Results 1 to 2 of 2

Math Help - continuous uniform distributions

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    560

    continuous uniform distributions

    Can someone please help with part 'c' and 'd'?

    for part 'a' I got:  1.6 . \frac{1}{20} = 0.08

    'b': mean =10
    variance =  \frac{(20-0)^{2}}{12} = \frac{100}{3}

    for part 'c', I cant understand what the question is asking, is it that they want to find the probability of the 'dot appearing in a square, of side 4cm, in the middle of the screen? I know 'X' represents the distance of the dot from the left edge of the screen, and Y the distance from the bottom edge, so would it just be multiplying the both?

    Also need help with part 'd', the question says 'within 2cm of the edge of the screen', should it not also tell me what 'edge' they are talking about?

    Any help appreciated,
    thank you.
    Attached Thumbnails Attached Thumbnails continuous uniform distributions-tt.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Tweety View Post
    Can someone please help with part 'c' and 'd'?

    for part 'a' I got:  1.6 . \frac{1}{20} = 0.08

    'b': mean =10
    variance =  \frac{(20-0)^{2}}{12} = \frac{100}{3}

    for part 'c', I cant understand what the question is asking, is it that they want to find the probability of the 'dot appearing in a square, of side 4cm, in the middle of the screen? I know 'X' represents the distance of the dot from the left edge of the screen, and Y the distance from the bottom edge, so would it just be multiplying the both?

    Also need help with part 'd', the question says 'within 2cm of the edge of the screen', should it not also tell me what 'edge' they are talking about?

    Any help appreciated,
    thank you.
    Since the distribution of the dot in each direction is uniform and the directions are independent you can do it geometrically: Probability = (Area of required region)/(Total area).

    By the way, it would be more helpful if you typed the question rather than attaching it (there is nothing in the question that makes that difficult to do).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. uniform distributions
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: April 16th 2011, 09:59 PM
  2. Exponential and Uniform distributions
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: January 27th 2011, 05:34 PM
  3. continuous uniform distributions
    Posted in the Statistics Forum
    Replies: 3
    Last Post: April 7th 2010, 04:49 PM
  4. Uniform Distributions.. plz help
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: October 26th 2009, 03:58 AM
  5. Uniform Continuous and Uniform Convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2007, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum