# When using Normal approximations of human height, what's the probability two people of unknown but approximately equal height are both tall?

Basically, I saw a photo of two people with unknown approximately equal heights and was struck with the thought "How likely is it that they're both tall?". That's the problem I want to answer.

I concede it's possible that I'm making an error in my thinking but I think it's just I don't know how to go about the last step.

The way I thought about that question mathematically is that we can assume Normals for height, i.e. $$X \sim N(\mu_X,\sigma^2_X), Y \sim N(\mu_Y,\sigma^2_Y)$$, and thus my question is:

$$\mathbb{P}(X = x_1 = Y = y_1 \pm \frac{\epsilon}{2} \space \cap \space x_1 > \mu_X \space \cap \space y_1 > \mu_Y)$$

where $$\epsilon$$ is some small positive constant which represents the limits of visual equivalence, e.g. if John's 181cm and Henry's 182cm are you really going to be able to tell they're a different height when standing next to each other? I think not, therefore $$\epsilon$$.

Now, while I suspect two people being photographed together don't have actually independent heights, I'm fine with assuming that they are, so therefore:

$$\mathbb{P}(X = x_1 = Y = y_1 \pm \frac{\epsilon}{2}) \cdot\mathbb{P}(x_1 > \mu_X)\cdot \mathbb{P}(y_1 > \mu_Y)$$

I believe those last two can be calculated simply as $$\mathbb{P}(x_1 > \mu_X) = \mathbb{P}(X > \mu_X)$$ = 1 - rnorm(mu, mu, sigma) (using R) but I'm not sure how to do the first one, especially in the situation where $$X = Y$$, which is an assumption I'd make if the two people were both male or both female. In particular, it seems that the starting point is something like:

$$\mathbb{P}(X > Y) => X - Y \sim N(\mu_X - \mu_Y, \sigma_X^2 - \sigma_Y^2)$$

but that will obviously be $$X - Y \sim N(0, 0)$$ when $$X = Y$$, which I'm not sure how to work with. I have even less of an idea about how to incorporate my $$\epsilon$$ wide interval of visual equivalence.

• I would have thought "approximately equal heights" and "both tall" require some quantification. In addition, is it possible to have them approximately equal heights but one tall and the other not? Commented May 16 at 13:18
• How do you assume two different distributions of height?? // To be answerable, your question needs to make a (strong) assumption about independence or lack thereof in the heights of pairs of people in photographs. If you assume independence and model the heights with a single common distribution, your question can be framed as "when $(X,Y)$ is a standard binormal variable and $c$ is the threshold of tallness, what is the conditional probability that $X\gt c$ and $Y\gt c$ (ie., "tall") conditioned on $X=Y$ (equal height) or, alternatively on $|X-Y|\lt \epsilon$ (approximately equal height)?"
– whuber
Commented May 16 at 13:36
• Sorry, I forgot about this. Henry, I would approach approximate equalness as whuber does i.e. the |X−Y|<ϵ idea. If that was unclear in the question, my bad. If the issue is the fact ϵ could be anything; that's because I was interested in a generalisable answer. As to the question of tallness itself. I just went with the means of the Normals above but my real thinking is again as whuber suggests, with X>c and Y>c, or rather X>c and Y>k, where c can be but is not necessarily equal to k. Tallness is a rather subjective concern. Commented Jun 12 at 5:27
• Also apologies to you, whuber for forgetting about this. The reason I have two different distributions of height was to allow for the case that we're interested in, say, a man and woman of approximately equal height. This is also why c and k wouldn't necessarily be equal. In my ignorance I assumed having two Normals wouldn't pose a threat to the, er, answerableness of the question. I would agree that the assumption of independence is a strong assumption, but as I said for an idle concern (I see no use case to this question) I am comfortable making it. Commented Jun 12 at 5:32

I assume you define "tall" as height being in 50th percentile or higher. The chance that the first person is tall is 50% The second person has nearly the same height, so the probability of being tall (if the first is tall) is nearly 100%. So the chance that a pair of near-equal height people are "tall" is nearly 50%.

The joint probability density of the two heights is $$f_X(x)f_Y(y)$$, assuming that the two parsons are unrelated (e.g., they are not relatives). Note that it makes sense to take different parameters (same $$\mu,\sigma^2$$) for the two, only if we believe that their heights are governed by different distributions - e.g., if one is Asian or the other is Scandinavian.

We also need a definition of a person being tall - e.g., that they are taller than some specified height h (which could be related to them belonging to a certain percentile - e.g., being among the tallest 20%.) We now need probability that $$|X-Y|<\epsilon, \frac{X+Y}{2}>h,$$ where $$\epsilon$$ is the allowed difference in heights - e.g., no more than one or two inches.

The required probability is then $$\int\int f_X(x)f_Y(y)\theta(\frac{x+y}{2}-h)\theta(\epsilon-|x-y|)dxdy,$$ where $$\theta$$ is the Heavyside step function. For normal distribution this integral can be evaluated in terms of error function.

• The evaluation is more complicated than that: it's a definite integral of the error function, not the error function itself.
– whuber
Commented May 16 at 15:44
• @whuber infinite integration limits are assumed, of course - the step functions do the rest. Commented May 16 at 15:57
• I don't think they do. This is twice the integral of a bivariate Normal distribution over a strip terminated by a triangle. It's not expressible exactly as an error function.
– whuber
Commented May 16 at 20:16