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    Hi everyone,

    a) k = 0.1
    b) E(X) = 3
    c) E(X^2) = 10
    d) Var(2-5X) = 25
    e) Seems like it's unrelated to parts abcd. No idea what to do.
    f) Clueless.
    g) Cannot do without part e + f
    Thank you all in advance.

    Zacken
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    (Original post by Wolfram Alpha)
    X

    Ah, statistics. Why not

    So you're told that two independent observations are made of X and you want P(X_1 + X_2 = 4)

    So think of all the ways in which that can happen, i.e (3,1), (1,3) and (2,2)

    Notice, for example, that

    P(X_1 = 3, X_2 = 1) = P(X_1 = 3)P(X_2 = 1)

    Just add them all up and you're done.
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    (Original post by Indeterminate)
    Ah, statistics. Why not

    So you're told that two independent observations are made of X and you want P(X_1 + X_2 = 4)

    So think of all the ways in which that can happen, i.e (3,1), (1,3) and (2,2)

    Notice, for example, that

    P(X_1 = 3, X_2 = 1) = P(X_1 = 3)P(X_2 = 1)

    Just add them all up and you're done.
    Hi Indeterminate, thanks for your response. I am still confused though. Firstly, I don't understand the actual question. I don't understand how X1 + X2 can = 4 which can then = 0.1...

    Also. how do you know X1 = 3 and X2 = 1?
    Thank you in advance...
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    (Original post by Wolfram Alpha)
    Hi Indeterminate, thanks for your response. I am still confused though. Firstly, I don't understand the actual question. I don't understand how X1 + X2 can = 4 which can then = 0.1...

    Also. how do you know X1 = 3 and X2 = 1?
    Thank you in advance...


    It's the *probability* that the sum of the two X's are equal to 4
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    (Original post by Wolfram Alpha)
    Hi Indeterminate, thanks for your response. I am still confused though. Firstly, I don't understand the actual question. I don't understand how X1 + X2 can = 4 which can then = 0.1...

    Also. how do you know X1 = 3 and X2 = 1?
    Thank you in advance...
    With regards to your first Q, it's simply what the user above pointed out.

    There is more than one way in which the sum of the 2 Xs can equal 4. You want to consider the probabilities of each case individually and then add them up.

    (3,1) is just one possibility and since the observations are taken independently you multiply them together to get the probability of it turning out that way.

    Do the same for the other 2 possibilities and then add all the results together to get the stated result.

    Let us know if you're ok with this.
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    (Original post by h3rmit)
    It's the *probability* that the sum of the two X's are equal to 4
    Thanks, I understand now!
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    (Original post by Indeterminate)
    With regards to your first Q, it's simply what the user above pointed out.

    There is more than one way in which the sum of the 2 Xs can equal 4. You want to consider the probabilities of each case individually and then add them up.

    (3,1) is just one possibility and since the observations are taken independently you multiply them together to get the probability of it turning out that way.

    Do the same for the other 2 possibilities and then add all the results together to get the stated result.

    Let us know if you're ok with this.
    I understand it now, thank you! But I have another Q that's related to this stuff; is it alright if I post it?
 
 
 
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