p sample, june 30

See here.

585: a continuous density function

The random variable $Y$ has the density function $$ f(y) = \begin{cases} \frac{y}{6} - \frac{y^2}{36} & \text{for } 0 < y < 6; \\ 0 & \text{elsewhere.} \end{cases} $$ Calculate $P(Y\in(1,3) | Y\in(2,4))$.

I was in discrete-probability mode and I just didn't understand that this was a continuous question. (I thought initially that there was an error in the question; a dangerous assumption.) By the time I got it I was several minutes in, and I did the integral wrong.

202: more than three phone calls

A representative of a market research firm contacts consumers by phone in order to conduct surveys. The specific consumer contacted by each phone call is randomly determined. The probability that a phone call produces a completed survey is 0.25.

Calculate the probability that more than three phone calls are required to produce one completed survey.

Oh, I made a fencepost error. I computed the probability for "three or more calls," but the question asked for "more than three calls."

251: independent fire cost estimates

Two independent estimates are to be made on a building damaged by fire. Each estimate is normally distributed with mean $10b$ and variance $b^2$.

Calculate the probability that the first estimate is at least 20 percent higher than the second.

This is some error-function integral, which I guesstimated at a hair under 0.24, so I chose 0.221 (without confidence).

Apparently the difference between two normally-distributed variables is also normally distributed (which I should have remembered, but okay). The scenario where $X > 1.2Y$ corresponds to the scenario where $W = X-1.2Y$ is positive. The random variable $W$ has

• mean $-2b$
• variance $V(X) + (1.2)^2 V(Y) = 2.44 b^2$

The scenario $W>0$ therefore has significance $z=\frac{+2b}{\sqrt{2.44} \cdot b}$, which apparently has probability $P(z>1.28) = 0.100$. There's a reference table of these someplace that I didn't have access to, but I would have looked in the wrong place anyway.

Let's find that reference table today: it's here.

459: a pool of Poisson-distributed claims

For its group life policies, an insurer models the number of claims per group as independent Poisson random variables with common mean 16.

The insurer randomly selects 64 of its groups.

Calculate the probability that the average number of claims per group is between 15 and 18.

From my physics approach. Each group should have $16\pm4$ claims, so the standard error on the mean of all groups is $\frac{4}{\sqrt{64}} = \frac 12$. That means the range 15–17 is already a 95\% confidence interval, but I guessed that the probability didn't round all the way to 1.00.

Again, a better answer from the table of $z$-scores.