\$p(mathrmheight|mathrmmale) = 1.5789\$ (A probability distribution over 1 is OK. It is the area under the bell curve the is equal to 1.)

How can a value \$>1\$ be OK? I assumed all probability values were expressed in the range \$0 leq ns leq 1\$. Furthermore, offered that the is possible to have such a value, how is that value acquired in the example presented on the page?

That Wiki web page is abusing language by introduce to this number together a probability. You space correct that it is not. It is actually a probability every foot. Specifically, the worth of 1.5789 (for a elevation of 6 feet) means that the probability that a height between, say, 5.99 and 6.01 feet is close to the complying with unitless value:

\$\$1.5789, <1/ extfoot> imes (6.01 - 5.99), < extfeet> = 0.0316\$\$

This worth must not exceed 1, as you know. (The little range that heights (0.02 in this example) is a critical part the the probability apparatus. The is the "differential" that height, i beg your pardon I will abbreviate \$d( extheight)\$.) Probabilities per unit of miscellaneous are called densities by analogy to other densities, like mass every unit volume.

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Bona fide probability densities can have arbitrarily large values, also infinite ones.

This instance shows the probability density role for a Gamma distribution (with form parameter that \$3/2\$ and scale that \$1/5\$). Because most the the density is much less than \$1\$, the curve needs to rise greater than \$1\$ in bespeak to have actually a complete area that \$1\$ as forced for every probability distributions.

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This density (for a beta distribution with parameters \$1/2, 1/10\$) becomes boundless at \$0\$ and also at \$1\$. The full area quiet is finite (and equals \$1\$)!

The worth of 1.5789 /foot is acquired in that instance by estimating the the heights the males have actually a normal circulation with average 5.855 feet and also variance 3.50e-2 square feet. (This have the right to be found in a previous table.) The square root of that variance is the conventional deviation, 0.18717 feet. Us re-express 6 feet together the variety of SDs indigenous the mean:

\$\$z = (6 - 5.855) / 0.18717 = 0.7747\$\$

The department by the typical deviation produce a relation

\$\$dz = d( extheight)/0.18717\$\$

The typical probability density, by definition, equals

\$\$frac1sqrt2 piexp(-z^2/2)dz = 0.29544 d( extheight) / 0.18717 = 1.5789 d( extheight).\$\$

(Actually, ns cheated: I merely asked Excel come compute NORMDIST(6, 5.855, 0.18717, FALSE). Yet then i really did inspect it versus the formula, just to be sure.) as soon as we piece the essential differential \$d( extheight)\$ native the formula just the number \$1.5789\$ remains, favor the Cheshire Cat"s smile. We, the readers, need to recognize that the number needs to be multiplied by a small difference in heights in order to produce a probability.