I am performing a study on individuals heights, comparing 4 different areas of my city. I want to calculate the probability of each of the 4 Areas being the tallest:

  • Example: What is the probability of A1 being the tallest Area ?

All of the 4 areas have a normal distribution on their heights

Areas = A1, A2, A3, A4

A1 = Mean 180 with SD 7.2
A2 = Mean 178 with SD 7
A3 = Mean 176 with SD 9
A4 = Mean 182 with SD 8.4

I don't know now, how to go forward from here. I do know how to compare two Areas I saw an example here:

For example, if I wanted to know the probability of A1 being taller than A2, I would get the Normal distribution of the difference of $A1-A2$, like:

$$D = A1 - A2 = 180 - 178 = 2\quad\text{ ----- this for the mean}$$

And for the SD, I get $(7.2^2 + 7^2)^{0.5} = 10.041$

So using the distribution D with mean 2 and SD 10.041, I just calculate the cumulative distribution for D>0. But, how to do the same analysis for the 4 areas?

$$A1 > (A2\ \&\ A3\ \&\ A4)$$

Thank you.

  • $\begingroup$ Could you explain what "being the tallest" means? Under your model assumptions, there is no such thing as a tallest person in any of the groups; moreover, when you are given "a" random person, you have one person: to whom are they being compared?? $\endgroup$
    – whuber
    May 24 '20 at 16:42
  • $\begingroup$ @whuber - I want to know what is the probability of any of the areas is the highest. Similar to this question khanacademy.org/math/ap-statistics/random-variables-ap/… but in my case I have 4 distributions. I have edited the question, I hope its more clear now and can be reopen. Thank you $\endgroup$
    – Dedes
    May 25 '20 at 5:58

First you need to set a threshold to decide what probability is enough to declare a winner. Say you fix this to 0.5, then an area $A_i$ is "taller" than $A_j$ if $P(A_i>A_j) > 0.5$.

You just do a pair-wise comparison (in your case 3 pairs) and keep the tallest at each step.

  • $\begingroup$ The question is unanswerable until a prior distribution for the four areas is provided. Therefore your reply cannot be correct. Note, too, that the question appears to be asking about the mean heights per area, not individual heights. $\endgroup$
    – whuber
    May 25 '20 at 15:22
  • $\begingroup$ @whuber, can I assume from your comment that "the prior distribution for the fours areas" is a combination for the individual distributions ? And, as the areas are all normal distributed, doesn't this mean that the distribution that combines the fours is also normal ? And, assuming I have that distribution, how do I proceed ? Thank you $\endgroup$
    – Dedes
    May 25 '20 at 15:52
  • $\begingroup$ None of that is correct. The combination is a mixture, which will not be Normal. The general method of proceeding requires a Bayesian analysis. $\endgroup$
    – whuber
    May 25 '20 at 16:46
  • $\begingroup$ @whuber Of course this can be made more rigorous, but for this specific case I dont see why this simple solution is wrong (or better, why the question in unanswerable) since the sum of two normally distributed and independent random variables is normally distributed. $\endgroup$
    – xi45
    May 25 '20 at 16:57
  • $\begingroup$ The question is unanswerable because there cannot even be a probability until a prior is specified. $\endgroup$
    – whuber
    May 25 '20 at 17:39

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