Chapter 12, Problem 12.46E

(a)

To determine

To find out what distribution does X follow and also give the mean and standard deviation of X .

Answer to Problem 12.46E

X follows Poisson distribution and the mean and standard deviation is $1.67$ and $1.2923$ respectively.

Explanation of Solution

In the question it is given that the average number of salmonellosis cases per month in South Dakota is $\text{1}\text{.67}$ . Now, X is the monthly count of salmonellosis cases in South Dakota. Thus, we can see that the probability is small compared to the population size that is large. Thus, we can say that X approximately follows Poisson distribution with parameter $\text{1}\text{.67}$ . Thus, the mean and standard deviation is calculated as:

  $\begin{array}{l}\text{Mean}=\mu \\ =1.67\\ \text{St}\text{.Dev}\text{.}=\sigma \\ =\sqrt{\mu }\\ =\sqrt{1.67}\\ =1.2923\end{array}$

(b)

To determine

To find out the probabilities $P(X=0),P(X\le 1),P(X\le 2),P(X\le 3)\text{ and P(X}\le \text{4)}$ and also find the probability that there would be more than four cases of salmonellosis in South Dakota in a given month.

Answer to Problem 12.46E

The probabilities are:

$P(X=0)$	0.1882
$P(X\le 1)$	0.5026
$P(X\le 2)$	0.7651
$P(X\le 3)$	0.9112
$P(X\le 4)$	0.9723

And the probability that there would be more than four cases of salmonellosis in South Dakota in a given monthis $0.0277$ .

Explanation of Solution

In the question it is given that the average number of salmonellosis cases per month in South Dakota is $\text{1}\text{.67}$ . X approximately follows Poisson distribution with parameter $\text{1}\text{.67}$ . Thus, to calculate the probabilities we will use the Poisson excel function as:

  $=\text{POISSON}\text{.DIST(}x,\text{mean, cumulative)}$

In the cumulative, the FALSE gives the exact value and TRUE gives the less than value.

Thus, the probabilities $P(X=0),P(X\le 1),P(X\le 2),P(X\le 3)\text{ and P(X}\le \text{4)}$ can be calculated as:

$P(X=0)$	=POISSON.DIST(0,1.67,FALSE)
$P(X\le 1)$	=POISSON.DIST(1,1.67,TRUE)
$P(X\le 2)$	=POISSON.DIST(2,1.67,TRUE)
$P(X\le 3)$	=POISSON.DIST(3,1.67,TRUE)
$P(X\le 4)$	=POISSON.DIST(4,1.67,TRUE)

The result will be as:

$P(X=0)$	0.1882
$P(X\le 1)$	0.5026
$P(X\le 2)$	0.7651
$P(X\le 3)$	0.9112
$P(X\le 4)$	0.9723

And the probability that there would be more than four cases of salmonellosis in South Dakota in a given month can be calculated as:

  $\begin{array}{l}P(X>4)=1-P(X\le 4)\\ =1-0.9723\\ =0.0277\end{array}$

(c)

To determine

To find what is the probability that $\text{14}$ or more cases would arise in one month and what does this unusual report suggest and explain what do you think would be your next action if you worked at the CDC at that time.

Answer to Problem 12.46E

The probability is zero.

Explanation of Solution

In the question it is given that the average number of salmonellosis cases per month in South Dakota is $\text{1}\text{.67}$ . X approximately follows Poisson distribution with parameter $\text{1}\text{.67}$ . Thus, to calculate the probabilities we will use the Poisson excel function as:

  $=\text{POISSON}\text{.DIST(}x,\text{mean, cumulative)}$

In the cumulative, the FALSE gives the exact value and TRUE gives the less than value.

Thus, theprobability that $\text{14}$ or more cases would arise in one month can be calculated as:

$P(X\ge 14)=$

=1-POISSON.DIST(13,1.67,TRUE)

The result will be:

$P(X\ge 14)=$

0

This suggest that it is almost impossible to have number of salmonellosis cases greater than or equal to $14$ in a month.