Question

Using a long rod that has length , you are going to lay out a square plot in which the length of each 2. side is . Thus the area of the plot will be However, you do not know the value of , so you decide to make n independent measurements X1;X2; :::;Xn of the length. Assume that each Xi has mean 2. (unbiased measurements) and variance

1. A large insurance agency services a number of customers who have purchased both a homeowner's policy and an automobile policy from the agency. For cach type of policy, a deductible amount must be specified. For an automobile policy, the choices are $100 and $250, whereas for a homeowner's policy, the choices are $0, $100, and $200. Suppose an individual with both types of policies is selected at random from the agency's files. Let X denote the deductible amount on the auto policy and Y denote the deductible amount on the homeowner's policy. The following joint pmf specifies the probabilities associated with each of these pairs: p(x, y) y 100 0 200 X 100 250 .20 .05 10 .15 .20 .30 a) Find the probability of a $100 deductible on both policies. [1] b) Find the probability of a deductible amount > 100 for a homeowner's policy (2) c) Find the probability of a $100 deductible on the auto policy given a $100 deductible on the homeowner's policy d) Find COV(X,Y) (2]

Answer 1

Let, X be the deductible amount on the auto policy & Y be the deductible amount on the home owner's policy.

Probability distibution table -

	0	100	200	P(x)
100	0.20	0.10	0.20	0.50
250	0.05	0.15	0.30	0.50
P(y)	0.25	0.25	0.50	1

a) -

Probability that $100 deductible on both the policies - P(x=$100, y=$100) =

P(x=$100, y=$100) = 0.10

Hence, probability that $100 deductible on both the policies is 0.10.

b) -

Probability of a deductible amount $\geq$ 100 for homeowner's policy - P(y $\geq$ 100) =

P(y $\geq$ 100) = P(y = 100) + P(y = 200) = 0.25 + 0.50 = 0.75

Probability of a deductible amount $\geq$ 100 for homeowner's policy is 0.75.

c) -

Probability that $100 deductible on auto policy given a $100 deductible on homeowner's policy - P(X=100|Y=100) =

By conditional probability,

P(AB) = P(ANB) P(B)

So,

$P(X=100|Y=100) = \frac{P(X=100,Y=100)}{P(Y=100)} = \frac{0.10}{0.25} = 0.4$

Probability that $100 deductible on auto policy given a $100 deductible on homeowner's policy is 0.4.

d) -

Formula for Cov(X,Y) is as follows -

Cov(X,Y) = E(XY) - E(X)E(Y)

Where,

$E(X) = \sum xP(x)$

$E(Y) = \sum yP(y)$

$E(XY) = \sum_{ij} x_{i}y_{j}P(x,y)$

So,

Cov(X,Y) = E(XY) - E(X)E(Y) = 23750 - (175)(125) = 23750 - 21875 = 1875

Hence, Cov(x,y) = 1875.