Question 576057: 8.The probability function for the number of insurance policies John will sell to a customer is given by f(x) .5 - (x/6) for x = 0, 1, or 2
a.Is this a valid probability function? Explain your answer.
b.What is the probability that John will sell exactly 2 policies to a customer?
c.What is the probability that John will sell at least 2 policies to a customer?
d.What is the expected number of policies John will sell?
e.What is the variance of the number of policies John will sell?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 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
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