http://family-health-plan.com/employees-of-a-large-company-all-choose-1-of-3-levels-of-health-insurance-coverage-for-which-premiums-denote.html Fri, 25 May 2012 21:53:34 +0000 hourly 1 http://wordpress.org/?v=3.3.2 http://family-health-plan.com/employees-of-a-large-company-all-choose-1-of-3-levels-of-health-insurance-coverage-for-which-premiums-denote.html/comment-page-1#comment-2391 Hahaha Mon, 07 Sep 2009 02:24:30 +0000 http://family-health-plan.com/employees-of-a-large-company-all-choose-1-of-3-levels-of-health-insurance-coverage-for-which-premiums-denote.html#comment-2391 First, find p(x,y): p(1,0) = 1/31; p(2,0) = 4/31; p(3,0) = 9/31; p(1,1) = 2/31; p(2,1) = 5/31; p(3,1) = 10/31. The problem does not give the statistical weights, such as how many people smoke and how many do not. Since Sum(p) = 1, every possible case is to be treated as equally weighted (1/6). The expected value is 1/6. So the variance is easily calculated to be: [Sum(over all different x and y) (p - 1/6)^2]/6 = [Sum(p^2 - p/3 + 1/36)]/6 = [(1+16+81+4+25+100)/31^2 - 1/3 + 1/6]/6 = (227/961 - 1/6)/6 = 401/34596 First, find p(x,y):
p(1,0) = 1/31; p(2,0) = 4/31; p(3,0) = 9/31;
p(1,1) = 2/31; p(2,1) = 5/31; p(3,1) = 10/31.
The problem does not give the statistical weights, such as how many people smoke and how many do not. Since Sum(p) = 1, every possible case is to be treated as equally weighted (1/6). The expected value is 1/6. So the variance is easily calculated to be:
[Sum(over all different x and y) (p - 1/6)^2]/6
= [Sum(p^2 - p/3 + 1/36)]/6
= [(1+16+81+4+25+100)/31^2 - 1/3 + 1/6]/6
= (227/961 – 1/6)/6
= 401/34596

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