A biologist captures 25 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 111 grizzly bears in the population.
(a) What is the probability that of the 15 grizzly
bears observed, 3 had radio collars? Use four decimals in your
answer.
?(?=3)=P(X=3)=
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(b) Find the probability that between 5 and 7
(inclusive) of the 15 grizzly bears observed were radio
collared?
?(5≤?≤7)=P(5≤X≤7)=
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(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)=
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(use two decimals)
??(?)=SD(X)=
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(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 8 (inclusive)
".
Given this information, what is the probability that the biologist
saw 7 radio-collared grizzly bears?
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(use four decimals in your answer)
Sample size, n = 15
No. of events of interest in population , k
= 25
Population size , N = 111
a)
P(X=3)= C(25,3)*C (86,12)/C(111,15) =0.2587
b) P(5≤X≤7)= 0.2171
c) E(X) = mean = E(X) = n(k/N) = 3.38
variance = VAR(X) = nk/N(1 - k/N)((N-n)/(N-1))
2.2843
variance = nk(N-k)(N-n)/(N²(N-1)) = 2.2843
SD(X) = std dev = √variance =
1.51
d) P(5≤x≤8) = 0.2213
P(X=7)= C(25,7)*C (86,8)/C(111,15) = 0.0188
Given this information, the probability that the biologist saw 7 radio-collared grizzly bears = 0.0188/0.2213 = 0.0850
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