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Assume we have a population of N men and women such that exactly $N/2$ people are men (set $M$) and $N/2$ people are women (set $W$).

Assume that the standard deviations for height between both groups is the same $\sigma$ however their averages are different with $\mu_M > \mu_W$.

We pick $n$ people at random from the population such that $n/2$ people are men and $n/2$ people are women.

Knowing $n$ what's the probability that the tallest person in the selected group is a man?

Note: this is not homework, I came up with this question on my own. Note 2: all distributions are normal distributions.

EDIT: My current reasoning is that if $Z = M-W$:

$E(Z) = E(M-W) = E(M) - E(W)$

And

$V(Z) = V(M-W) = V(M) + V(W) = \sqrt2\sigma$

Thus the probability that one man is taller than one woman is:

$P(Z>0)$

And so the probability that all women are taller than all men in the sample is: $(1 - P(Z>0))^{n/2}$:

However this result has lead me to some very odd conclusions. So I suspect I am wrong.

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  • $\begingroup$ Hint: At least one of the N/2 men has to be greater than all woman. We can rephrase it with the converse probability: $1-P("\textrm{No one of the N/2 men has to be greater than all woman}")$ I´m not sure if it helps. $\endgroup$ – callculus Jan 26 at 22:46
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Let \begin{align*} M_1, \cdots, M_{N/2} &\overset{\text{iid}}{\sim} N(\mu_M, \sigma^2) \\ W_2, \cdots, W_{N/2} &\overset{\text{iid}}{\sim} N(\mu_W, \sigma^2) \\ S_M, S_W &\overset{\text{iid}}{\sim} \text{SRSWOR}(\{1, \cdots, N/2\}) \\ \end{align*} where SRSWOR meaning simple random sample without replacement. Define \begin{align*} \textbf{A}_{S} = \{A_i\}_{i \in S} \end{align*} So we are looking at computing \begin{align*} P(\max(\textbf{M}_{S_M}) > \max(\textbf{W}_{S_W})) \end{align*} This is equal to \begin{align*} \sum_{s_M, s_W \in \binom{\{1, \cdots, N/2\}}{n/2}}P(\max(\textbf{M}_{S_M}) > \max(\textbf{W}_{S_W})|S_M=s_M, S_W=s_W)P(S_M=s_M)P(S_W=s_W) \end{align*} But since \begin{align*} P(S_M=s_M) = P(S_W=s_W) = 1/\binom{N/2}{n/2} \end{align*} due to SRSWOR and \begin{align*} P(\max(\textbf{M}_{S_M}) > \max(\textbf{W}_{S_W})|S_M=s_M, S_W=s_W) = P(\max(\textbf{M}_{S_M}) > \max(\textbf{W}_{S_W})) \end{align*} due to i.i.d, we could just consider computing \begin{align*} P(\max(M_1, \cdots, M_{n/2}) > \max(W_1, \cdots, W_{n/2})) \end{align*} Working directly with the distribution of the maximum of normal random variables would be straightforward to compute numerically, although I don't think a closed-form would exist.

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First off, for a normal distribution to be applicable, $N$ needs to be sufficiently large, so let's assume it is and we can then ignore $N$ in our analysis. For small $N$, we would need additional information to solve this problem, preferably the exact discrete distribution of heights.

What you need to solve this is "the distribution of the nth order statistic". Or as Wolfram language calls it: the OrderDistribution

Let Dm be the distribution of the tallest male and Dw likewise for females

Dm = OrderDistribution[{NormalDistribution[\[Mu]m, \[Sigma]], n/2}, n/2]
Dw = OrderDistribution[{NormalDistribution[\[Mu]w, \[Sigma]], n/2}, n/2]

You want to know the probability that x>y given that x is distributed as Dm and y is distributed as Dw.

Mathematica can do the math for us as follows.

Probability[x > y, {x \[Distributed] Dm, y \[Distributed] Dw}, Method -> "Trace"]

yields:

$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\frac{2^{-n-1} n^2 \text{erfc}\left(\frac{\text{$\mu $1}-x}{\sqrt{2} \sigma }\right)^{\frac{n-2}{2}} \text{erfc}\left(\frac{\text{$\mu $2}-y}{\sqrt{2} \sigma }\right)^{\frac{n-2}{2}} e^{-\frac{\text{$\mu $1}^2+\text{$\mu $2}^2+x^2-2 \text{$\mu $1} x+y^2-2 \text{$\mu $2} y}{2 \sigma ^2}} \text{Boole}[x>y]}{\pi \sigma ^2}dydx$

which is the answer as an improper double integral. It will only evaluate symbolically for special cases like these where 1) our random selection has just one man and woman each or 2) men and women have the same average height:

Mathematica input and output

Now WolframAlpha will tell us that the average man's height is 5'6" and 5'2" for a woman. Standard deviation of height yields 7.9".

With these numbers we can plot the answer corresponding to the real world:

ListPlot[Table[{2^nn,NProbability[x>y,{x\[Distributed]Dm,y\[Distributed]Dw}/.{n->2^nn,\[Mu]m->5+6/12,\[Mu]w->5 +2/12,\[Sigma]->7.9/12}]},{nn,1,8}],Joined->True,PlotMarkers->Automatic,AxesLabel->{"n","p"}]

Note that I used NProbability to essentially compute the value of the above improper double integral numerically. I just chose to plot powers of 2 for n.

Obviously, this becomes asymptotic to 1 but quite a bit more slowly than I would have guessed.

enter image description here

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Here is an idea. First, you divide population into two groups, then independently select $n/2$ people from men sub-population and $n/2$ people from women sub-population. This way you select sub-samples of men and women independently from each other. Next, denote $Y$ as a random variable that records the maximum women's height in the sample, and $X_{i}$ as the order statistic such that $X_{1}$ is the minimum height of a man and $X_{n/2}$ is the maximum height of a man. The tallest person is a man if the following event happens: $$(Y \leq X_{1}) \cup (Y > X_{1},Y \leq X_{2}) \cup (Y > X_{2},Y \leq X_{3}) \cup ... \cup (Y > X_{n/2 - 1},Y \leq X_{n/2})$$ This union event actually simplifies to $Y \leq X_{n/2}$ event, so the target probability is $\mathbb{P}[Y \leq X_{n/2}]$. Next steps seem to requite some additional assumptions. First, it seems a bit strange to assume that heights are normally distributed, since normal distribution has support over all reals, and heights cannot be negative. Assuming your probability density function of men's heights is $f_m(x)$ and probability density function of women's heights is $f_w(y)$, then you can write: $$\mathbb{P}[Y \leq X_{n/2}] = \int_{0}^{u}\int_{y}^{u}f_m(x)f_w(y)dxdy$$ where $u$ is whatever upper bound you wish to impose on height, can be even infinity.

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  • $\begingroup$ Many probabilities that are not true normal distributions can be easily aproxiamted with a normal distribution with great accuracy, think binomial distributions for large $n$ for example. $\endgroup$ – Makogan Jan 26 at 23:36
  • $\begingroup$ I know that, I just added this point for the sake of precision. Technically, you can use normal pdfs with the formula above as well. $\endgroup$ – Evgeny Jan 26 at 23:40

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