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Consider the following analogy:

I observe a man who is 175 cm tall. Independent of height, there is an equal chance that he could be from either Country A, or Country B.
The height of men in Country A is normally distributed with a mean of 190 cm and a standard deviation of 12 cm. The height of men in Country B is normally distributed with a mean of 170 cm and a standard deviation of 8 cm. What is the likelihood that the observed man is from Country A?

Pat B
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  • And how do I get the answer? – Pat B Jan 15 '20 at 16:25
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    Welcome to CV. Since you’re new here, you may want to take our tour, which has information for new users. Since this looks like homework, tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. If this is self-study rather than homework, let us know, and... it's still a good idea to show us what you've tried. – jbowman Jan 15 '20 at 16:30
  • This is actually for a workplace analytical application - not homework. (The metric is actually a rock quality measure rather than height, but the same idea.) I tried to phrase it so the situation was clear. I believe that I can simply take the probability value for 175 cm from each of the pdfs. The value from the Country A pdf divided by the sum of the values from Country A and Country B should give me my probability the man is from Country A. But I'm not sure if it is as simple as that. – Pat B Jan 15 '20 at 18:00
  • It is that simple if your prior probabilities are 50% for each country! – jbowman Jan 15 '20 at 18:07
  • Thanks @jbowman. I am not sure what you mean by prior probabilities, or how they could be a single value (50%). The two populations have a distribution of heights, each height having a different probability. What am I missing? – Pat B Jan 15 '20 at 19:27
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    What is the probability that someone comes from country A, given that you don't know his height? That is your prior probability. Note that if you just have the likelihoods, you can't make probability statements such as the one in your comment above; you need prior probabilities (see Bayes' Rule) in order to do that. – jbowman Jan 15 '20 at 19:45
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    For intuition, suppose Country B is remote and has a tiny population that never travels while Country A is enormous and everyone there travels to Paris in the summer. You are in Paris in the summer and observe someone 175 cm tall--so they are closer to the average height of Country B than Country A. By the clothes they are wearing you are sure they must be either from Country A or Country B. What odds would you give that they are just a short person from Country A rather than from Country B? – whuber Jan 15 '20 at 23:16
  • I have a similar question here: https://stats.stackexchange.com/questions/444952/choosing-between-two-normal-distributions Can anyone explain why it is ok to simply divide the pdfs? This is something I asked in my question. Thanks! @jbowman – adn bps Jan 16 '20 at 00:08
  • https://stats.stackexchange.com/questions/7351/bayesian-statistics-tutorial might be of help - to you too, @adnbps – jbowman Jan 16 '20 at 01:16
  • I am having a hard time understanding how you assign a probability to a continuous variable being equal to 175 cm. Normally, I'd say the answer is P(A|175) = P(175|A)/(P(175|A) + P(175|B) ) as the 1/2 cancels out because either country is equally likely to be sample. Though how would you calculate P(X=175| X ~ N(u, sigma) )? – ma83 Jan 16 '20 at 02:28
  • @M.Austin - it's not a probability, its a probability density. See https://stats.stackexchange.com/questions/17000/probability-of-continuous-random-variable and https://stats.stackexchange.com/questions/20689/probability-of-a-measurement-under-a-continuous-model for more information. – jbowman Jan 16 '20 at 02:46
  • Just to make sure I am using prior probabilities correctly, assume that, independent of height, it is twice as likely that the observed person is from Country A vs. Country B. (0.67, 0.33). I could then get my answer as follows: At 175 cm, the probability density value for Country A is 0.015. I would then weight that by the prior probability to get 0.010 [i.e. 0.015 * 0.67]. For Country B, the same calculation would give 0.014 [i.e. 0.041 * 0.33]. The likelihood of the man being from Country A would then be 0.43 [i.e. 0.010/(0.010+0.014)]. Does that look right? – Pat B Jan 16 '20 at 15:26

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