Q. 8.15 - A First Course in Probability

Question

An insurance company has $10,000$automobile policyholders. The expected yearly claim per policyholder is$240$ a standard deviation of$800$. Approximate the probability that the total yearly claim exceeds $2.7$a million.

Step-by-Step Solution

Answer

Approximate the probability that the total yearly claim exceeds $2.7$a million$0.0000884\approx 0$.

Step 1 Given Information.

An insurance company has $10,000$automobile policyholders. The expected yearly claim per policyholder is$240$, with a standard deviation of$800$.

Step 2 Explanation.

Let $X_i$represents the yearly claim perish policyholder,$i=1,2,$\dots,$10, 000$ It is given that and${\sigma }^2=Var\left(X_i\right)={800}^2=640,000$.

Let's $\times$denote the total yearly claim$X=\sum _1^{10,000}{X}_i$.

Because of the independence of random variables${\mathrm{X}}_{-}$, using the corresponding properties of expectation and variance we get$E\left[X\right]=(10,000)\mu =2,400,000$, $Var\left(X\right)=(10,000){\sigma }^2=6,400,000,000$

The probability that the total yearly claim exceeds $2.7$million dollars is

$P\{X>2,700,000\}.$ To approximate this probability we use the central limit theorem and in that case, we get$$\begin{gathered} :P\{X>2,700,000\}==1-P\{X\le 2,700,000\}=1-P\left\{\frac{X-E\left[X\right]}{\sqrt{Var\left(X\right)}}\le \frac{2,700,000-E\left[X\right]}{\sqrt{Var\left(X\right)}}\right\} \\ =1-P\left\{\frac{X-2,400,000}{\sqrt{6,400,000,000}}\le \frac{2,700,000-2,400,000}{\sqrt{6,400,000,000}}\right\}\approx 1-\Phi (3.75) \\ \text{ software package }1-.9999116=\mathbf{0}\mathbf{.}\mathbf{0000884}\mathbf{\approx }\mathbf{0} \end{gathered}$$