## Related questions with answers

The probability distribution for damage claims paid by the Newton Automobile Insurance Company on collision insurance follows.

$\begin{matrix} \text{Payment (\$)} & \text{Probability}\\ \text{0} & \text{0.85}\\ \text{500} & \text{0.04}\\ \text{1000} & \text{0.04}\\ \text{3000} & \text{0.03}\\ \text{5000} & \text{0.02}\\ \text{8000} & \text{0.01}\\ \text{10000} & \text{0.01}\\ \end{matrix}$

Use the expected collision payment to determine the collision insurance premium that would enable the company to break even.

Solutions

VerifiedThe expected collision payment is equal to:

$E(x)=\sum x\cdot f(x)$

In this exercise, $x$ is equal to payment and $f(x)$ is equal to probability.

The expected collision payment is $E(x)=\mu$:

$\begin{align*} E(x)=&\sum xf(x)\\ =&0\times 0.85+500\times 0.04+1000\times 0.04+3000\times 0.03\\ &+5000\times 0.02+8000\times 0.01+10000\times 0.01\\ =&0+20+40+90+100+80+100\\ =&430 \end{align*}$

The collision insurance premium that would enable the company to break even is $\$430$.

In this exercise, we determine the expected value of the given probability distribution.

*What is the expected value? How can it be derived?*

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