If two populations are approximately normally distributed, such heights for men and women, and the means and standard deviations are known, can it be determined how often a man will be taller than a woman if one member from each population is randomly selected a large number of times?
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ If you ignore the fact that they are only approximately normal by knowing the means & variances exactly you then know the joint density of heights for men & women & just integrate it over the region where X>Y when X is the height for a man. $\endgroup$– Michael R. ChernickSep 18, 2019 at 22:01
-
$\begingroup$ Your question seems to be about probability rather than likelihood. Likelihood has a specific, technical meaning in statistics that is quite distinct from the ordinary English usage where it essentially just means 'probability'; in statistics it doesn't mean probability, though it is connected to it. I have edited to remove the term, replacing it with probability so as to reduce confusion. $\endgroup$– Glen_bSep 18, 2019 at 23:10
-
$\begingroup$ I believe this kind of question has been answered several times already; I'll see if I can locate a duplicate $\endgroup$– Glen_bSep 18, 2019 at 23:15
-
1$\begingroup$ What could you say about the distribution of the difference between the heights of a randomly selected man and woman? $\endgroup$– Sextus EmpiricusSep 18, 2019 at 23:48
-
$\begingroup$ math.stackexchange.com/questions/261073/finding-probability-pxy $\endgroup$– Sextus EmpiricusSep 18, 2019 at 23:55
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Specific case: Let $X \sim \mathsf{Norm}(\mu_X = 69, \sigma_X = 4)$ and, independently, $Y \sim \mathsf{Norm}(\mu_Y = 66, \sigma_Y = 3).$ I'm not saying this is exactly correct for any populations of men and women (in inches), but it can serve as an example.
Then, following @MartijnWettering's suggestion, let $D = X - Y,$ so that $\mu_D = 3$ and $\sigma_D = \sqrt{25} = 5$ and thus $D \sim \mathsf{Norm}(\mu_D = 3, \sigma_D = 5).$
Finally, $P(X > Y) = P(D > 0),$ which you should be able to evaluate (upon standarization) using printed tables of the standard normal distribution. Alternatively, from R statistical software, one has $P(X > y) \approx 0.726.$
1 - pnorm(0, 3, 5)
[1] 0.7257469
What area in the figure below represents the answer?
For the general case, show how to justify each step, and then how to express the result in terms of the standard normal CDF $\Phi.$
Note: Simulation in R of a million choices of a man from the first
population and a woman from the second. Results should be accurate to
2 or 3 decimal places. [The vector x > y is logical with a million
TRUEs and FALSEs; its mean is the fraction of its TRUEs.]
set.seed(918)
x = rnorm(10^6, 69, 4); y = rnorm(10^6, 66, 3)
mean(x > y)
[1] 0.725808
