Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By chiro

Math Help - Local Weather Forecast

  1. #1
    Junior Member
    Joined
    Oct 2013
    From
    Canada
    Posts
    31
    Thanks
    8

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,694
    Thanks
    618

    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.
    Thanks from JohnDoe2013
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2013
    From
    Canada
    Posts
    31
    Thanks
    8

    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 0 \le x < \frac{1}{12}

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

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

    which I then combine into 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. proving local maximum and local minimum
    Posted in the Calculus Forum
    Replies: 8
    Last Post: October 27th 2013, 03:27 PM
  2. Replies: 4
    Last Post: March 21st 2011, 01:23 AM
  3. Replies: 6
    Last Post: January 5th 2011, 02:34 AM
  4. best forecast
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 24th 2010, 09:59 PM
  5. forecast and fine
    Posted in the Statistics Forum
    Replies: 0
    Last Post: January 24th 2010, 01:47 PM

Search Tags


/mathhelpforum @mathhelpforum