Animal behaviour is often influenced by environmental signals. Suppose we are tr
ID: 3053216 • Letter: A
Question
Animal behaviour is often influenced by environmental signals. Suppose we are trying to model how a certain species of bird selects a suitable nesting site; these birds evaluate the quality of a potential nesting site based on the density of the surrounding vegetation. For simplicity, let’s assume that there are three levels of vegetation (denoted as types A, B, and C) that our birds can distinguish between:
A- Very sparse vegetation
B- Average density vegetation
C- Very dense vegetation
The probability a bird encounters each type of nesting site is given by:
Is this a probability density function? (Please enter "Y" for yes, or "N" for no) .
Given that a bird encounters a particular type of nesting site, the probability that they choose to nest there is given by:
Is this a probability density function? (Please enter "Y" for yes, or "N" for no) .
What is the probability that a bird nests in a type B site? (ie. that they find a type B site and choose to nest there?) (Enter your answer as a decimal with four decimal places)
Now, suppose that birds who choose better nesting sites will have more offspring. The number of offspring associated with each type of nesting site is as follows:
A- 2 offspring
B- 3 offspring
C- 4 offspring
Assuming the probabilities of encountering a particular site and choosing a particular site are as given above, what is the expected number of offspring a bird will have? (Enter a decimal number with four decimal places)
Type Probability of Encounter A 0.62 B 0.33 C 0.05Explanation / Answer
(a)
Type Probability of Encounter
A 0.62
B 0.33
C 0.05
--------------------------------
Total 1.00
Since the Total Probability = 0:
Answer:
Y
(b)
Type Probability of Nesting
A 0.25
B 0.75
C 1.00
----------------------------------
Total 2.00
Since the Total Probability is not 1:
Answer is:
N
(c)
By Multiplication Theorem:
P(find Type B site & Choose to nest) = P(find Type B site) X P(Choose to nest) = 0.33 X 0.75 = 0.2475
(d)
Expected Number of offspring = 3 X 0.2475 =0.7425
=