Answer to Question #105631 in Statistics and Probability for Christine Palapuz

Question

The Winnipeg Health Region states that the number of people per year in the Winnipeg area who contract flesh-eating disease is a random variable with a mean of 4.6. Let us suppose, the random variable is denoted by X.
(a) What is the distribution of X. Give the mean and standard deviation of X.
(b) What is the probability that six people will contract flesh-eating disease in the Winnipeg region?
(c) What is the probability that between two and four people, inclusive, will contract the flesh-eating disease in a four-month period?

Answer 1

a) the distribution of X is Poisson with $\lambda$ =4.6

The mean =4.6

The standard deviation =$\sqrt {4.6}$ =2.1447

b) P(X=6)

P(X=x)=$\frac {\lambda^x e^{-\lambda}} {x!}$

P(X=6) =$\frac {4.6^6 e^{-4.6}} {6!}$

=0.1323

c) p$(2\le X \le 4)$

The mean for four month period is 4.6÷3=1.533

$\lambda =1.533$

$P(2\le X \le 4)=\displaystyle \sum_{x=2}^4 \frac {\lambda^x e^{-\lambda}} {x!}$

=$\displaystyle \sum_{x=2}^4 \frac {1.533^x e^{-1.533}} {x!}$

=0.4331