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Chapter 5: Problem 29

Suppose that \(10 \%\) of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred 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 probability 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.

Short Answer

Expert verified
In summary: a. On average, 10 fields are expected to be infested with sweet potato whiteflies. b. We would expect to find the number of infested fields between 4.12 and 15.88 with approximately 95% probability. c. If we found that 25 fields were infested, it might suggest that the binomial characteristics may not be satisfied, as the infestation rate or independence of trials might not be met.

Step by step solution

01

Understanding the Binomial Distribution

A binomial distribution can be described for a random variable that represents the number of successes in a fixed number of trials, each with a constant probability of success. In this case, the "success" represents finding a field infested with whiteflies. There are two outcomes, infested or not infested, and the trials (fields) are independent of each other.
02

Calculate the Average (Expected Value)

In a binomial distribution, the average (also called the expected value) is given by the formula: $$ \mu = n \times p, $$ where \(n\) is the total number of trials (fields) and \(p\) is the probability of success (infestation). For this problem, \(n = 100\) and \(p = 0.10\). Thus, the average number of infested fields is: $$ \mu = 100 \times 0.10 = 10. $$
03

a. Answer

On average, 10 fields are expected to be infested with sweet potato whiteflies.
04

Calculate the Standard Deviation (Approximation)

To find the limits for the approximate 95% probability interval, we will use the normal distribution approximation of the binomial distribution. First, let's calculate the standard deviation, which is given by: $$ \sigma = \sqrt{n \times p \times (1-p)}, $$ So calculating, $$ \sigma = \sqrt{100 \times 0.10 \times (1-0.10)} = \sqrt{9} = 3. $$
05

Approximate the 95% Probability Interval

Using the normal approximation, for a 95% interval, we can use the formula: $$ \mu - 1.96 \times \sigma \leq X \leq \mu + 1.96 \times \sigma. $$ Plugging in the values, $$ 10 - 1.96 \times 3 \leq X \leq 10 + 1.96 \times 3 \\ 10 - 5.88 \leq X \leq 10 + 5.88 \\ 4.12 \leq X \leq 15.88. $$
06

b. Answer

We would expect to find the number of infested fields between 4.12 and 15.88 with approximately 95% probability.
07

Conclusion and Characteristics of a Binomial Experiment

If we found that 25 fields were infested, that would be significantly outside the expected 95% range calculated in part b. This would suggest that the assumption of a constant probability of infestation (\(p = 0.10\)) might not be correct. There might be a factor affecting the infestation rate, such as clustering of whiteflies or other environmental factors. It's possible that one of the characteristics of the binomial experiment is not satisfied, as the trials may not be independent (independence is violated if the presence of whiteflies in one field affects the probability of their presence in another field).
08

c. Answer

If 25 fields were found infested, we might conclude that the binomial characteristics may not be satisfied, as the infestation rate or independence of trials might not be met.

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