1. ## Local Weather 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 to break up the hour into the following intervals:

0, (0, 10), 10, (10, 25), 25, (25, 30), 30, (30, 45), 45, (45, 55), 55, (55, 60)
The largest interval is 15 minutes, so the wait time can range from 0 to 15 minutes.
Not sure how to proceed from here on. Can someone help?

2. ## Re: Local Weather Forecast

Hey JohnDoe2013.

One thing I suggest you add is the time of the fore-cast since it won't be instantaneous. Your idea of breaking up the time is definitely the right approach.

You should take wait all the time to distributions and average over them for each waiting time to get your final waiting time distribution.

As an example if you had say two distributions (0,5) and (0,10) the total would be 15 and averaging over (0,5) gives (5+5)/15 = 10/15 for (0,5) and 5/15 for (5,10) resulting in a probability density function P(0 < T < 5) = 10/15 = 2/3 and P(10 < T < 15) = 1/3.

3. ## Re: Local Weather Forecast

Hi Chiro,

I figured it out. The wait intervals are 10, 15, 5, 15, 10 and 5 minutes respectively. So, in hours, we have:

2 intervals of 1/12 hour
2 intervals of 1/6 hour (2 times as likely as the interval of 1/12 hour)
2 intervals of 1/4 hour (3 times as likely as the interval of 1/12 hour)

So I end up with 3 uniform distributions:

f(x) = 6 for $\displaystyle 0 \le x < \frac{1}{12}$

f(x) = 4 for $\displaystyle \frac{1}{12} \le x < \frac{1}{6}$

f(x) = 2 for $\displaystyle \frac{1}{6} \le x < \frac{1}{4}$

which I then combine into 1.