Chapter 12, Problem 12.18AYK

(a)

To determine

To use technology to find the cumulative probabilities of Xequals $0,1,2,3,\text{ to }15$ .

Answer to Problem 12.18AYK

X	cumulative probabilities
1	0.00394
2	0.017364
3	0.051819
4	0.118145
5	0.220287
6	0.351369
7	0.49556
8	0.634343
9	0.75308
10	0.844508
11	0.908507
12	0.949573
13	0.973897
14	0.987275
15	0.994143

Explanation of Solution

It is given in the question that the CDC receives reports of $7.7$ cases of typhoid fever per week, on average, from all over the United States, although most cases were acquired during international travel. Let Xbe the number of cases of typhoid cases received. Xhave a Poisson distribution with average of $7.7$ cases of typhoid fever per week. To find the cumulative probabilities of X equals $0,1,2,3,\text{ to }15$ we will use excel function. The excel function is:

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

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

Thus, the calculation is as:

X	cumulative probabilities
1	=POISSON.DIST(AJ116,7.7,TRUE)
2	=POISSON.DIST(AJ117,7.7,TRUE)
3	=POISSON.DIST(AJ118,7.7,TRUE)
4	=POISSON.DIST(AJ119,7.7,TRUE)
5	=POISSON.DIST(AJ120,7.7,TRUE)
6	=POISSON.DIST(AJ121,7.7,TRUE)
7	=POISSON.DIST(AJ122,7.7,TRUE)
8	=POISSON.DIST(AJ123,7.7,TRUE)
9	=POISSON.DIST(AJ124,7.7,TRUE)
10	=POISSON.DIST(AJ125,7.7,TRUE)
11	=POISSON.DIST(AJ126,7.7,TRUE)
12	=POISSON.DIST(AJ127,7.7,TRUE)
13	=POISSON.DIST(AJ128,7.7,TRUE)
14	=POISSON.DIST(AJ129,7.7,TRUE)
15	=POISSON.DIST(AJ130,7.7,TRUE)

The result will be as:

X	cumulative probabilities
1	0.00394
2	0.017364
3	0.051819
4	0.118145
5	0.220287
6	0.351369
7	0.49556
8	0.634343
9	0.75308
10	0.844508
11	0.908507
12	0.949573
13	0.973897
14	0.987275
15	0.994143

(b)

To determine

To find out what is the probability that the CDC receives reports of more than $15$ cases of typhoid fever next month and if this happen what would you conclude.

Answer to Problem 12.18AYK

  $0.003346$ .

Explanation of Solution

It is given in the question that the CDC receives reports of $7.7$ cases of typhoid fever per week, on average, from all over the United States, although most cases were acquired during international travel. Let Xbe the number of cases of typhoid cases received. Xhave a Poisson distribution with average of $7.7$ cases of typhoid fever per week. To find the probability that the CDC receives reports of more than $15$ cases of typhoid fever next month we will use excel function. The excel function is:

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

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

Thus, the calculation is as:

  $P(X>15)=1-P(X<15)$

P(X>15)=

=1-POISSON.DIST(15,7.7,TRUE)

The result is as:

P(X>15)=

0.005857

Thus, the probability that the CDC receives reports of more than $15$ cases of typhoid fever next monthis $0.003346$ . If this happens that then we conclude that the city have most cases for typhoid fever and it will be increasing more and the people must take precaution for the same by avoiding unhygienic foods but the probability calculated above is less so the chance for this is to happen is low.