are interested in that part of your n = 10 sample that does not include defected trout. How many different X regular trout from the total number of regular trout, without replacement? ways can you sample n- d. (4 points) Now we'll put it all together! What is the probability of having X 3 detected trout in your sample of n = 10 trout? Hint: Think Pdt "parf will consist of the ways you can configure the sample. That will consist of your part of the sample containing the defective trout times the possible ways your sample where your "whole is the fotal possible samples and your whole might contain regular trout e. (4 points) You've just derived the Hypergeometric Distribution all by yourself! State the pmf of this distribution by replacing all of the numbers in this example by the corresponding arbitrary letters. (And don't forget what happens "otherwise")

Explanation / Answer

N = 200, n = 10. Let X be the number of defected Trouts in the sample (Without replacement)

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Part a.

In any case of N distinct objects, the number of ways to choose n objects (Withotu replacement) is: NCn

200C10 = 22,451,004,309,013,280 [Answer (a)]

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Part b.

Number of defective Trouts = 20

Number of ways in which we choose X defective trouts out of the 20 = 20CX [Answer (b)]

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Part c.

Number of regular Trouts = 180

Number of ways in which we choose (10-X) regular trouts out of the 180 = 180C(10-X) [Answer (c)]

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Part d.

Number_of_ways(X defective trouts out of 200) = Number_of_ways(X defective from 20)*Number_of_ways(10-X regular from 180)

=> Number_of_ways(X defective trouts out of 200) = [ 20CX ] * [ 180C(10-X) ]

Total number of ways sample of 10 can be gotten (part a.) = 200C10

=> Probability of X defective Trouts in the sample = [ 20CX ] * [ 180C(10-X) ] / [ 200C10 ]

For X = 3,

Probability of 3 defective Trouts in the sample = [ 20C3 ] * [ 180C7 ] / [ 200C10 ]

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Part e.

Probability Mass Function: As calculated part d. itself, substituting the corresponding 'indices' to get a general probability mass function,

P(X defective trouts in sample of 10 trouts from a total of 200) = [ 20CX ] * [ 180C(10-X) ] / [ 200C10 ] [Answer (e)]

Note: for X = 0 to 10. For X > 10, the function gets a value of zero (0)

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