1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Continuous random variable

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose the distance X between a point target and a shot aimed at the point in a coin-operated target game is a continuous random variable with pdf

    f(x) = { k(1−x^2), −1≤x≤1
    0, otherwise.

    (a) Find the value of k.

    (b) Find the cdf of X.

    (c) Compute P (−.5 < X ≤ .5).

    (d) Find the expected distance between a point target and a shot aimed.

    3. The attempt at a solution
    a) [itex] k\int_{-1}^1(1-x^2)dx [/itex]

    [itex]= k[\int_{-1}^1dx-\int_{-1}^1x^2dx] [/itex]

    [itex]= k[x\Big|_{-1}^1-\frac{1}{3}x^3\Big|_{-1}^1] [/itex]

    = k(2-2/3) = 1

    k(4/3) = 1

    k = 3/4

    b) [itex] \frac{3}{4} \int_{-1}^X(1-x^2)dx [/itex]

    c) [itex] \frac{3}{4}[x\Big|_{-.5}^{.5}-\frac{1}{3}x^3\Big|_{-.5}^{.5}] [/itex]

    = (3/4)(1-(1/3)[2(1/3)(1/8)])

    = (3/4)(1-1/36)
    = .7292

    d) [itex] \frac{3}{4}\int_{-1}^1x(1-x^2)dx [/itex]

    [itex] =\frac{3}{4}\int_{-1}^1(x-x^3)dx [/itex]

    [itex]= \frac{3}{4}[\int_{-1}^1xdx-\int_{-1}^1x^3dx] [/itex]

    [itex]= \frac{3}{4}[\frac{1}{2}x^2\Big|_{-1}^1-\frac{1}{4}x^4\Big|_{-1}^1] [/itex]

    = 0

    am I doing this right?
  2. jcsd
  3. Oct 19, 2015 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    For b), instead of expressing the cdf as an integral, you should actually carry out the integration and express the cdf as a function of X.
  4. Oct 19, 2015 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Not for (d), no. You computed ##E X##, but what is wanted is ##E |X|##. Also, I get a different answer for (c).
  5. Oct 19, 2015 #4
    for part b)
    [itex] F(X) = \frac{3}{4}[x\Big|_{-1}^X-\frac{1}{3}x^3\Big|_{-1}^X] [/itex]

    [itex] F(X) = \frac{3}{4} [(X+1) - \frac{1}{3} (X^3 +1)] [/itex]

    [itex] F(X) = \frac{3}{4}[X + 1 - \frac{X^3}{3} - \frac{1}{3}] [/itex]

    [itex] F(X) = \frac{3}{4}[X - \frac{X^3}{3} + \frac{2}{3}] [/itex]

    [itex] F(X) = \frac{3X}{4} - \frac{X^3}{4} + \frac{1}{2} [/itex]

    part c) I made a arithmetic error. it comes down to (3/4)(11/12) = .6875

    part d)
    I am not quite sure how to do this. plugging in |x| whereever there is an x gives me the same answer. should I switch the limits of integration from 0 to 2 instead of -1 to 1?
  6. Oct 19, 2015 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No. Without doing any calculations you can see why ##EX = 0##: it is because ##x f(x)## is an odd function on ##[-1,1]## so integrates to zero automatically. That is NOT the case for ##|x| f(x)##, because this not now an odd function on ##[-1,1]##. I won't say any more.
  7. Oct 19, 2015 #6
    d) [itex] \frac{3}{4}\int_{-1}^1|x|(1-x^2)dx [/itex]

    [itex] =\frac{3}{4}\int_{-1}^1(|x|-|x|x^2)dx [/itex]

    [itex]= \frac{3}{4}[\int_{-1}^1|x|dx-\int_{-1}^1|x|x^2dx] [/itex]

    [itex]= \frac{3}{4}[-\int_{-1}^0xdx + \int_0^1xdx-\int_{-1}^0x^3dx + \int_0^1x^3dx] [/itex]

    [itex]= \frac{3}{4}[-\frac{1}{2}x^2\Big|_{-1}^0 +\frac{1}{2}x^2\Big|_0^1-\frac{1}{4}x^4\Big|_{-1}^0 +\frac{1}{4}x^4\Big|_0^1] [/itex]

    = (3/4) [(1/2) + (1/2) + (1/4) + (1/4)]
    = (3/4)(3/2) = 9/8 = 1.125
  8. Oct 19, 2015 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    This cannot possibly be right: ##|X| \leq 1## for all non-zero probability values, so ##E|X| \leq 1## (and, in fact, ##E|X| < 1## strictly).
  9. Oct 19, 2015 #8
    in that case I am lost =[
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted