In: Statistics and Probability
i) The probability that any individual has the disease = 1.2% = 0.012.
The pdf of a binomial distribution is given by
The probability that at least 20 people test positive for the disease is given by
The probability that at least 20 people test positive for the disease is = 0.021
ii) Using a normal approximation to the binomial distribution, we have
The average number of individuals getting the disease = 1.2% of 1000 = 12
The variance of the number of individuals getting the disease
= np(1-p)
= 1000 * 0.012 * 0.988
= 11.856
The probability that at least 20 individuals will test positive for the disease is given by
Using the normal approximation, the probability that at least 20 individuals will test positive for the disease is= 0.01
iii) Using a Poisson's distribution
The average number of individuals getting the disease = 1.2% of 1000 = 12
The pdf of a Poisson's distribution is given by
The probability that at least 20 individuals will get the disease
= 1 - The probability that at most 19 people will get the disease
Hence, using a Poisson's distribution, we have
The probability that at least 20 individuals will get the disease = 0.021
iv) Using normal approximation to Poisson's distribution
The average number of individuals getting the disease = 1.2% of 1000 = 12
The variance of the number of individuals getting the disease= 12
The probability that at least 20 individuals will test positive for the disease is given by
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