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Homework Help: Median of f(x) - quadratic

  1. Sep 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Given that:

    f(x) = 2(1-x) for 0 < x < k
    ...........0 for otherwise

    sketch the graph of f(x) and find:

    (i) the value of k,

    (ii) the median of x

    2. Relevant equations



    3. The attempt at a solution

    (i) I sketched the graph and got a line from (0,2) to (1,0) which has an area of 1, so k = 1.

    (ii) [tex]^{}M[/tex][tex]_{}0[/tex][tex]\int[/tex](2 - 2x) = 1/2
    The code is messing up but intergation of (2 - 2x) with m (median) as the top limit and 0 being the bottom one and it = 1/2

    [2x - x^2]m0 = 1/2

    2m - m2 = 1/2

    I'm stuck on getting m separated as it's a negative square

    Help me please :(
     
  2. jcsd
  3. Sep 30, 2010 #2

    statdad

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    Homework Helper

    Try arranging the equation so all terms are on one side.
     
  4. Sep 30, 2010 #3

    HallsofIvy

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    Science Advisor

    You don't really need to integrate to get this. y= 2(1- x) is a straight line from (0, 2) to (k, 2(1- k)).

    If k= 1, the area under it is the area of the triangle with height 2 and base 1 and so area (1/2)(2)(1= 1. That is what we need to have in order to have a valid probability distribution.

    As a check, if k< 1, the area under the line is the area of the trapezoid with bases 2 and 2(1- k) and height k. It's area is k(2+ 2(1- k))/2= k(4- 2k)/2= k(2- k) and we want that equal to 1. [itex]k(2- k)= 2k- k^2= 1[/itex] is the same as [itex]k^2- 2k+ 1= (k- 1)^2= 0[/itex]. As before, it reduces to k= 1, the triangle case.
    (You say "I sketched the graph and got a line from (0,2) to (1,0)". Well, you couldn't know it would be a line from (0, 2) to (1, 0) until after you knew that k= 1.)

    Now the median is where the area both left and right is equal to 1/2. If we take that to be at "x", then we have a triangle on the right with base 1- x and height 2(1- x) which has area [itex](1/2)(1-x)(2(1-x))= (1- x)^2[/itex] and we want that equal to 1/2: [itex](1- x)^2= 2[/itex]. Solve that for x.

    If you really want to integrate, you are integrating
    [tex]\int_0^x 2(1- t)dt= 2 \int_0^x 1- t dt= 2\left(x- (1/2)x^2)= 1/2[/itex]] which gives the same result.

    You can simplify that integral by letting u= 1- t so that du= -dt, when t= 0, u= 1 and when t= x, u= 1- x. Now the integral is [itex]2\int_0^{1- x}u^2du= 1/2[/=itex], giving, again, the same result.
     
  5. Sep 30, 2010 #4
    Thanks for the input to both of you :)

    I understand now thanks, HallsOfIvy :smile: woo!
     
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