Question

[4F-13] Suppose that waiting time until the next claim is exponentially distributed with mean of 4 days. For last 2 days, there have been no claims. Calculate the probability that waiting time until the next claim will exceed 4 more days (6 days in total). (Round to 4 decimals)

Answer 1

Answer:

The probability that waiting time until the next claim will exceed 4 more days i.e. P(X>6/X>2) = 0.3679

Step-by-step explanation:

From the given information:

Let consider X to be a random variable that follows an exponential distribution fo the waiting time.

Then; $X \simeq Exp \bigg (\dfrac{1}{4} \bigg )$ since the mean is distributed between 1 - 4 days

Thus:

$F_x(x) = \left \{ {{\dfrac{1}{4}e^{-x/4} \atop {0}} \right ; x > 0 \ or \ otherwise$

The objective is to determine:

$P(X > 6/ X > 2) = \dfrac{P(X >6 \cap X >2)}{P(X > 2)}$

$P(X > 6/ X > 2) = \dfrac{P(X >6)}{P(X > 2)}$

Recall that:

$P(X>a)=\int ^{\infty}_{a}\dfrac{1}{4}e^{-x/4} \ dx$

$P(X>a)= \big ( -e ^{-x/4} \big ) ^{\infty}_{a}$

$\implies P(X>6/X>2) = \dfrac{e^{-6/4}}{e^{-2/4}}$

$P(X>6/X>2) = {e^{-(6/4 - 2/4)}}$

$P(X>6/X>2) = {e^{-4/4}$

$P(X>6/X>2) = {e^{-1}$

$P(X>6/X>2)$ = 0.3678794412

To four decimal places; we have:

The probability that waiting time until the next claim will exceed 4 more days i.e. P(X>6/X>2) = 0.3679