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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.

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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 $157.89\%$.

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It looks like in this case we should use Normal distribution for the trainig set, therefore start to calculate all you need for distribution

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

  2. next find the variance $\sigma ^ 2$

and substitute to the formula of Normal distribution with f(x=6) as a testing value

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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.

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It is the use of the word "probability" in the wikipedia article instead of "likelihood" (as you correctly call it) that is the cause of the confusion. In fact, earlier in the article, it says "Then, the probability (emphasis added) of some value given a class, $P(x = v \mid c)$, can be computed by plugging $v$ into the equation for a Normal distribution parameterized by $\mu_c$ and $\sigma^2_c$" which is nonsensical from a probability theory viewpoint but not uncommon in applied statistical circles. – Dilip Sarwate Oct 5 '11 at 13:43

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