If the probability that an individual suffers a bad reaction from an injection of a given serum is 0.001, determine the probability that out of 1500 individuals i) more than 2 individuals suffer from bad reaction , ii) mean and variance of distribution.

Number of individual suffers a bad reaction.

n = 1500

probability that an individual suffers a bad reaction.

p = 0.001

Calculation λ.

λ = np = 1500(0.001) = 1.5

Poisson distribution is written as.

X~P(λ) ⇒ X~P(1.5)

Probability that more than 2 individuals suffer from bad reaction.

P(X>2)

Poisson distribution P(X=x) = \$(e^(-λ) λ^x)/x!\$

P(X>2).

1 - P(X≤2)

P(X>2).

1 - [P(X=0) + P(X=1) + P(X=2)]

P(X>2).

1 - \$(e^(-1.5) 1.5^0)/(0!) + (e^(-1.5) 1.5^1)/(1!) + (e^(-1.5) 1.5^2)/(2!)\$

P(X≤2).

1 - \$2.718^(-1.5) + 2.718^(-1.5)×1.5 + (2.718^(-1.5) 1.5^2)/2\$

P(X≤2).

1 - \$0.2231 + 0.2231×1.5 + 0.2231(2.25/2)\$

i)Probability that more than 2 individuals suffer from bad reaction P(X>2).

1 - 0.8087 = 0.1913

Mean and variance of poisson distribution.

mean = variance = λ

ii)Mean and variance

λ = 1.5

Answer

A