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	115 49 70 59 3

(a) For each age group, compute the probability that a moose in that age group is killed by a wolf. (Use 3 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. (Use 2 decimal places.)

μ
σ

Answer 1

Solution :

(a) Given that age of moose in years 0.5 ,1 - 5,6 - 10,11 - 15,16 - 20

Number killed by wolves 115 , 49 , 70 , 59 , 3

=> sum of Number killed by wolves is 115 + 49 + 70 + 59 + 3 = 296

Age Probability P(x)

0.5 115/296 = 0.389

1 - 5 49/296 = 0.166

6 - 10 70/296 = 0.236

11 - 15 59/296 = 0.199

16 - 20 3/296 = 0.010

(b) midpoints of Age

x as 0.5 = 1/2 , (1 + 5)/2 = 3 , (6 + 10)/2 = 8 , (11 + 15)/2 = 13 , (16 + 20)/2 = 18

P(x) as 0.389 , 0.166 , 0.236 , 0.199 , 0.010

=> Expectation μ = sum of x*P(x)

= 0.5*0.389 + 3*0.166 + 8*0.236 + 13*0.199 + 18*0.010

= 5.35

=> Standard deviation σ = 5.00

Explanation :-

=> sum of x^2*P(x) = 0.5^2*0.389 + 3^2*0.166 + 8^2*0.236 + 13^2*0.199 + 18^2*0.010

= 53.57

=> standard deviation σ = sqrt(∑ x^2*P(x) − μ^2)

= sqrt(53.57 - (5.35)^2)

= 4.9947

= 5.00