Question

The lifespans of wild wolves vary dramatically. Although the average lifespan is between six and eight years, many will die sooner, and some can reach thirteen. Assume that the lifespan of a wild wolf is normally distributed with an average of 80 months and a variance of 800 months^2. In addition, assume that fifteen wild wolves are randomly selected and radio tracked by a group of biologists working for the International Wolf Center (IWC).

1. What is the probability that only three of them live more than 9 years?
2. What is the probability that the average lifespan does not exceed 10 years?

Answer 1

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 80

S.d = √800

1)

P(x>9) = P(x>108 months)

Z = (108 - 80)/(√800) = 0.99

From z table, P(z>0.99) = 0.1611

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.1611

N = number of trials = 15

R = desired success = 3

P(3) = 0.23111085842

2)

P(x<120)

Z = (120 - 80)/√800) = 1.41

From z table, P(z<1.41) = 0.9207