Answer:
<em>Part a) Probability that a moose in that age group is killed by a wolf</em>
<em>Part b)</em>
- <em>Expected age of a moose killed by a wold</em>
μ = 5.61 years
- <em>Stantard deviation of the ages</em>
σ = 4.97 years
Explanation:
1) Start by arranging the table to interpret the information:
<u>Age of Moose in years Number killed by woolves</u>
Calf (0.5) 107
1 - 5 47
6 - 10 79
11 - 15 60
16 - 20 3
You can now verify the total number of moose deaths identified as wolf kills: 107 + 47 + 79 + 60 + 3 = 296, such as stated in the first part of the question.
2) <u>First question: </u>a) For each age, group, compute the probability that a moos in that age group is killed by a wolf.
i) Formula:
Probability = number of positive outcomes / total number of events.
ii) Probability that a moose in an age group is killed by a wolf = number of moose killed by a wolf in that age group / total number of moose deaths identified as wolf kills.
iii) P (0.5 year) = 107 / 296 = 0.361
iv) P (1 - 5) = 47 / 296 = 0.159
v) P (6 - 10) = 79 / 296 = 0.267
vi) P (11 - 15) = 60 / 296 = 0.203
vii) P (16 - 20) = 3 / 296 = 0.010
3) <u>Second question:</u> b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages.
i) Class midpoint
0.5 0.5
1 - 5 (1 + 5) / 2 = 3
6 - 10 (6 + 10) / 2 = 8
11 - 15 (11 + 15) / 2 = 13
16 - 20 (16 + 20) = 18
ii) Expected age of a moose killed by a wolf = mean of the distribution = μ
μ = Sum of the products of each probability times its age (mid point)
μ = 0.5 ( 0.361) + 3 ( 0.159) + 8 ( 0.267) + 13 ( 0.203) + 18 ( 0.010) = 5.61 years
μ = 5.61 years ← answer
iii) Stantard deviation of the ages = σ
σ = square root of the variace
variance = s = sum of the products of the squares of the differences between the mean and the class midpoint time the probability.
s = (0.5 - 5.61)² (0.361) + (3 - 5.61)² ( 0.159) + (8 - 5.61)² ( 0.267) + (13 - 5.61)² ( 0.203) + (18 - 5.61)² ( 0.010) = 24.65
σ = √ (24.65) = 4.97 years ← answer
