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Math Help - Variance of multiple

  1. #1
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    Sampling of a General Population Probability question

    Here's my homework question, my work/thoughts on it thus far are below the question.

    A life insurance application asks the question "Are you a smoker?" The percentage of smokers in the general population is 15%. Furthermore, 40% of applicants who are smokers lie on the application, and say that they are non-smokers, but none of the non-smokers lie on the application. What proportion of applicants who say they are non-smokers are actually non-smokers?

    Now, if the general population were the entire applicant pool, this would be really simple:
    X = general population
    .15X = Number of actual smokers
    .4(.15X) = Number of smokers who said they were non smokers
    .85X-(.4(.15X)) = Number of non-smokers polled who are actually non smokers.

    But, seeing as this is for my college-level probability class and it doesn't say that the pool of applicants is in fact the general population, I feel like I'm missing something here. Does this have to do with conditional distribution and expectation values? Or am I overthinking this? Any hints or nudges in the right direction would be greatly appreciated!
    Last edited by puggles; September 19th 2011 at 10:32 AM. Reason: Title doesn't make sense
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  2. #2
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    Re: Sampling of a General Population Probability question

    But, seeing as this is for my college-level probability class and it doesn't say that the pool of applicants is in fact the general population,
    Good point, and in most practical situations its very likely that smokers will be over-represented in applications for life cover. This is because compared to the general population, they are more likely to die and hence have more to gain from the policy (depending on how it is priced).

    However for the purpose of a probability class, i think you are expected to assume that 15% of applications are in fact from smokers.




    .85X-(.4(.15X)) = Number of non-smokers polled who are actually non smokers.
    You need a proportion, rather than a number. And im not sure you have calculated your number properly either. The number of people who are actually non smokers is 0.85X, and the number of people claiming to be nonsmokers is 0.85X plus 0.4(0.15X).


    Its likely you were taught conditional probability formula before being set this question, try using that.
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  3. #3
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    Re: Variance of multiple

    Thanks for your help. Here's what I got with your hints:
    let A = applicants who claim they are non smokers
    let B = actual non smokers

    So, P(A|B) = Probability (~proportion) that the people that claim to be non smokers are in fact non smokers.
    P(A|B)=P(A<intersect>B)/P(A) = .85X/(.85X + .4(.15X)) = .85/.91

    So the proportion of people claiming to be non smokers that are in fact non smokers would be 85/91
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