Let x denote the distance that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose that x has an exponential distribution with parameter lambda = 0.01386.
a. What is the probability that the distance is at most 100m?
b. What is the probability that distance exceeds the mean distance by more than 2 standard deviations?
The part in bold is where I am having struggles. I've tried the following.
mean and standard deviation both = 72.15
$P(X > \mu\text{ by more than two }\sigma) = 1 – P(X > \mu + \sigma) = 1 – (72.15*2)$
I get the feeling this is wrong however. Can someone help me?
Best Answer
The mean of $X$ is $\frac{1}{\lambda}$, and the variance of $X$ is $\frac{1}{\lambda^2}$. So $X$ has standard deviation $\frac{1}{\lambda}$.
To say that $X$ exceeds the mean by more than $2$ standard deviation units is to say that $X\gt \frac{1}{\lambda}+2\cdot \frac{1}{\lambda}=\frac{3}{\lambda}$.
Finally, $$\Pr\left(X\gt \frac{3}{\lambda}\right)=\int_{3/\lambda}^\infty \lambda e^{-\lambda x}\,dx.$$ Integrate. You should get $e^{-3}$.