Find the density function of the wait time of a local forecast A weather channel has the local forecast on the hour and at 10, 25, 30, 45, and 55 minutes past. Suppose that you wake up in the middle of the night and turn on the TV and let X be the time you have to wait to see the local forecast, measured in hours. Find the density function of X. My thoughts are: Since the largest interval is 15 minutes or 0.25 hour, $0 \le X \le 0.25$ The average number of arrivals (i.e. forecasts) are 6 per hour or $\frac{6}{4}$ every 0.25 hour. Is this a Poisson Distribution?

No it is not a Poisson distribution, which is discrete, takes integer values, and is potentially unlimited in value. This question is a continuous distribution and cannot take a value greater than a $\frac14$ of an hour.

Suppose the forecast was only on the hour. Then the distribution of waiting times to the start of the next forecast would essentially be uniform on the interval $[0,1)$.

The intervals are in fact $\frac16, \frac14, \frac{1}{12}, \frac14, \frac16, \frac{1}{12}$ of hours. So the distribution is uniform on each of the equal-length waiting time intervals $\left[0, \frac{1}{12}\right), \left[\frac{1}{12}, \frac16\right), \left[\frac16, \frac14\right)$ but you are three times as likely to have a waiting time in the first as in the third, and twice as likely in the second than the third.

I will leave it to you to adjust the three densities to give a total probability of $1$; it is not difficult, especially if you sketch the densities.