p sample, july 12

$ \gdef\and{\cap} \gdef\or{\cup} \gdef\bar#1{\overline{#1}} $

See here.

Well, that's pleasing.

349. Passing two exams.

Each student in a group will take an exam in January and another in February. While 70% of the students will pass the January exam, only 50% will pass the February exam. Students who pass the January exam are twice as likely to pass the February exam as those who fail the January exam.

Calculate the probability that a randomly selected student will pass both exams.

This threw me briefly. I wanted to write

$$\begin{aligned} P(F|J) + P(F|\bar J) &\neq P(F) \end{aligned}$$

instead of the correct

$$\begin{aligned} P(F|J)P(J) + P(F|\bar J)P(\bar J) &= P(F). \end{aligned}$$

Fortunately my answer wasn't among the distractors.

416. Many dental policies.

A company has 1000 dental insurance policies. The number of claims filed by a policyholder during one year is a Poisson random variable with variance 1. The number of claims filed by these policyholders are mutually independent.

The payment for each dental claim is 100 and the annual premium for each policy is 103% of the total expected claim payments for that policy.

Calculate the probability, using the normal approximation, that the total claim payments on the 1000 policies exceeds the total premium collected.

I'm really flummoxed by this. With a variance of one claim-squared per year, the mean should be one claim per year, the expected payout should be $\$100$, and the premium should be $\$103$. This gives $1000\pm\sqrt{1000}$ claims per year across the entire pool. The hypothesis that the actual claims are under $\$1030$ corresponds to a $z$-score of $z=\frac{30}{31.6}$,, and we have $F(0.9487)\approx 0.8289.$ So the probability of excess should be $1-F = 0.1711.$ But the options are

$$\begin{aligned} 0.001 && 0.159 && 0.167 && 0.488 && 0.500 \end{aligned}$$

The closest approximation is right, but I don't follow the reasoning.

Solution manual: the continuity correction.

The difference is the continuity correction: the probability that there are at least thirty extra claims has $z$-score $\frac{30.5}{\sqrt{1000}}$, which accounts for the difference.

Let's reverse-engineer some of these answers:

$$\begin{aligned} 1 - F(z): && 0.001 && 0.159 && 0.167 && 0.488 && 0.500 \\ F(z): && 0.999 && 0.841 && 0.833 && 0.512 && 0.500 \\ z : && 3.62 && 1.00 && 0.965 && 0.03 && 0.00 \\ \end{aligned}$$

I'm sure each of these has some logic, but I don't know what it is.

393. Mode of several payments

A homeowner purchases a policy from an insurance company covering losses from hurricanes and fires. Under the policy, the insurance company pays 1000 for each loss.

In each year, the number of hurricanes is Poisson distributed, with a common mean for all years. Similarly, for each year, the number of fires is also Poisson distributed, with a common mean for all years. A hurricane occurs on average once every 10 years, while a fire occurs on average once every 50 years. The numbers of hurricanes and numbers of fires in different years are all mutually independent.

Let T be a random variable representing the total payments made by the insurance company to the homeowner over the next 40 years.

Calculate the mode of T.

I wound up programming my calculator to make a table of Poisson probabilities, with a loop like

    0 → A
    e^(-4) * 4^A / A!
    A+1 → A

and then cycling through the up arrow and taking notes. I ended up with

Then I found some products, ending up with

I was confused by the two identical rows. But after I started writing this down, I realized that I needed to take sums of the probabilities along the diagonals of the table:

I had guessed that four payments would be more common than three, but I didn't find the reasoning until later.

394. An exponential loss, with a deductible.

The loss, X, subject to reimbursement under an insurance policy, has density function

$$\begin{aligned} f(x) &= \begin{cases} \frac{1}{\beta} e^{-(x-d)/\beta} & \text{for } x \geq d \\ 0 & \text{otherwise} \end{cases} \end{aligned}$$

where d is the deductible, and β is a positive constant.

Determine the absolute value of the difference between the mode of X and the 10th percentile of X.

Oh, I see what I've done here. We have $\text{mode}(x) = d$, because of the exponential. The integral for the tenth percentile gives

$$\begin{aligned} \frac {9}{10} &= e^{-(x_{10}-d)/\beta} &&= \int_{x_{10}}^\infty \frac{1}{\beta} e^{-(x-d)/\beta} \mathrm dx \\ \ln\frac{9}{10} &= -\frac{x_{10}-d}{\beta} \\ d + \beta\ln\frac{10}{9} &= x_{10} \end{aligned}$$

But I just picked the tenth percentile $x_{10}$ from the list of answers, rather than subtracting off the mode.