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Homework Help: Uniform distrobution

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data

    The proportion of time x that an industrial robot is in operation during a 40 hour work week is a random variable wth probability density function
    f(x)= 2x 0≤x≤1

    I already found my E(x) to be 2/3 " Mean" and V(x) to be 1/18 " variance"

    Here is where things get confusing for me.

    For the robot under study, the profit Y for a week is given by


    Find E(Y) and V(Y).

    So I found E(Y) to be just y=200(E(x))-60

    which makes sense.

    But for V(Y)

    we have to ignore the constant and sqareroot the first term so it would be

    V(Y)=v[200x-60]=2002*v(x) "ignoring the constant"=2002*1/18

    I dont understand the logic to computing the variance. Why would we ignore the constant and just square the first term?


    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 5, 2013 #2


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    I presume that you know that [itex]E(x)= \int xf(x)dx[/itex] and [itex]V= \int (x- x_0)^2f(x)dx[/itex]. Given that y= 2x- 60, yes, you can just put E(x) in for x and get [itex]E(y)= 2(2/3)- 60= 4/3- 60= -176/3[/itex] or -58 and 2/3.

    If you were to replace x by 2x- 60 and [itex]E(x)[/itex] by that mean, you should be able to see that in [itex]((2x- 60)- (4/3- 60))^2[/itex] you can cancel the two "-60" terms. to find the variation of Y, integrate [itex](2x- 4/3)^2[/itex].
  4. Mar 5, 2013 #3


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    Gold Member

    In general, if ##y = ax + b##, then ##E[y] = aE[x] + b##, so
    y - E[y] &= (ax + b) - (aE[x] + b)\\
    &= ax - aE[x]\\
    &= a(x - E[x])
    var(y) &= E\bigl[(y - E[y])^2\bigr]\\
    &= E\bigl[\bigl(a(x - E[x])\bigr)^2]\\
    &= E\bigl[a^2(x - E[x])^2\bigr]\\
    &= a^2E\bigl[(x - E[x])^2\bigr]\\
    &= a^2 var(x)
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