p sample, july 11

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

See here.

Better today than yesterday.

596. Calculate a normal distribution's variance from some values of $F(x)$.

An insurance company’s annual profit is normally distributed. The probability that the annual profit exceeds 3.50 is 0.3264 and the probability that it exceeds 3.62 is 0.2743.

Calculate the variance of the insurance company’s annual profit.

My try

My table of $F(x)$ is for values above 0.5, so I did the lookup for $1-F(x)$:

$$\begin{aligned} 0.3264 &= 1-0.6736 &&= F(z=+0.45) \\ 0.2743 &= 1-0.7257 &&= F(z=+0.60) \\ \end{aligned}$$

These $z$-scores correspond to

$$\begin{aligned} +0.45 &= \frac{3.50-\mu}{\sigma} \\ +0.60 &= \frac{3.62-\mu}{\sigma} \\ \end{aligned}$$

Multiply both sides and subtract the two equations:

$$\begin{aligned} +0.60\sigma &= +(3.62-\mu) \\ -0.45\sigma &= -(3.50-\mu) \\ \hline 0.15\sigma &= 0.12 \\ \sigma &= \frac{0.12}{0.15} = 0.8 \end{aligned}$$

This was option C, but the correct answer was $0.64$ — which is of course $\sigma^2$, the variance. Argh.

597. Uniformly distributed losses with deductible.

A theft insurance company offers two policies, A and B. Under each policy, a loss is uniformly distributed on the interval [a, 19].

For a loss under policy A, the deductible is 0 and the expected value of the claim payment is 10.

For a loss under policy B, the deductible is 4.

Calculate the expected value of the claim payment for a loss under policy B.

My try

With zero deductible, the payment and the loss are the same. In order to have $E(\text{payment}) = 10$, must have $a=1$. If the deductible is 4, the expected loss is still 10, so the payment should be six.

But I suspect there's some conditional magic here, because small claims don't get filed.

Solution manual

I'm correct that $a=1$.

For policy $B$, the expected value is

$$\begin{aligned} \int_4^{19} \frac{x-4}{18} \mathrm dx &= \left.\frac{(x-4)^2}{36}\right|_4^{19} \\&= \frac{15^2-0^2}{36} = 6.25. \end{aligned}$$

I guess what I did was to integrate starting from $a=1$ rather than from $d=4$.

215. Median of exponential losses.

Losses under an insurance policy are exponentially distributed with mean 4. The deductible is 1 for each loss.

Calculate the median amount that the insurer pays a policyholder for a loss under the policy.

My take

I did

$$\begin{aligned} \frac12 &= \int_1^\infty \frac{1}{\tau} e^{-t/\tau} \mathrm dt \\ -\ln 2 &= -\frac{1}{\tau} \\ \tau &= \frac{1}{\ln 2} = 1.44, \end{aligned}$$

which was not an option. I guessed correctly at 1.77, the smallest option.

The solution manual.

Let $X$ represent the amount of the loss. The median loss is

$$\begin{aligned} \frac12 &= \int_m^\infty \frac14e^{-x/4}\mathrm dx &&= e^{-m/4} \\ \ln 2 &= \frac m4 \\ m &= 4\ln 2 = 2.77 \\ \end{aligned}$$

The loss exceeds 2.77 if and only if the claim payment exceeds 1.77, which is therefore the median claim payment.

Worth learning, apparently:

If I have an exponential distribution $\mathrm dP = \frac{1}{x_0} e^{-x/x_0} \mathrm dx$ with mean $x_0$, the median is $x_0 \ln 2$.

This is the same as the difference between half-life versus mean life.