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To estimate the population of horned lizards in a national park, a tea

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To estimate the population of horned lizards in a national park, a tea  [#permalink]

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New post 18 Apr 2018, 18:26
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To estimate the population of horned lizards in a national park, a team from the park service captures and tags 25 horned lizards and then releases the tagged lizards back into the park. Two weeks later, the team returns and captures 20 horned lizards, 4 of which bear tags that identify them as being among the previously captured lizards. If all the tagged lizards have dispersed throughout the park an are still active when the second group of lizards is captured, what is the best estimate of the horned lizard population in the park, based on the information obtained through this capture-tagr-ecapture strategy?

(A) 250
(B) 125
(C) 100
(D) 80
(E) 75

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Re: To estimate the population of horned lizards in a national park, a tea  [#permalink]

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New post 18 Apr 2018, 20:02
I just want to give it a try.

1. Total Tagged horned lizard = 25
2. Two weeks after team captures 20 lizards, of which 4 lizards are already tagged (from previously captured lizards)
=> \(\frac{{Untagged Horned Lizard}}{{tagged Horned Lizard}}\) = \(\frac{16}{4}\) = \(\frac{4}{1}\)

=> Total Untagged Horned Lizard = \(\frac{4}{1}\)* Total Tagged horned lizard
= \(\frac{4}{1} * 25\)
= 100

Hence Total Horned Lizard = Tagged Lizard + Untagged Lizard = 25 + 100 = 125
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To estimate the population of horned lizards in a national park, a tea  [#permalink]

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New post 19 Apr 2018, 15:57
Bunuel wrote:
To estimate the population of horned lizards in a national park, a team from the park service captures and tags 25 horned lizards and then releases the tagged lizards back into the park. Two weeks later, the team returns and captures 20 horned lizards, 4 of which bear tags that identify them as being among the previously captured lizards. If all the tagged lizards have dispersed throughout the park an are still active when the second group of lizards is captured, what is the best estimate of the horned lizard population in the park, based on the information obtained through this capture-tagr-ecapture strategy?

(A) 250
(B) 125
(C) 100
(D) 80
(E) 75

Of 20 lizards caught the second time, 4 are tagged, which constitutes
\(\frac{4}{20}=.20*100=20\) percent of the lizards caught

That's a sample of the whole lizard population.

The total number of tagged lizards, now dispersed, should be represented in the whole population in the same percent, roughly, as the percent of tagged lizards sample (the second catch.)

25 tagged lizards = .20 of all lizards

All lizards = \(\frac{25}{.2}= 125\) lizards

Answer B
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Re: To estimate the population of horned lizards in a national park, a tea  [#permalink]

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New post 23 Apr 2018, 16:57
Bunuel wrote:
To estimate the population of horned lizards in a national park, a team from the park service captures and tags 25 horned lizards and then releases the tagged lizards back into the park. Two weeks later, the team returns and captures 20 horned lizards, 4 of which bear tags that identify them as being among the previously captured lizards. If all the tagged lizards have dispersed throughout the park an are still active when the second group of lizards is captured, what is the best estimate of the horned lizard population in the park, based on the information obtained through this capture-tagr-ecapture strategy?

(A) 250
(B) 125
(C) 100
(D) 80
(E) 75


Let’s let x = the total population of horned lizards in the park. We know that 25 of those x lizards were tagged. Thus, after the tagging, the ratio of tagged lizards to total population is 25/x. Two weeks later, we know that 4 of the 20 lizards captured were tagged. To estimate the total population, it is assumed that the ratio of tagged lizards to total lizards in the group of 20 captured lizards (which is 4/20) is the same as the earlier ratio of 25/x. Thus, we can create the proportion:

4/20 = 25/x

4x = 500

x = 125

Answer: B
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Re: To estimate the population of horned lizards in a national park, a tea &nbs [#permalink] 23 Apr 2018, 16:57
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