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 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
x_{1} and x_{2} be random
variables representing Joel's and David's life expectancies. It is
reasonable to assume x_{1} and
x_{2} are independent.
Joel, x_{1}: 49.0; σ_{1} =
10.3
David, x_{2}: 49.0; σ_{1} =
10.3
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 = 0.5x_{1} + 0.5x_{2}. Compute the mean, variance, and standard deviation of W. (Use 2 decimal places.)
μ | |
σ^{2} | |
σ |
a)
µ = 0.5* E(x_{1} ) + 0.5 E(x_{2} ) = 0.5*49+0.5*49 = 49
σ² = 0.5² * V(X1) + 0.5²* V(X2) = 0.5²*10.3² + 0.5² * 10.3² = 53.05
σ = √σ² = √53.045 = 7.28
b)
the means are same
c)the std 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 = 49.3 years and σ = 11.1 years (Vital Statistics Section of the Statistical Abstract of the United States, 116th Edition). Suppose...