Question 1197206: Suppose that 10% of the fields in a given agricultural**a) Average Number of Infested Fields**
* **Binomial Distribution:** This situation fits the binomial distribution:
* **Fixed number of trials:** 100 fields (n = 100)
* Two possible outcomes: infested or not infested
* Probability of success (infestation) is constant: p = 0.10
* Trials are independent
* **Mean of Binomial Distribution:**
* The average number of infested fields (mean) is given by: μ = n * p
* μ = 100 * 0.10 = 10 fields
**b) 95% Confidence Interval**
* **Standard Deviation of Binomial Distribution:**
* σ = √(n * p * (1 - p))
* σ = √(100 * 0.10 * 0.90) = √9 = 3
* **95% Confidence Interval:**
* For a normal distribution, approximately 95% of the data falls within 2 standard deviations of the mean.
* Lower Limit: μ - 2σ = 10 - 2 * 3 = 4 fields
* Upper Limit: μ + 2σ = 10 + 2 * 3 = 16 fields
* **Therefore, we would expect to find the number of infested fields within the range of 4 to 16 fields with a probability of approximately 95%.**
**c) Finding 25 Infested Fields**
* **25 infested fields is significantly higher than the expected range (4-16 fields).**
* **Possible Explanation:**
* **Violation of the assumption of independence:**
* If the infestation in one field increases the likelihood of infestation in nearby fields (e.g., due to pest migration), the trials would not be independent, and the binomial distribution might not accurately model the situation.
* **Change in infestation rate:**
* The actual infestation rate in the agricultural area might have increased significantly since the initial estimate of 10%.
* **Sampling bias:**
* The 100 fields selected might not be a truly random sample and may be more likely to be infested than the overall population of fields.
**In conclusion:**
Finding 25 infested fields in the sample suggests that the observed number of infested fields is significantly higher than expected under the assumption of a binomial distribution with a 10% infestation rate. This discrepancy could indicate a violation of the independence assumption or a change in the underlying infestation rate. Further investigation would be necessary to determine the cause of this unexpected result.
area are infested with the sweet potato whitefly. 100 fields in this area are randomly selected and checked for whitefly. a) What is the average number of fields sampled that are infested with whitefly? b) Within what limits would you expect to find the number of infested fields, with robability approximately 95%? c) What might you conclude if you found that x = 25 fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.
Found 2 solutions by ElectricPavlov, ikleyn:
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **a) Average Number of Infested Fields**
* **Binomial Distribution:** This situation fits the binomial distribution:
* **Fixed number of trials:** 100 fields (n = 100)
* Two possible outcomes: infested or not infested
* Probability of success (infestation) is constant: p = 0.10
* Trials are independent
* **Mean of Binomial Distribution:**
* The average number of infested fields (mean) is given by: μ = n * p
* μ = 100 * 0.10 = 10 fields
**b) 95% Confidence Interval**
* **Standard Deviation of Binomial Distribution:**
* σ = √(n * p * (1 - p))
* σ = √(100 * 0.10 * 0.90) = √9 = 3
* **95% Confidence Interval:**
* For a normal distribution, approximately 95% of the data falls within 2 standard deviations of the mean.
* Lower Limit: μ - 2σ = 10 - 2 * 3 = 4 fields
* Upper Limit: μ + 2σ = 10 + 2 * 3 = 16 fields
* **Therefore, we would expect to find the number of infested fields within the range of 4 to 16 fields with a probability of approximately 95%.**
**c) Finding 25 Infested Fields**
* **25 infested fields is significantly higher than the expected range (4-16 fields).**
* **Possible Explanation:**
* **Violation of the assumption of independence:**
* If the infestation in one field increases the likelihood of infestation in nearby fields (e.g., due to pest migration), the trials would not be independent, and the binomial distribution might not accurately model the situation.
* **Change in infestation rate:**
* The actual infestation rate in the agricultural area might have increased significantly since the initial estimate of 10%.
* **Sampling bias:**
* The 100 fields selected might not be a truly random sample and may be more likely to be infested than the overall population of fields.
**In conclusion:**
Finding 25 infested fields in the sample suggests that the observed number of infested fields is significantly higher than expected under the assumption of a binomial distribution with a 10% infestation rate. This discrepancy could indicate a violation of the independence assumption or a change in the underlying infestation rate. Further investigation would be necessary to determine the cause of this unexpected result.
Answer by ikleyn(52770)  (Show Source):
You can put this solution on YOUR website! .
In this post, the problem's description is corrupted by other person (not intentionally, I hope).
Therefore, it is not possible to understand the problem and the solution,
as well as is not possible to check the solution.
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