Question 576057
a)

It's only a valid probability function if all the individual probabilities add to one.


f(0) = 0.5 - 0/6 = 0.5, so f(0) = 0.5 or f(0) = 1/2


f(1) = 0.5 - 1/6 = 1/3, so f(1) = 1/3


f(2) = 0.5 - 2/6 = 1/6, so f(2) = 1/6


Now add up the individual probabilities:


f(0) + f(1) + f(2) = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = (3+2+1)/6 = 6/6 = 1


Since they all add to 1, this is a valid probability function


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b)

P(X = 2) = 1/6 = 0.1667 and this was found in part a)

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c)

P(At least 2) = 1 - P(None)


P(At least 2) = 1 - 1/2


P(At least 2) = 1/2


P(At least 2) = 0.5


So the probability of selling at least two policies to a customer is 0.5 ( which is 50% chance)

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d)


Expected number = Expected value = Sum of values*probabilities = (0)*(1/2)+(1)*(1/3)+(2)*(1/6) = 2/3 = 0.667


So the expected number is 0.667, which means that he expects to sell somewhere between 0 and one policy (with more weight/chance towards selling 1 policy)

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e)


E(X^2) = Sum(X^2*probability)


E(X^2) = (0^2)*(1/2) + (1^2)*(1/3) + (2^2)*(1/6)


E(X^2) = 0 + 1/3 + 2/3


E(X^2) = 1


Variance


sigma^2 = E(X^2) - (E(X))^2


sigma^2 = 1 - (2/3)^2


sigma^2 = 1 - 4/9


sigma^2 = 5/9


sigma^2 = 0.556


So the variance of the number of policies John will sell is roughly 0.556