# Exponential distribution problem

1. Oct 14, 2015

### toothpaste666

1. The problem statement, all variables and given/known data
The amount of time that a surveillance camera will run without having to be reset is a random variable having the exponential distribution with beta = 50 days. find the probabilities that such a camera will
a) have to be reset in less than 20 days
b) not have to be reset in at least 60 days

3. The attempt at a solution
first we integrate the function
f(x) = (1/beta)e^(-x/beta) = (1/50)e^(-x/50)
after integrating we get
-e^(-x/50) evaluated from a to b
for part a)
we are evaluating from 0 to 20 so we get
-(e^(-20/50) - 1) = 1 - e^(-20/50) = .3297

for part b) we evaluate from 60 to ∞
-(0 - e^(-60/50)) = e^(-60/50) = .3012

is this method correct?

2. Oct 14, 2015

### andrewkirk

It looks right to me.

3. Oct 14, 2015

### Staff: Mentor

It looks OK to me as well.
Another way to do part b is this: $Pr(x \ge 60) = 1 - Pr(0 \le x < 60)$
So $\frac 1 {50}\int_{60} ^{\infty} e^{-x/50}dx = \frac 1 {50}\left(1 - \int_0^{60}e^{-x/50}dx\right)$

4. Oct 17, 2015

### toothpaste666

ok great thank you!

5. Oct 17, 2015

### Ray Vickson

Yes.

Here are a small number of handy facts about the exponential distribution---worth committing to memory (and pretty easy to derive for yourself).

If $X$ has distribution $\text{Exp}(\lambda)$ (so that its pdf on $\{ x > 0 \}$ is $\lambda e^{-\lambda x}$) then:
$$\begin{array}{cl} 1)& EX = \displaystyle \frac{1}{\lambda} \\ 2)&\text{Var} X = \displaystyle \frac{1}{\lambda^2} \\ 3)& \text{Coefficient of variation} \equiv \displaystyle \frac{\text{standard deviation}}{\text{mean}} = 1 \\ 4) & P(X > x) = e^{- \lambda x} , \; x \geq 0 \end{array}$$
From 4) it follows that $P(X \leq x) = 1 - e^{- \lambda x}$ for $x \geq 0$. Also, remembering 3) is a convenient way of remembering 2).