Answer:
a) [tex]f(x)=\frac{2}{3}e^{-\frac{2}{3}x}[/tex] when [tex]x\geq 0[/tex]
[tex]f(x)=0[/tex] otherwise
b) [tex]P(X<2)=0.2636[/tex]
Step-by-step explanation:
First of all we have a Poisson process with a mean equal to :
μ = λ = [tex]\frac{2}{3}[/tex] (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.
[tex]f(x)=a(e)^{-ax}[/tex]
With
[tex]x\geq 0[/tex]
and
[tex]a>0[/tex]
[tex]f(x)=0[/tex] Otherwise
In this exercise λ= a = [tex]\frac{2}{3}[/tex] ⇒
[tex]f(x)=(\frac{2}{3})(e)^{-\frac{2}{3}x}[/tex]
[tex]x\geq 0[/tex]
[tex]f(x)=0[/tex] Otherwise
That's incise a)
For b) [tex]P(X>2)[/tex] We must integrate between 2 and ∞ to obtain the probability or either use the cumulative probability function of the exponential
[tex]P(X\leq x)=0[/tex]
when [tex]x<0[/tex]
and
[tex]P(X\leq x)=1-e^{-ax}[/tex] when [tex]x\geq 0[/tex]
For this exercise
[tex]P(X\leq x)=1-e^{-\frac{2}{3}x}[/tex]
Therefore
[tex]P(X>2)=1-P(X\leq 2)[/tex]
[tex]P(X>2)=1-(1-e^{-\frac{2}{3}.2})=e^{-\frac{4}{3}}=0.2636[/tex]
