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2. Suppose that wildlife workers capture 328 penguins on an island, mark them, a

ID: 135993 • Letter: 2

Question

2.         Suppose that wildlife workers capture 328 penguins on an island, mark them, and allow them to mix with the rest of the population. Later, they capture 200 penguins, 64 of which are marked.What is the estimate for the number of penguins on the island?

3.         You have been tasked to estimate the number of rattlesnakes using a series of hibernacula along a ridge in western Texas. To accomplish your assigned task, you capture 37 rattlesnakes and mark them with tags. You return to the site one week later and resample the population by capturing 52 rattlesnakes. Of the 52, 30 are marked from your first sample. How many rattlesnakes are in the population?

Example 1: Using the data from Communities 1 and 2 above and calculate Simpson’s Index.

For Community 1: 18 of 20 individuals are Species A So for Species A, (pi)2 = (18/20)2= 0.81

         1 of 20 individuals are Species B For Species B, (pi)2 = (1/20)2= 0.025

         1 of 20 individuals are Species C For Species C, (pi)2 = (1/20)2 = 0.025

         ? (pi)2 = 0.815?

Simpson's Index is 1 - 0.815, or 0.185 for Community 1

For Community 2:

D = 1 - [ (7/20)2 + (8/20)2 + (5/20)2] Simpson's Index = 1 - [0 .1225 + 0.16 + 0.0625 ]

Simpson's Index = 1 - [0.345]? =  0.655

So Community 2 has a higher biodiversity.

Now, compare the diversity of these two kinds of mixed nuts using Simpson’s Index:

Species

Number in “Mixed Nuts”

Number in “Deluxe Mixed Nuts”

Brazil nut

1

8

Cashew

8

15

Pecan

1

7

Almond

15

18

Peanut

85

0

We will set up a table to record the calculations:

Species

Number in “Mixed Nuts”

(pi)2

Species

Number in “Deluxe Mixed Nuts”

(pi)2

Brazil nut

1

Brazil nut

8

Cashew

8

Cashew

15

Pecan

1

Pecan

7

Almond

15

Almond

18

Peanut

85

Peanut

0

Total

Total

“Mixed Nuts”:        ? (pi)2   =    ______________     and           1- ? (pi)2 ___________

                        So D = ______________

“Deluxe Mixed Nuts”:       ? (pi)2   =    ______________     and           1- ? (pi)2 __________

So D = ______________

1.         Which of the “populations” of nuts has the higher biodiversity?

Species

Number in “Mixed Nuts”

Number in “Deluxe Mixed Nuts”

Brazil nut

1

8

Cashew

8

15

Pecan

1

7

Almond

15

18

Peanut

85

0

Explanation / Answer

Answer:

2) Based on the given information:

3)

4)

For Mixed Nuts, sum(pi^2) = 0.621, 1 - sum(pi^2) = 1 - 0.621 = 0.379  and  

For Delux Mixed Nuts, sum(pi^2) = 0.287, 1 - sum(pi^2) = 1 - 0.287 = 0.713

So population of Delux Mixed Nuts has higher biodiversity.

Species Number in “Mixed Nuts” (pi)2 Species Number in “Deluxe Mixed Nuts” (pi)2 Brazil nut 1 =(1/110)^2 Brazil nut 8 =(8/48)^2 Cashew 8 =(8/110)^2 Cashew 15 =(15/48)^2 Pecan 1 =(1/110)^2 Pecan 7 =(7/48)^2 Almond 15 =(15/110)^2 Almond 18 =(18/48)^2 Peanut 85 =(85/110)^2 Peanut 0 0.000 Total =(1+8+1+15+85) Total =(8+15+7+18+0)
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