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An insurance company found that 2.5% of male drivers between the ages of 18 to 25 are involved in serious accidents annually. To simplify the analysis, assume (artificially) the following: 1) every such accident costs the insurance company \$50,000; 2) a driver can only have one of these accidents in a year; 3) the insurer charges $2,000 for a policy; 4) the accidents occur independently of each other.

Suppose that the company writes 1,000 such policies to a collection of drivers as described above. Describe the total payments from these policies by the random variable Y. What are the expected value and SD of Y?

So far I have that E(Y) = (25)(-50,000) + (975)(2000) = 700,000. Is this right, and how do I find the SD? Also, from the insurer's point of view, is the expected value of an individual policy = (2000) - (50,000 x .025) = 750?

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  • $\begingroup$ You should add the self-study tag $\endgroup$ – kjetil b halvorsen Dec 1 '17 at 4:56
  • $\begingroup$ The number of policies out of 1000 that result in a claim follows a binomial distribution (assuming all policies are independent). The total payout is a simple transformation of the number of claims, and the mean and SD of a binomial distribution are known, so the mean and SD of the payout is easily derived. $\endgroup$ – Gordon Smyth Dec 1 '17 at 5:03
  • $\begingroup$ Your expected value for an individual policy is right but your E(Y) is not. Hint: the latter should be 1000 times the former! $\endgroup$ – Gordon Smyth Dec 1 '17 at 5:19
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In my lay perspective of probability, well the expectation and variance are given by (this question is discrete case) $$E(Y) = \sum_xxp(x), ~~~~~~V(Y) = E\big[[Y-E(Y)]^2\big]$$

Given accident occurs independently, we could calculate the expectation by $$E(Y)=1000 \cdot(0.025)\cdot(-50000)+1000\cdot(2000)=750000$$ , since all the 1000 people, including the 25 people expected to have accident, and the 975 people expected to be safe, purchase the policy.

Then, for the variance, we first have the expected value of an individual policy is 750. Therefore, $$V(Y) = E\big[[Y-750]^2\big] = (0.025)\cdot(-50000+750)^2+(0.975)\cdot750^2=61187500$$ Taking the square root, $$SD(Y) = \sqrt{V(Y)}=7822.244$$

This variance is for individual, not for the whole 1000 people, we have to be extremely careful about it. And, double check the result since I am not 100% sure. Finally, sometime we have an alternative method to write variance:

$$V(Y) = E(Y^2)-[E(Y)]^2=\sum_xx^2p(x)-\Big[\sum_xxp(x)\Big]^2$$

Hope it helps.

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