Unformatted text preview: m .m x /x! Where p(x) is the probability that a sample (one corpsyear) will include x events. Applying this to the Prussian cavalry, we get: Expected Observed p = em = 0.543 Multiply by N to get n , etc n = 109 109 p 1 = m.em = 0.331 n 1 = 66.3 65 p 2 = m 2 /2.em = 0.101 n 2 = 20.2 22 p 3 = m 3 /6.em = 0.021 n 3 = 4.1 3 p 4 = m 4 /24.em = 0.003 n 4 = 0.6 1 Pretty good agreement. For the neuromuscular junction, the each stimulus to the nerve provides a sample. We analyze the number of responses in which we get an event the same size as 1 mepp, the same size as 2 mepp’s, 3 mepp’s, etc....
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 Fall '08
 Stretton
 Poisson Distribution, Probability theory, total number, AirTrain Newark, Pretty good agreement, 20 army corps

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