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    Question a) ii
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    The mean is E (X) which is the summation of the X values multiplied by their corresponding probabilities. The variance is
    E (X^2 ) - (E (X))^2 )
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    (Original post by Radioactivedecay)
    The mean is E (X) which is the summation of the X values multiplied by their corresponding probabilities. The variance is
    E (X^2 ) - (E (X))^2 )
    Apparently the mean= 10/6 and variance= 50/36

    but I couldn't get that from using those formulas
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    Ive tried using mean = np and variance = np(1-p)
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    (Original post by aceplayyt)
    Apparently the mean= 10/6 and variance= 50/36

    but I couldn't get that from using those formulas
    X here is the number of times you get the number 6.
    Probability is 1/6.
    But since here is only one value, E (X) would simply be the number of trials times the probability, which is 10 times 1/6.

    For the Variance you essentially do the mean of the squares minus the square of the mean.

    In this case Var (X)= E (X^2 ) - (10/6)^2
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    (Original post by Radioactivedecay)
    X here is the number of times you get the number 6.
    Probability is 1/6.
    But since here is only one value, E (X) would simply be the number of trials times the probability, which is 10 times 1/6.

    For the Variance you essentially do the mean of the squares minus the square of the mean.

    In this case Var (X)= E (X^2 ) - (10/6)^2
    This is not going into my head aha

    So to get E(X^2) you do (10^2)*1/6 - (10/6)^2 ?
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    Dont worry i got it now
 
 
 
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