Chapter 12, Problem 12.47E

(a)

To determine

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

Answer to Problem 12.47E

X follow Poisson distribution and the mean and standard deviation are $15.58$ and $3.947$ respectively.

Explanation of Solution

In the question, it is given that the CDC reports that the average number of salmonellosis per month in Wisconsin is $15.58$ . Now, X is the monthly count of salmonellosis cases in Wisconsin. 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 $15.58$ . Thus, the mean and standard deviation is calculated as:

  $\begin{array}{l}\text{Mean}=\mu \\ =15.58\\ \text{St}\text{.Dev}\text{.}=\sigma \\ =\sqrt{\mu }\\ =\sqrt{15.58}\\ =3.947\end{array}$

(b)

To determine

To find the probabilities $P(X=0),P(X\le 5),P(X\le 15),P(X\le 25)$ and also find the probability that there would be more than $25$ cases of salmonellosis in Wisconsin in a given month.

Answer to Problem 12.47E

The probability is:

$P(X=0)$	1.71 $\times {10}^{-7}$
$P(X\le 5)$	0.00186
$P(X\le 15)$	0.50889
$P(X\le 25)$	0.99031

And the probability that there would be more than $25$ cases of salmonellosis in Wisconsin in a given month is $0.0097$ .

Explanation of Solution

In the question, it is given that the CDC reports that the average number of salmonellosis per month in Wisconsin is $15.58$ . Now, X is the monthly count of salmonellosis cases in Wisconsin. X approximately follows Poisson distribution with parameter $15.58$ . 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 $P(X=0),P(X\le 5),P(X\le 15),P(X\le 25)$ can be calculated as:

$P(X=0)$	=POISSON.DIST(0,15.58,FALSE)
$P(X\le 5)$	=POISSON.DIST(5,15.58,TRUE)
$P(X\le 15)$	=POISSON.DIST(15,15.58,TRUE)
$P(X\le 25)$	=POISSON.DIST(25,15.58,TRUE)

The result will be as:

$P(X=0)$	1.71 $\times {10}^{-7}$
$P(X\le 5)$	0.00186
$P(X\le 15)$	0.50889
$P(X\le 25)$	0.99031

And the probability that there would be more than $25$ cases of salmonellosis in Wisconsin in a given month is:

  $\begin{array}{l}P(X>25)=1-P(X\le 25)\\ =1-0.9903\\ =0.0097\end{array}$

(c)

To determine

To find the probability that $48$ or more cases would arise in one month and what do you make of the fact that a salmonellosis outbreak occurred in two states at the same time.

Answer to Problem 12.47E

Explanation of Solution

In the question, it is given that the CDC reports that the average number of salmonellosis per month in Wisconsin is $15.58$ . Now, X is the monthly count of salmonellosis cases in Wisconsin. X approximately follows Poisson distribution with parameter $15.58$ . 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 probability that $48$ or more cases would arise in one month is calculated as:

  $P(X\ge 48)=1-P(X<47)$

$P(X\ge 48)=$

=1-POISSON.DIST(47,15.58,TRUE)

The result is as:

$P(X\ge 48)=$

$3.53\times {10}^{-11}$

Thus, from this we can say that this is too unlikely to be due to random isolated cases and it points to an epidemic. The contaminated food must have been distributed at least in South Dakota and Wisconsin.