Understanding conditional probabilities in Bayes classifiers in the wikipedia page example I'm using the wikipedia page example:
Why is $P(\text{height} \mid \text{male}) = 1.5789$? This means the probability of height given male? The talk page has a similar question, unanswered.

the example added last august about sex classification is puzzling me,
  could anyone tell me how to compute the $P(\text{height} \mid \text{man})$?
  The author gave the value 1.5789 with a note stating that "probability
  distribution over one is OK. It is the area under the bell curve that
  is equal to one" which also puzzle me.

 A: The value $1.5789$ that is calculated is the density of the probability mass
at $x = 6$ with units mass/foot; it is not the probability that a randomly selected male has height exactly $6$ feet.  To get a probability, you have to multiply the value of the probability density by a length.  In other words, 
the probability that a randomly selected male has height between $5$ feet, $11\frac{1}{2}$ inches and $6$ feet, $\frac{1}{2}$ inches is approximately
$1.5789 \times \frac{1}{12}$ (because the unit of height ($x$ axis) is a foot and
thus the length in question is  $1$ inch = $\frac{1}{12}$ foot).  Note that
the value of the probability that we thus obtain is an approximation, but a very good approximation in this instance.  (To get an exact value, we would 
need to compute the value of an integral, but $1.5789/12 = 0.1315\ldots$ is 
good enough for gummint purposes).
As a practical matter, heights are often recorded to the nearest inch, and so
when someone says a particular male is $6$ feet tall, people usually take it to
mean that the person is between $5$'$11\frac{1}{2}$" and $6$'$\frac{1}{2}$"
anyway.  But in this sense of the phrase, the probability that a randomly chosen
male is $6$ feet tall is $13.15\%$, not $157.89\%$.
A: It looks like in this case we should use Normal distribution for the trainig set, therefore start to calculate all you need for distribution


*

*$\mu = \frac{6 + 5.92 + 5.58 + 5.92}{4} = 5.855 $ - mean value

*next find the variance $\sigma ^ 2$
and substitute to the formula of Normal distribution with f(x=6) as a testing value
A: The male height distribution is approximated from the training set to be a normal distribution with mean $\mu = 5.855$ and variance $\sigma^2 = 0.035$. The likelihood of seeing a height of 6 feet given that the sample is male is therefore
$$P(6 | \textrm{male}) = \frac{1}{\sqrt{2\sigma^2}} \exp\left( \frac{ (6 - \mu)^2}{2\sigma^2} \right) \approx 1.579$$
which is the quoted likelihood in the article.
