Determine probabilities involving exponential distribution

[s3a]

Homework Statement

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.

Correct answers:
(a) 0.3679
(b) 0.1353
(c) 0.9502
(d) 29.96

Homework Equations

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

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!

 

[LCKurtz]

Homework Statement

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.

Correct answers:
(a) 0.3679
(b) 0.1353
(c) 0.9502
(d) 29.96

Homework Equations

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

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!

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

 

[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?

 

[Ray Vickson]

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?

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.