Question 920902: The following question is a Hypergeometric Distribution.
x~Hypergeometric(N=110, r=24, n=15) where "N" is the population size, "r" is the number of successes in population, and "n" is the sample size. I have answered (a), (b), and (c), using their respected formulas, the answers are provided. I am struggling with part (d). I am not looking for the exact numerical answer, but rather the way to go about answering it.
A biologist captures 24 grizzly bears during the spring, and fits each with a radio collar. At the end of summer, the biologist is to observe 15 grizzly bears from a helicopter, and count the number that are radio collared. This count is represented by the random variable X.
Suppose there are 110 grizzly bears in the population.
(a) What is the probability that of the 15 grizzly bears observed, 5 had radio collars? Use four decimals in your answer.
P(X=5)= 0.128
(b) Find the probability that between 3 and 7 (inclusive) of the 15 grizzly bears observed were radio collared?
P(3≤X≤7)= P(X≤7)-P(X≤2) = 0.6806 (use four decimals)
(c) How many of the 15 grizzly bears observe from the helicopter does the biologist expect to be radio-collared? Provide the standard deviation as well.
E(X)= 3.27 (use two decimals)
SD(X)= 1.49 (use two decimals)
(d) The biologist gets back from the helicopter observation expedition, and was asked the question: How many radio collared grizzly bears did you see? The biologist cannot remember exactly, so responds " somewhere between 5 and 9 (inclusive) ".
- I interpreted this information to be : P(5≤X≤9), which is P(X≤9)-P(X≤4) = 0.20014
Given this information, what is the probability that the biologist saw 7 radio-collared grizzly bears?
- As for calculating X=7, how is the above information P(5≤X≤9) relevant?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! d) observation,
7 is halfway between 5 and 9, therefore
probability that the biologist saw 7 radio-collared grizzly bears = 0.20014 / 2 = 0.10007
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