Insurance companies know the risk of insurance is
greatly reduced if the company insures not just one person, but
many people. How does this work? Let x be a random
variable representing the expectation of life in years for a
25-year-old male (i.e., number of years until death). Then the mean
and standard deviation of x are μ = 49.3 years
and σ = 11.1 years (Vital Statistics Section of the
Statistical Abstract of the United States, 116th
Edition).
Suppose Big Rock Insurance Company has sold life insurance policies
to Joel and David. Both are 25 years old, unrelated, live in
different states, and have about the same health record. Let
x1 and x2 be random
variables representing Joel's and David's life expectancies. It is
reasonable to assume x1 and
x2 are independent.
Joel, x1: 49.3; σ1 =
11.1
David, x2: 49.3; σ1 =
11.1
If life expectancy can be predicted with more accuracy, Big Rock will have less risk in its insurance business. Risk in this case is measured by σ (larger σ means more risk).
(a) The average life expectancy for Joel and David is W = 0.5x1 + 0.5x2. Compute the mean, variance, and standard deviation of W. (Use 2 decimal places.)
μ | |
σ2 | |
σ |
(b) Compare the mean life expectancy for a single policy (x1) with that for two policies (W).
The mean of W is smaller.The mean of W is larger. The means are the same.
(c) Compare the standard deviation of the life expectancy for a
single policy (x1) with that for two policies
(W).
The standard deviations are the same.The standard deviation of W is smaller. The standard deviation of W is larger.
a)
E(W) =μ=0.5E(x1)+0.5E(x2)=49.3 |
Var(W) =σ2=(0.5)^2*Var(x1)+(0.5)^2*Var(x2)=61.61 |
SD(W)=σ=√Var(W)=7.85 |
b)
The means are the same.
c)
The standard deviation of W is smaller.
Insurance companies know the risk of insurance is greatly reduced if the company insures not just...
Insurance companies know the risk of insurance is greatly reduced if the company insures not just one person, but many people. How does this work? Let x be a random variable representing the expectation of life in years for a 25-year-old male (i.e., number of years until death). Then the mean and standard deviation of x are μ = 49.0 years and σ = 10.3 years (Vital Statistics Section of the Statistical Abstract of the United States, 116th Edition). Suppose...
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6. Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. They want to know if their clients this year have a longer life expectancy, on average, so the company randomly sampled 20 policyholders to see if the mean life expectancy of policyholders has increased. The mean life expectancy for the random sample was 78.6 years with a sample standard deviation of 5.12 years. The insurance company will only change their premium structure if...
Show all work: Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. Last year the average life expectancy of all policyholders was 77 years. ABI Insurance wants to determine if their clients now have a longer life expectancy, on average, so they randomly sample some of their recently paid policies. Suppose ABI samples 100 recently paid policies. This sample yields a mean of 77.7 years and a standard deviation of 3.6 years. Find...
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