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    http://mathsathawthorn.pbworks.com/f/S2.pdf
    Q3b?
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    What is the probability that 1 egg weighs less than 49 grams? Call this probability p. Then the probability that exactly three eggs weight less than 49 grams is \mathbb{P}(X = 3) where X \sim \text{B}(6, p).
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    (Original post by Zacken)
    What is the probability that 1 egg weighs less than 49 grams? Call this probability p. Then the probability that exactly three eggs weight less than 49 grams is \mathbb{P}(X = 3) where X \sim \text{B}(6, p).
    Can i show what ive done first?
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    (Original post by Ayaz789)
    Can i show what ive done first?
    Why didn't you put that in the OP?
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    So ive done P(0>3Y-X)
    3Y-X -N(54-50, 1.2^2 +2^2)
    E(3Y-X)=(4,5.44)
    Is that right so far?
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    (Original post by Zacken)
    Why didn't you put that in the OP?
    Because i think you have done it a different way to me
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    (Original post by Ayaz789)
    So ive done P(0>3Y-X)
    3Y-X -N(54-50, 1.2^2 +2^2)
    E(3Y-X)=(4,5.44)
    Is that right so far?
    Ah, my bad - I was answering (a) (ii) instead.

    What you've done is fine - yes (except check your variance! It's wrong). Although I have no clue why you're finding E(3Y-X) in the last line.


    So you have (I'll call it W to make it more obvious) 3Y - X = W \sim N(4, \text{variance}) and you want \mathbb{P}(3Y-X < 0) = \mathbb{P}(W < 0).

    Well, that's a simple normal distribution problem now. Can you see what to do from here?
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    (Original post by Zacken)
    Ah, my bad - I was answering (a) (ii) instead.

    What you've done is fine - yes. Although I have no clue why you're finding E(3Y-X) in the last line.


    So you have (I'll call it W to make it more obvious) 3Y - X = W \sim N(4, 1.2^2 + 2^2) and you want \mathbb{P}(3Y-X < 0) = \mathbb{P}(W < 0).

    Well, that's a simple normal distribution problem now. Can you see what to do from here?
    The thing is in the mark scheme it says E(3Y-X)= (4,16.96) Where do they get 16.96 from? And do you use z=x-mean/ (standard deviation)
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    (Original post by Ayaz789)
    The thing is in the mark scheme it says E(3Y-X)= (4,16.96) Where do they get 16.96 from? And do you use z=x-mean/ (standard deviation)
    If it says that, then it's wrong. E(3Y-X) = 3E(Y) - E(X) = 4

    \text{Var}(3Y-X) = \text{Var}(3Y) + \text{Var}(X) = 9\text{Var}(Y) + \tetx{Var}(X) = 16.96.

    So W \sim N(4, 19.69). Then yeah, use z = x - mean / standard dev.
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    (Original post by Zacken)
    If it says that, then it's wrong. E(3Y-X) = 3E(Y) - E(X) = 4

    \text{Var}(3Y-X) = \text{Var}(3Y) + \text{Var}(X) = 9\text{Var}(Y) + \tetx{Var}(X) = 16.96.

    So W \sim N(4, 19.69). Then yeah, use z = x - mean / standard dev.
    Okay thanks
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    (Original post by Ayaz789)
    Okay thanks
    Sure.
 
 
 
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