Probability?

It is claimed that 95% of the population in a certain village are right-handed. A random sample of 80 villagers are tested to see if they are right or left-handed. Use a poisson approximation to estimate the probability that the number who are right-handed is (a)80, (b)79, (c) at least 78

Probability?
This is an example of how the Poisson distribution is a limiting case of the Binomial. The more accurate calculations for this problem are done with the binomial. I will walk through these and the calculations for with the Poisson.





Let X be the number of people who are right handed in the village.





X has the binomial distribution with n = 80 trials and success probability p = 0.95 .





In general, if X has the binomial distribution with n trials and a success probability of p then


P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n


P[X = x] = 0 for any other value of x.





The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.


Or, to be more accurate, the binomial is the sum of n independent and identically distributed Bernoulli trials.





X ~ Binomial( n , p )





the mean of the binomial distribution is n * p = 76


the variance of the binomial distribution is n * p * (1 - p) = 3.8


the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.949359





P( X = 78 ) = 0.1445667120682631


P( X = 79 ) = 0.0695384184632151


P( X = 80 ) = 0.01651537438501358





P( X ≥ 78 ) =


P(X = 78) + P(X = 79) + P(X = 80)


= 0.2306205049164921





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To use the Poisson approximation we need to have a very large n and a very small p.





In general, if X has the Poisson distribution with a rate of λ:





X ~ Poisson(λ)


P(X = x) = (λ)^x * exp(-λ) / x! for x = 0, 1, 2, 3, 4, ...


P(X = x) = 0 otherwise





the mean of the Poisson distribution is the parameter, λ


the variance of the Poisson distribution is the parameter, λ





here λ = n * p = 76 for right handed people. p is not small, it's large, so lets look at solving for the number of left handed people and then use that to find the number of right handed people.





Let X be the number of right handed people


Let Y be the number of left handed people





Y ~ Poisson(n * (1 - p)) = Poisson( 80 * 0.05)


Y ~ Poisson( 4 )





P(X = 80) ≈ P( Y = 0 ) = 0.01831564


P(X = 79) ≈ P( Y = 1 ) = 0.07326256


P(X = 78) ≈ P( Y = 2 ) = 0.14652511





P(X ≥ 78) ≈ P( Y ≤ 2)


= P(Y = 0) + P(Y = 1) + P(Y = 2)


= 0.2381033
Reply:The general Poisson equation is:


P(X=x) = (µ^xe^(-µ))/x!


Because this equation involves factorials and most calculators won't calculate factorials greater than 69!, we cannot solve the problem as stated.


But we can calculate the probability for LEFT-handed persons and subtract that from 1 to find the probability of Right-handed persons.


The Poisson parameter is


µ = np = (80)(0.05) = 4 (for lefties)


The Poisson approximation to the binomial is


(a) P(x=0) = 4^0e^(-4)/0! = 0.0183 for lefties, and,


1-0.183 = 0.9817 for righties.


(b) P(x=1) = 4^1e^(-4)/1! = 0.0733 for lefties and


1-0.0733 = 0.927 for righties


(c) P(x=2) + P(x=1) +P(x=0)


= 4^2e^(-4)/2! + 0.0733 + 0.0183 = 0.2381 for lefties


and, 1-0.2381 = 0.762 for righties.