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

Question

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Answer 1

Question

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 .

(a)

Expert Solution

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 1.67 . 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 1.67 . Thus, the mean and standard deviation is calculated as:

Mean=μ=1.67St.Dev.=σ=μ=1.67=1.2923

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 12.46E

The probabilities are:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,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≤1),P(X≤2),P(X≤3) and P(X≤4) can be calculated as:

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

The result will be as:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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:

P(X>4)=1−P(X≤4)=1−0.9723=0.0277

(c)

To determine

To find what is the probability that 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.

(c)

Expert Solution

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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,mean, cumulative)

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

Thus, theprobability that 14 or more cases would arise in one month can be calculated as:

P(X≥14)=

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

The result will be:

P(X≥14)=

0

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

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Answer 2

Question

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 .

(a)

Expert Solution

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 1.67 . 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 1.67 . Thus, the mean and standard deviation is calculated as:

Mean=μ=1.67St.Dev.=σ=μ=1.67=1.2923

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 12.46E

The probabilities are:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,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≤1),P(X≤2),P(X≤3) and P(X≤4) can be calculated as:

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

The result will be as:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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:

P(X>4)=1−P(X≤4)=1−0.9723=0.0277

(c)

To determine

To find what is the probability that 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.

(c)

Expert Solution

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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,mean, cumulative)

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

Thus, theprobability that 14 or more cases would arise in one month can be calculated as:

P(X≥14)=

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

The result will be:

P(X≥14)=

0

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

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

Answer 3

Question

Chapter 12, Problem 12.46E

(a)

To determine

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

(a)

Expert Solution

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 1.67 . 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 1.67 . Thus, the mean and standard deviation is calculated as:

Mean=μ=1.67St.Dev.=σ=μ=1.67=1.2923

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 12.46E

The probabilities are:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,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≤1),P(X≤2),P(X≤3) and P(X≤4) can be calculated as:

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

The result will be as:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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:

P(X>4)=1−P(X≤4)=1−0.9723=0.0277

(c)

To determine

To find what is the probability that 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.

(c)

Expert Solution

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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,mean, cumulative)

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

Thus, theprobability that 14 or more cases would arise in one month can be calculated as:

P(X≥14)=

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

The result will be:

P(X≥14)=

0

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

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

Answer 4

Question

Chapter 12, Problem 12.46E

(a)

To determine

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

(a)

Expert Solution

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 1.67 . 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 1.67 . Thus, the mean and standard deviation is calculated as:

Mean=μ=1.67St.Dev.=σ=μ=1.67=1.2923

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 12.46E

The probabilities are:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,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≤1),P(X≤2),P(X≤3) and P(X≤4) can be calculated as:

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

The result will be as:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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:

P(X>4)=1−P(X≤4)=1−0.9723=0.0277

(c)

To determine

To find what is the probability that 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.

(c)

Expert Solution

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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,mean, cumulative)

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

Thus, theprobability that 14 or more cases would arise in one month can be calculated as:

P(X≥14)=

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

The result will be:

P(X≥14)=

0

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

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

Answer 5

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 .

(a)

Expert Solution

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 1.67 . 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 1.67 . Thus, the mean and standard deviation is calculated as:

Mean=μ=1.67St.Dev.=σ=μ=1.67=1.2923

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 12.46E

The probabilities are:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,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≤1),P(X≤2),P(X≤3) and P(X≤4) can be calculated as:

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

The result will be as:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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:

P(X>4)=1−P(X≤4)=1−0.9723=0.0277

(c)

To determine

To find what is the probability that 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.

(c)

Expert Solution

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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,mean, cumulative)

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

Thus, theprobability that 14 or more cases would arise in one month can be calculated as:

P(X≥14)=

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

The result will be:

P(X≥14)=

0

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

Answer 6

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 .

(a)

Expert Solution

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 1.67 . 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 1.67 . Thus, the mean and standard deviation is calculated as:

Mean=μ=1.67St.Dev.=σ=μ=1.67=1.2923

Answer 7

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 8

(a)

Expert Solution

Answer to Problem 12.46E

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

In the question it is given that the average number of salmonellosis cases per month in South Dakota is 1.67 . 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 1.67 . Thus, the mean and standard deviation is calculated as:

Mean=μ=1.67St.Dev.=σ=μ=1.67=1.2923

Answer 13

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 12.46E

The probabilities are:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,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≤1),P(X≤2),P(X≤3) and P(X≤4) can be calculated as:

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

The result will be as:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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:

P(X>4)=1−P(X≤4)=1−0.9723=0.0277

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

Answer to Problem 12.46E

The probabilities are:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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 .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

In the question it is given that the average number of salmonellosis cases per month in South Dakota is 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,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≤1),P(X≤2),P(X≤3) and P(X≤4) can be calculated as:

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

The result will be as:

P(X=0)	0.1882
P(X≤1)	0.5026
P(X≤2)	0.7651
P(X≤3)	0.9112
P(X≤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:

P(X>4)=1−P(X≤4)=1−0.9723=0.0277

Answer 20

(c)

To determine

To find what is the probability that 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.

(c)

Expert Solution

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 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,mean, cumulative)

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

Thus, theprobability that 14 or more cases would arise in one month can be calculated as:

P(X≥14)=

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

The result will be:

P(X≥14)=

0

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

Answer 21

(c)

To determine

To find what is the probability that 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 22

(c)

Expert Solution

Answer to Problem 12.46E

The probability is zero.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

In the question it is given that the average number of salmonellosis cases per month in South Dakota is 1.67 . X approximately follows Poisson distribution with parameter 1.67 . Thus, to calculate the probabilities we will use the Poisson excel function as:

=POISSON.DIST(x,mean, cumulative)

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

Thus, theprobability that 14 or more cases would arise in one month can be calculated as:

P(X≥14)=

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

The result will be:

P(X≥14)=

0

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

Answer 27

Ch. 12 - Prob. 12.1AYK Ch. 12 - Prob. 12.2AYK Ch. 12 - Prob. 12.3AYK Ch. 12 - Prob. 12.4AYK Ch. 12 - Prob. 12.5AYK Ch. 12 - Prob. 12.6AYK Ch. 12 - Prob. 12.7AYK Ch. 12 - Prob. 12.8AYK Ch. 12 - Prob. 12.9AYK Ch. 12 - Prob. 12.10AYK

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

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

Answer to Problem 12.46E

Explanation of Solution

To find out the probabilities P(X=0),P(X≤1),P(X≤2),P(X≤3) and P(X≤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

Explanation of Solution

To find what is the probability that 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

Explanation of Solution

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