To use technology to find the cumulative probabilities of Xequals 0 , 1 , 2 , 3 , to 15 .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 12, Problem 12.18AYK

(a)

To determine

To use technology to find the cumulative probabilities of Xequals 0,1,2,3, to 15 .

(a)

Expert Solution

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, to 15 we will use excel function. The excel function is:

=POISSON.DIST(x,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.

(b)

Expert Solution

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:

=POISSON.DIST(x,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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

The table below was compiled for a middle school from the 2003 English/Language Arts PACT exam. Grade 6 7 8 Below Basic 60 62 76 Basic 87 134 140 Proficient 87 102 100 Advanced 42 24 21 Partition the likelihood ratio test statistic into 6 independent 1 df components. What conclusions can you draw from these components?

What is the value of the maximum likelihood estimate, θ, of θ based on these data? Justify your answer. What does the value of θ suggest about the value of θ for this biased die compared with the value of θ associated with a fair, unbiased, die?

Show that L′(θ) = Cθ394(1 −2θ)604(395 −2000θ).

Answer 2

Question

Chapter 12, Problem 12.18AYK

(a)

To determine

To use technology to find the cumulative probabilities of Xequals 0,1,2,3, to 15 .

(a)

Expert Solution

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, to 15 we will use excel function. The excel function is:

=POISSON.DIST(x,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.

(b)

Expert Solution

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:

=POISSON.DIST(x,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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 12, Problem 12.18AYK

(a)

To determine

To use technology to find the cumulative probabilities of Xequals 0,1,2,3, to 15 .

(a)

Expert Solution

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, to 15 we will use excel function. The excel function is:

=POISSON.DIST(x,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.

(b)

Expert Solution

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:

=POISSON.DIST(x,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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 12, Problem 12.18AYK

(a)

To determine

To use technology to find the cumulative probabilities of Xequals 0,1,2,3, to 15 .

(a)

Expert Solution

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, to 15 we will use excel function. The excel function is:

=POISSON.DIST(x,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.

(b)

Expert Solution

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:

=POISSON.DIST(x,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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 12, Problem 12.18AYK

(a)

To determine

To use technology to find the cumulative probabilities of Xequals 0,1,2,3, to 15 .

(a)

Expert Solution

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, to 15 we will use excel function. The excel function is:

=POISSON.DIST(x,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.

(b)

Expert Solution

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:

=POISSON.DIST(x,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.

Answer 6

Chapter 12, Problem 12.18AYK

(a)

To determine

To use technology to find the cumulative probabilities of Xequals 0,1,2,3, to 15 .

(a)

Expert Solution

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, to 15 we will use excel function. The excel function is:

=POISSON.DIST(x,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

Answer 7

Chapter 12, Problem 12.18AYK

(a)

To determine

To use technology to find the cumulative probabilities of Xequals 0,1,2,3, to 15 .

Answer 8

(a)

Expert Solution

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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, to 15 we will use excel function. The excel function is:

=POISSON.DIST(x,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

Answer 13

(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.

(b)

Expert Solution

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:

=POISSON.DIST(x,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.

Answer 14

(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 15

(b)

Expert Solution

Answer to Problem 12.18AYK

0.003346 .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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:

=POISSON.DIST(x,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.