Question

The lifespans of wild wolves vary dramatically. Although the average lifespan is
between six and height 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 months2. 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

1.

9yrs = 108 months

So,

P(X>108) = P(Z>(108-80)/28.284))

= P(Z>0.98995)

= 0.1611

Now, this becomes a case of binomial distribution with n = 15 and p = 0.1611.

We know that,

P(X=x) = (nCx)*p^x(1-p)^n-x

P(X=3) = 0.2311

2.

= = 80

= = 28.28/3.873

= 7.302

P(<10) = P(Z<(120-80)/7.302)

= P(Z<5.47795)

= 1

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