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A random sample of n = 20 salmon taken from a particular Scottish river were examined and the number of parasites recorded as follows:

xi : 2 0 4 0 1 0 3 2 0 5 1 2 0 3 0 1 6 2 2 7

Using these data, estimate ø giving approximate 95% confidence limits for the true value.

Show that the probability of one or more parasites being found on a fish is given by

P(X >= 1) = 1 - e^-ø

(ø is theta)

I assume that your θ is the mean of the values given, in which case θ = (1/20)∑xi. For a Poisson distribution, the mean = variance = θ, so

The 95-percent confidence interval is θ±1.96√(θ/n).

The second part of the question uses a "trick" that is very important to know; i.e., the probabiity od ANY event happening is 1 - NO events happening. For the Poisson distribution, the probability of 0 events happening is just (θ^x)exp(-θ)/x! where x = 0 = exp(-θ) so P(x>1) = 1-P(x=0) = 1 - exp(-θ).

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college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related **Publications**

J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane**Education/Credentials**

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