Suppose the random variable $X$ represents the amount of damage done to a rusty Pontiac Sunfire during the 2007-2007 year. Ninety-two percent of the time, the vehicle will sustain no damage. Otherwise we model the damage to be uniformly distributed on the interval [500,3000].
If the insurance were purchased with a $1000$ deductible, calculate the average insurance payment.
My thinking was this, there is a $.08$ chance that damage will occur so that means $Y=.08X$. But $X=\dfrac{1}{2500} \;\text{if}\; 500\leq x\leq 3000$. So the expected amount should be:
$0.08\cdot \bigg[\displaystyle\int_{500}^{3000}\dfrac{x}{2500} – \displaystyle\int_{500}^{1000}\dfrac{x}{2500}\bigg]$ but this isn't the answer. Whats wrong with my logic?
Best Answer
The insurance company payout is $0$ if damage is from $0$ to $1000$, and $X-1000$ if $1000\le X\le 2000$.
The probability of a payout is $(0.08)\frac{3000-1000}{2500}$. Thus with probability $0.064$ there is a payout $Y$, uniformly distributed between $0$ and $2000$. It follows that $$E(Y)=(0.064)\int_{0}^{2000}\frac{y}{2000}\,dy.$$
Remark: The calculation in the OP does not take account of the fact that if, for example, the damage is $1400$, then the insurance company only pays out $400$.