Question

Isle Royale, an island in Lake Superior, has provided an important study site of wolves and their prey. Of special interest is the study of the number of moose killed by wolves. In the period from 1958 to 1974, there were 296 moose deaths identified as wolf kills. The age distribution of the kills is as follows.

Age of Moose in Years	Number Killed by Wolves
Calf (0.5 yr) 1-5 6-10 11-15 16-20	105 55 74 58 4

(a) For each age group, compute the probability that a moose in that age group is killed by a wolf. (Round your answers to three decimal places.)

0.5
1-5
6-10
11-15
16-20

(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. (Round your answers to two decimal places.)

μ	=
σ	=

Answer 1

Solution:
Probabilities can be calculated as follows:

Age Of mose	No. Of killed by Wolves	Probability
0.5	105	105/296 = 0.35
1-5	55	55/296=0.19
6-10	74	74/296=0.25
11-15	58	58/296=0.2
16-20	4	4/296=0.01

Solution(b)
After calculating Midpoints probabilities are

Age Of mose	No. Of killed by Wolves	Probability
0.5	105	0.35
3	55	0.19
8	74	0.25
13	58	0.2
18	4	0.01

Expected mean = summation(Xi*P(Xi)) = (0.5*0.35) + (3*0.19) + (8*0.25) + (13*0.2) + (18*0.01) = 5.53
Standard deviation = sqrt(Summation((Xi-mean)^2*P(Xi))

Age Of mose	No. Of killed by Wolves	Probability	(Xi-mean)	(Xi-mean)^2	(Xi-mean)^2*P(Xi)
0.5	105	0.35	-5.03	25.3009	8.855315
3	55	0.19	-2.53	6.4009	1.216171
8	74	0.25	2.47	6.1009	1.525225
13	58	0.2	7.47	55.8009	11.16018
18	4	0.01	12.47	155.5009	1.555009
					24.72

Standard deviation =sqrt(24.72) = 4.97