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Homework Help: Comparing random variables with a normal distribution

  1. Aug 25, 2011 #1
    1. The problem statement, all variables and given/known data

    You have 7 apples whose weight (in gram) is independent of each other and normally distributed, N([itex]\mu[/itex]= 150, [itex]\sigma[/itex]2 = 202).
    You also have a cabbage whose weight is independent of the apples and N(1000, 502)

    What is the probability that the seven apples will weigh more than the cabbage?

    2. Relevant equations

    Let X represent the weight of the seven apples combined, and Y the weight of the cabbage.
    X~N(1050, 2800)
    Y ~N(1000, 502)

    3. The attempt at a solution

    I have an easy time calculating the probability that a random variable will yield a number within a specific interval. For example I know how to get the probability that the 7 apples will weigh more that 1000g,
    p(X > 1000) =
    [itex]\varphi[/itex](X > (1000 - [itex]\mu[/itex]X)/[itex]\sigma[/itex]x) =
    [itex]\Phi[/itex](0.94) = 0.8264, which I got from a chart for [itex]\Phi[/itex](x).

    I am completely lost however on how to calculate the probability that a certain random variable will yield a bigger number that another random variable, both normally distributed but with different parameters: p(X > Y).

    Thank you.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 25, 2011 #2
    Let [itex]X_i[/itex] be the distribution of the apples. And let Y be the distribution of the cabbage.

    You know that [itex]X_i\sim N(150,20^2)[/itex] and [itex]Y\sim N(1000,50^2)[/itex].

    Can you find out the distribution of -Y??
    Can you use this to find out the distribution of [itex]X:=X_1+X_2+X_3+X_4+X_5+X_6+X_7-Y[/itex]??
    Can you use this to find [itex]P(X\geq 0)[/itex]??
     
  4. Sep 1, 2011 #3
    Thank you :)
    I like how stuff is obvious when someone tells you.

    -y ~ N(-1000, 502)

    [itex]X[/itex]i ~ N(1050 - 1000, 2800 + 2500)

    P( [itex]X[/itex][itex]\geq[/itex]0) = [itex]\Phi[/itex]([itex]\stackrel{50}{\sqrt{5300}}[/itex]) = [itex]\Phi[/itex](0.69) = 0.7549

    Thanks again :D
     
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