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Normal Distribution Problem

  1. Dec 2, 2006 #1
    The weight of a large loaf of bread is a normal variable with mean 420g and standard deviation 30g. The weight of a small loaf is a normal variable with mean 220g and standard deviation 10g.

    1) Find the probability that 5 large loaves of bread are heavier than 10 small loaves.

    My Working:
    Let X be the weight of a large loaf, Y be the weight of a small loaf

    X~N(420, 900) , Y~N(220,100)

    5X>10Y
    5X-10Y>0
    X-2Y>0

    E(X-2Y)=E(X) - 2E(Y)=420-440=-20
    Var(X-2Y)=Var(X)+4Var(Y)=900+400=1300

    P(X-2Y>0)=P(Z>20/(1300)^1/2)=P(Z>0.5547)=0.2896

    But the answer given is totally different. Is there anything I miss in my working?
     
  2. jcsd
  3. Dec 2, 2006 #2
    Are [tex] X,Y [/tex] normally distributed random variables? If so, then:

    [tex] Z_{X} = \frac{X-420}{30} [/tex], and [tex] Z_{Y} = \frac{Y-220}{10} [/tex]

    Also [tex] var(aX+bY) = a^{2}var(X) + b^{2}var(Y) + 2abcov(X,Y) [/tex]

    Since [tex] X,Y [/tex] are independent, then [tex] cov(X,Y) = 0 [/tex]
     
    Last edited: Dec 2, 2006
  4. Dec 3, 2006 #3
    You are working with different loaves of bread here, with EACH of their weights being normally distributed. It would not be correct to consider the case where 5X > 10Y, because this would imply you are thinking about the scenario where 5 times the weight of ONE large loaf is greater than 10 times the weight of ONE small loaf.

    Instead you should think about the distribution of [tex]X_{1} + X_{2} + X_{3} + X_{4} + X_{5}[/tex].

    Observe that [tex]5X \sim N(2100,22500)[/tex] but [tex]X_{1} + X_{2} + X_{3} + X_{4} + X_{5} \sim N(2100,4500)[/tex], so these 2 distributions are indeed different.
     
    Last edited: Dec 3, 2006
  5. Dec 4, 2006 #4
    Thanks, I understand it now.
     
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