# Standard deviation with exponential distribution

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?

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 $e^{-3}$.
• Well, you may have been told that the probability that $X\le x$ is $1-e^{-\lambda x}$. So the probability that $X\gt x$ is $1-(1-e^{-\lambda x})$, which is $e^{-\lambda x}$. Now put $x=\frac{3}{\lambda}$. So if you have been given the formula for $\Pr(X\le x)$, no integration is needed. – André Nicolas Apr 5 '14 at 18:36
• No mistake, at least on your part. Thank you! The rest is right, the answer is $e^{-3}$. – André Nicolas Sep 11 '14 at 1:53