Question

# 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 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: 49.0; σ1 = 10.3
David, x2: 49.0; σ1 = 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 = 0.5x1 + 0.5x2. Compute the mean, variance, and standard deviation of W. (Use 2 decimal places.)

 μ σ2 σ

a)

µ = 0.5* E(x1 ) + 0.5 E(x2 ) = 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.

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• ### 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...