Question

A continuous random variable, X, whose probability density function is given by f(x) = ( λe−λx , if x ≥ 0 0, otherwise is said to be an exponential random variable with parameter λ denoted by X ∼ Exp(λ) where λ > 0.

(a) Find the c.d.f of X.
(b) Suppose that the length of a phone call in minutes is an exponential random variable with parameter λ = 1 10 . If someone arrives immediately ahead of you at a public telephone booth, find the probability that you will have to wait between 10 and 20 minutes

Answer 1

Answer:

a) $F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x}$

b) $P(10 < X<20)=e^{-0.1*10} -e^{-0.1*20}=0.368-0.135=0.233$

Explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution".

Part a

Let X the random variable of interest. We know on this case that $X\sim Exp(\lambda)$

And we know the probability denisty function for x given by:

$f(x) = \lambda e^{-\lambda x} , x\geq 0$

In order to find the cdf we need to do the following integral:

$F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x}$

Part b

Assuming that $X \sim Exp(\lambda =0.1)$, then the density function is given by:

$f(x) = 0.1 e^{-0.1 x} dx , x\geq 0$

And for this case we want this probability:

$P(10 < X<20) = \int_{10}^{20} 0.1 e^{-0.1 x} dx = -e^{-0.1 x} \Big|_{10}^{20}$

And evaluating the integral we got:

$P(10 < X<20)=e^{-0.1*10} -e^{-0.1*20}=0.368-0.135=0.233$