Answer to Question #115089 in Statistics and Probability for Bernice

Question

8. A financial analyst has found out that policyholders are four times as likely to file two
claims as to file four claims. If the number of claims filed has a posson distribution,
what is the variance of the number of claims filed.

The lifetime of a machine is continuous on the interval (0, 40) with probability density
function f, where f(t) is proportional to (t + 10)) 2
, and t is the lifetime in years.
Calculate the probability that the lifetime of the machine part is less than 10 years.
Hint: Show that f(t) is legitimate and find the proportionality constant

Answer 1

1. We want to find $\lambda$ for the Poisson RV from $P(X=2)=4P(X=4)$

$$P(X=k)={e^{-\lambda}\lambda^k\over k!}$$

$$P(X=2)={e^{-\lambda}\lambda^2\over 2!}=4P(X=4)=4\cdot{e^{-\lambda}\lambda^4\over 4!}$$

$$\lambda^2={4!\over4\cdot2!}=3$$

$V(X)=\sigma^2=\lambda=\sqrt{3}$

2.

$$f(t) = \begin{cases} c(t+10)^2 & t\in[0,40] \\ 0 &otherwise \end{cases}$$

$$\displaystyle\int_{-\infin}^{\infin}f(t)dt=1$$

$$\displaystyle\int_{0}^{40}c(t+10)^2dt=c\big[{(t+10)^3\over 3}\big]\begin{matrix} 40 \\ 0 \end{matrix}=$$

$$={c\over 3}((40+10)^3-(0+10)^3)={124000\over 3}c=1$$

$$c={3\over 124000}$$

$$P(t<10)=\displaystyle\int_{0}^{10}{3\over 124000}(t+10)^2dt=$$

$$={3\over 124000}\big[{(t+10)^3\over 3}\big]\begin{matrix} 10 \\ 0 \end{matrix}={1\over 124000}((10+10)^3-(0+10)^3)=$$

$$={7\over 124}\approx0.056452$$