In: Statistics and Probability
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 = 48.7 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 x2be random
variables representing Joel's and David's life expectancies. It is
reasonable to assume x1 and
x2 are independent.
Joel, x1: 48.7; σ1 =
10.3
David, x2: 48.7; 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 |
(b) Compare the mean life expectancy for a single policy (x1) with that for two policies (W).
The mean of W is larger.The means are the same. The mean of W is smaller.
(c) Compare the standard deviation of the life expectancy for a
single policy (x1) with that for two policies
(W).
The standard deviation of W is smaller.The standard deviation of W is larger. The standard deviations are the same.
(a)
The mean of W is given by:
The variance of W is given by:
The standard deviation of W is given by:
(b)
The mean of x1 is 48.7 which is equal to the mean of W.
Thus, The means are same.
(c)
The standard deviation of x1 is 10.3 which is larger than the standard deviation of W which is 7.28.
Thus, The standard deviation of W is smaller.
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