Question

Telephone calls arrive at a doctor’s office according to a Poisson process on the average of two every 3 minutes. Let X denote the waiting time until the first call that arrives after 10 a.m.
(a) What is the pdf of X?
(b) Find P(X > 2).

Answer 1

Answer:

a) $f(x)=\frac{2}{3}e^{-\frac{2}{3}x}$ when $x\geq 0$

$f(x)=0$ otherwise

b) $P(X$

Step-by-step explanation:

First of all we have a Poisson process with a mean equal to :

μ = λ = $\frac{2}{3}$ (Two phone calls every 3 minutes)

Let's define the random variable X.

X : ''The waiting time until the first call that arrives after 10 a.m.''

a) The waiting time between successes of a Poisson process is modeled with a exponential distribution :

X ~ ε (λ)    Where λ is the mean of the Poisson process

The exponential distribution follows the next probability density function :

I replace λ = a for the equation.

$f(x)=a(e)^{-ax}$

With

$x\geq 0$

and

$a>0$

$f(x)=0$ Otherwise

In this exercise λ= a = $\frac{2}{3}$ ⇒

$f(x)=(\frac{2}{3})(e)^{-\frac{2}{3}x}$

$x\geq 0$

$f(x)=0$ Otherwise

That's incise a)

For b) $P(X>2)$ We must integrate between 2 and ∞ to obtain the probability or either use the cumulative probability function of the exponential

$P(X\leq x)=0$

when $x$

and

$P(X\leq x)=1-e^{-ax}$ when $x\geq 0$

For this exercise

$P(X\leq x)=1-e^{-\frac{2}{3}x}$

Therefore

$P(X>2)=1-P(X\leq 2)$

$P(X>2)=1-(1-e^{-\frac{2}{3}.2})=e^{-\frac{4}{3}}=0.2636$