(a)
To find out what distribution does X follow and also give the mean and standard deviation of X .
(a)
![Check Mark](/static/check-mark.png)
Answer to Problem 12.46E
X follows Poisson distribution and the mean and standard deviation is
Explanation of Solution
In the question it is given that the average number of salmonellosis cases per month in South Dakota is
(b)
To find out the probabilities
(b)
![Check Mark](/static/check-mark.png)
Answer to Problem 12.46E
The probabilities are:
0.1882 | |
0.5026 | |
0.7651 | |
0.9112 | |
0.9723 |
And the probability that there would be more than four cases of salmonellosis in South Dakota in a given monthis
Explanation of Solution
In the question it is given that the average number of salmonellosis cases per month in South Dakota is
In the cumulative, the FALSE gives the exact value and TRUE gives the less than value.
Thus, the probabilities
=POISSON.DIST(0,1.67,FALSE) | |
=POISSON.DIST(1,1.67,TRUE) | |
=POISSON.DIST(2,1.67,TRUE) | |
=POISSON.DIST(3,1.67,TRUE) | |
=POISSON.DIST(4,1.67,TRUE) |
The result will be as:
0.1882 | |
0.5026 | |
0.7651 | |
0.9112 | |
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:
(c)
To find what is the probability that
(c)
![Check Mark](/static/check-mark.png)
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
In the cumulative, the FALSE gives the exact value and TRUE gives the less than value.
Thus, theprobability that
=1-POISSON.DIST(13,1.67,TRUE) |
The result will be:
0 |
This suggest that it is almost impossible to have number of salmonellosis cases greater than or equal to
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Chapter 12 Solutions
ACHIEVE F/PRACT OF STAT IN LIFE-ACCESS
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