Probability result in well-known paper appears to use std deviation correction of sqrt(n/(n-2.5)) The below problem is cited in the 1972 paper "Subjective Probability: A Judgment of Representativeness" by Daniel Kahneman and Amos Tversky.  The paper provides an answer to the problem for the purpose of comparing that answer with the impressions of test subjects to whom the problem was posed.

The average heights of adult males and females in the US are, respectively, 5 ft 10 in. and 5 ft 4 in. Both distributions are approximately normal with a standard deviation of about 2.5 in.
An investigator has selected one population by chance and has drawn from it a random sample.
What do you think are the odds that he has selected the male population if
(i) the sample consists of a single person whose height is 5 ft 10 in.?
(ii) the sample consists of 6 persons whose average height is 5 ft 8 in.?
........ The correct odds are 16 in case (i) and 29 in case (ii).  

Bayes Theorem and the Gaussian PDF give odds of 17.8 and 317 for cases (i) and (ii), calculations below.  Since the problem provides a population standard deviation, using a sample standard deviation adjustor does not seem in order.  However, in case(ii) I checked a few adjustors $\sigma_{adj} = \sigma \cdot \sqrt{\frac{n}{n-k}}$ for $k = 1$ (Bessel's correction) and $k=1.5$ (alternative correction for normal distribution).  Neither returned the authors' given answer, although each moved the calculated odds in that direction.

The provided answer for case(ii) IS obtained using an adjustor based on k=2.5!  

The question: what method could the authors be using that would provide their stated answer?  Why would k=2.5 be used?

Solution using Bayes Theorem and the Gaussian PDF

Case (i)

Using Bayes Theorem $$P(A|B)=\frac{P(B|A)\cdot P(A)}{P(B)}$$
$$$$

$$P(male | 5'10)=\frac{P(5'10 | male)\cdot P(male)}{P(5'10)}$$
$$$$

$$P(female | 5'10)=\frac{P(5'10 | female)\cdot P(female)}{P(5'10)}$$
$$$$
Assuming $P(male)=P(female)$ which is not explicitly stated in the problem ...
$$$$
$$odds=\frac{P(male | 5'10)}{P(female | 5'10)}=\frac{P(5'10 | male)}{P(5'10 | female)}$$
$$$$

Using the Gaussian PDF $$f(x)={\frac 1{\sqrt{2\pi}\sigma}e^\frac {-(x-\mu)^2}{2\sigma^2}}$$
yields ...
$$\frac{P(5'10 | male)}{P(5'10 | female)}=\frac{{\frac 1{\sqrt{2\pi}\cdot 2.5}e^\frac {-(70-70)^2}{2\cdot 2.5^2}}}{{\frac 1{\sqrt{2\pi}\cdot 2.5}e^\frac {-(70-64)^2}{2\cdot 2.5^2}}}=\frac{{e^0}}{{e^{-2.88}}}=17.8$$
$$$$
Case (ii)

Applying the same methodology to case (ii) with an adjusted $\sigma=\frac{2.5}{\sqrt6}=1.0206$ gives $$odds=\frac{P(5'8\,avg | male)}{P(5'8\,avg | female)}=\frac{{\frac 1{\sqrt{2\pi}\cdot 1.0206}e^\frac {-(68-70)^2}{2\cdot 1.0206^2}}}{{\frac 1{\sqrt{2\pi}\cdot 1.0206}e^\frac {-(68-64)^2}{2\cdot 1.0206^2}}}=\frac{{e^{-1.92}}}{{e^{-7.68}}}=317.4$$

$$$$
So there is a small difference (17.8 vs. 16) from the given answer in case (i) and a very large difference (317 vs. 29)in case (ii).

What method could be arriving at the stated answers of 16 and 29?
Thanks for posting this. I've been bugged by my inability to come up with the answer to case ii. The whole population SD switch to sample SD seems so odd (along with the -2.5 correction). Also odd to me is using the height of normal dist (PDF) instead of the area under a range of scores (CDF) as the basis for the probability estimates given that the probability that a person is exactly 5'10" is, strictly speaking, zero; i.e., P(5'10"|male) = 0. Same for female of course.

The answers aren't any better when using CDF values either. So it seems clear that T&amp;K used a PDF in their calculation.
Thanks for posting this. I've been bugged by my inability to come up with the answer to case ii. The whole population SD switch to sample SD seems so odd (along with the -2.5 correction). Also odd to me is using the height of normal dist (PDF) instead of the area under a range of scores (CDF) as the basis for the probability estimates given that the probability that a person is exactly 5'10" is, strictly speaking, zero; i.e., P(5'10"|male) = 0. Same for female of course.

The answers aren't any better when using CDF values either. So it seems clear that T&amp;K used a PDF in their calculation.