# Determine probabilities involving exponential distribution

1. Feb 13, 2015

### s3a

1. The problem statement, all variables and given/known data
Problem(s):
Suppose that X has an exponential distribution with mean equal to 10.

Determine the following:
(a) P(X > 10)
(b) P(X > 20)
(c) P(X < 30)
(d) Find the value of x such that P(X < x) = 0.95.

(a) 0.3679
(b) 0.1353
(c) 0.9502
(d) 29.96

2. Relevant equations
Exponential distribution: f(x) = lambda * exp(-lambda*x) when x > 0 and 0 elsewhere (always assuming lambda > 0)

3. The attempt at a solution
To be honest, I'm extremely confused, and I'm stuck at part (a).

What I'm doing is

P(X > 10) = 1 - P(X <= x)
P(X > 10) = 1 - integral of lambda * exp(-lambda*x) from 0 to 10 (I am integrating because I want the probability density function to be a cumulative density function)
P(X > 10) = 1 - -[exp(-10*10) - exp(0)]
P(X > 10) = 1 - -[exp(-100) - 1]
P(X > 10) = 1 + [exp(-100) - exp(0)]
P(X > 10) = 1 + exp(-100) - exp(0)
P(X > 10) = 1 + exp(-100) - 1
P(X > 10) = exp(-100)

P(X > 10) = 3.72007597602083596296e-44 (which is not 0.3679)

Any help in solving this problem would be GREATLY appreciated!

2. Feb 13, 2015

### LCKurtz

Your problem is that if the mean is $10$, then $\lambda = \frac 1 {10}$, not $\lambda = 10$.

3. Feb 13, 2015

### s3a

Oh! So, my work was not complete nonsense! :D

Mu is often the letter used to represent the mean, but what does lambda represent, though?

4. Feb 13, 2015

### Ray Vickson

It is an average rate. If we have an "arrival process" whose times of arrivals are the random epochs $0 = T_0, T_1, T_2, T_3, \ldots$, and the successive interarrival times $X_1 = T_1 - T_0, X_2 = T_2 - T_1, X_3 = T_3 - T_2, \ldots$ are independent and exponentially distributed with parameter $\lambda$, then the (random) number of arrivals in a time interval of length $L$ is Poisson with mean $\lambda L$. That is, the expected number of arrivals in time $L$ is $\lambda L$, so $\lambda$ is the expected number of arrivals per unit time = expected arrival rate.