Question

1. (3 POINTS) An insurance policy pays a individual $500 per day for up to 3 days of hospitalization and $100 per day for each day of hospitalization thereafter. The number of days of hospitalization is a random variable X with

                                P(X=x) = (6-x)/15, if x = 1,2,3,4,5.

Calculate the expected payment for hospitalization under this policy.

2. (4 POINTS) An insurance policy reimburses a loss up to a benefit limit 0f $10. There is no deductible. The policyholder’s loss X (in $) has probability density function

                         

                             f(x) = 2/ x^3, if x > 1; and f(x) = 0, otherwise.

What is the expected value of the benefit paid under the insurance policy?

Hint: Find E(Y) where Y = min (X,10).

3. (4 POINTS) The probability distribution of claim sizes for an auto insurance policy is as follows:

Claim Amount (in $):   200       300       400       500       600       700       800

Probability:                   0.15     0.10      0.05      0.20      0.10      0.10      0.30

What percentage of the claims is within one standard deviation of the expected claim size?

4. (4 POINTS) Compute the expected value and the standard deviation of the number of spades in a poker hand.

Answer 1

(1). Let X denote the numbe of days in the hospital and Y the total payment, the given information can be summarized in the following table:

5/15 500 4/15 1000 4 2/15 1600 1/15 1700 P(X =X 3/15 1500

Hence, the expected payment is

$E(Y)= 500 \cdot \frac{5}{15} +1000 \cdot \frac{4}{15}+1500 \cdot \frac{3}{15} +1600 \cdot \frac{2}{15}+1700 \cdot \frac{1}{15}$

$E(Y)=1060$

(2). The amount paid under this policy is

$W = \left\{\begin{matrix} X, & 1<X\leq 10, \\ 10,& X>10\end{matrix}\right.$

The expected amount paid is

$E(W) = \int_{1}^{10}x \cdot \frac{2}{y^{3}}dy + \int_{10}^{\infty}10 \cdot \frac{2}{y^{3}}dy$

$=(-\frac{2}{y})|^{10}_{1} - (\frac{10}{y^{2}})|^{\infty}_{10} = 2-0.2+0.1 = 1.9$

(3).

The mean claim size is

The second moment of the claim size is

Therefore, the variance of the claim size is

The standard deviation is

$\sqrt{Var(X)} = \sqrt{47500} = 217.9$

The claim sizes within one standard deviation of the mean claim size of 550 are those claim sizes between

550 - 217.9 = 332.1 and 550 + 217.9 = 767.9

Those claim sizes are 400, 500, 600 and 700. The total probability of those claim sizes is

0.5 + 0.20 + 0.1 + 0.1 = 0.45

Thus the answer is 0.45 or 45 %

(4). Define a random variable X as the number of spades in a poker hand. We know that

$=\frac{^{n}C_{13} \ ^{5-n}C_{39}}{^{5}C_{52}}, \ 0\leq n\leq 5.$

$E(X)=\sum_{n=0}^{5} n P(X=n) = \sum_{n=0}^{5} n\frac{^{n}C_{13} \ ^{5-n}C_{39}}{^{5}C_{52}}$

$\approx 1.248$

$Var(X) = \sum_{n=0}^{5}[n-E(X)]^{2} P(X=n) \approx 0.866$

$\sigma = \sqrt{Var(X)} \approx 0.931$