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I usually find these sort of questions straightforward but this was worded pretty vaguely as I haven't really come across anything like it.

I answered all the questions but I'm not sure about the accuracy of the answers.

I was just wondering about the correct method to go about attempting such questions and if I was on the correct track.

The American Charted Insurance Company finds that 0.02% of the population (from which its policy holders are drawn) dies from a certain disease each year. The company insures 10000 people against this disease. Analysts in the company have decided that a Poisson distribution is appropriate to model the occurrence of the disease.

i) What is the mean number of people insured against this disease with British Chartered who will die from the disease in:

a) the next year?

$$10000 \times 0.0002 = 2$$

b) the next two years?

$$10000 \times 0.0004 = 4$$

ii) Find the probability that British Chartered will pay out on more than two cases of death in:

a) the next year

$$P(X=r) = m^r \times \frac{e^{-m}}{r!}$$ $$P(X=r) = 2^1 \times \frac{e^{-2}}{1!}$$ $$P(X=r) = 0.271$$

b) the next two years

$$P(X=r) = m^r \times \frac{e^{-m}}{r!}$$ $$P(X=r) = 4^2 \times \frac{e^{-4}}{2!}$$ $$P(X=r) = 0.147$$

c) the next five years

$$P(X=r) = m^r \times \frac{e^{-m}}{r!}$$ $$P(X=r) = 10^5 \times \frac{e^{-10}}{5!}$$ $$P(X=r) = 0.038$$

Thanks in advance!

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  • $\begingroup$ In (ii), you say you're calculating $P(X=r)$ (where for some unknown reason you seem to be taking $r$ to be the number of years). But the question asks about $P(X > 2)$. $\endgroup$ Mar 14, 2012 at 22:42

1 Answer 1

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The formula $$P(X=r)=m^r\cdot{e^{-m}\over r!}$$ gives the probability that exactly $r$ deaths occurred where $m$ is the average number of deaths over the time period considered.

In part b), you want to calculate the probability that more than 2 deaths occurred in two years. Here, $m=4$ (two years and on average two deaths per year). More than 2 could be exactly 3, or exactly 4, or ...

So, you need to find $P(X>2)=\sum\limits_{r=3}^\infty P(X=r)$. Using the formula to compute the probabilities that 3, 4, 5, ... deaths occurred and adding would be a pain.

The easiest way to find $P(X>2)$ is to use the formula $$ P(X>2) = 1-\bigl(\,P(X=0)+P(X=1)+P(X=2)\,\bigr), $$ which says that the probability sought is 1 minus the probability that $X$ is not more than 2.

To calculate $P(X=0)$ use the formula with $m=4$ and $r=0$: $$ P(X=0) = 4^0\cdot{e^{-4}\over 0!} =e^{-4} $$

To calculate $P(X=1)$ use the formula with $m=4$ and $r=1$: $$ P(X=1) = 4^1\cdot{e^{-4}\over 1!} =4e^{-4} $$ And for $m=4$, $r=2$ $$ P(X=2) = 4^2\cdot{e^{-4}\over 2!} =8e^{-4} $$ Thus $$ P(X>2) = 1-P(X=0)-P(X=1) = 1-e^{-4}-4e^{-4} - 8e^{-4}\approx.7619. $$

Can you handle the other two parts now?

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  • $\begingroup$ Yes I can. Thank you so much! Are my answers for i-a) and i-b) correct tho? $\endgroup$
    – methuselah
    Mar 14, 2012 at 23:14
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    $\begingroup$ @methusaleh You're welcome, glad to help. Yes, you did parts i-a) and b) correctly. I would calculate part i-b) as "two years and two deaths per year on average is 4 deaths on average over two years", though. For ii-c), your $m$ is 10 (five years, two per year), which you stated in part 11-c) correctly. $\endgroup$ Mar 14, 2012 at 23:18

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