Question

Suppose that 15% 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. Based on your knowledge of the empirical rule, within what limits would you expect to find the number of infested fields, with probability approximately 95%? (Round your answers to three decimal places.) 4.29 X fields to 25.71 x fields What might you conclude if you found that x = 45 fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.
A. Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent. O
B. Based on the limits above, it is unlikely that we would see x = 45, so it might be possible that the trials have more than two possible outcomes.
C. Based on the limits above, it is unlikely that we would see x = 45, so it might be possible that there are an indefinite number of trials.
D. Based on the limits above, it is likely that we would see x = 45, so all of the characteristics of a binomial experiment are satisfied.

Answer 1

Answer:

a.$\mu=15$

b.$\mu=7.8586 \ and \ \mu=22.1414$

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

$P(X=k)=C_k^n.p^k.(1-p)^{n-k}$

Let n be the number of trials,$n=100$

and p be the probability of success,$p=15\%$

The mean of a binomial distribution is the probability x sample size.

$\mu=np=100\times0.15=15$

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as:$\sigma=\sqrt npq\\q=1-p$

Therefore, $\sigma=\sqrt(100\times0.015\times0.85)=3.5707$

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

$sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414$

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c.  $x=45$ is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.