Please help! Thank you! 5.23 Monitoring the emerald ash borer. T ash borer is a
ID: 3045335 • Letter: P
Question
Please help! Thank you! 5.23 Monitoring the emerald ash borer. T ash borer is a beetle that poses a serio trees. Purple traps are often used to detect populations of this pest. In the counties of where the beetle is present, thousand to monitor the population. These traps arused odically. The distribution of beetle countse discrete and strongly skewed. A ma beetles, and only a few will have more than 1 hhave this exercise, assume that the mean numbeee. Fo trapped is 0.3 with a standard deviation of 0.8 (a er. The emerald parts to detect or monitor of traps are re checked peri- per trap is have no s es (a) Suppose that your state does not have the check all the traps, and so it plans to check of n = 100 traps. What are the mean and standard de he only an SRS evia- tion of the average number of beetles in 100 t traps? (b) Use the central limit theorem to find the probabil that the average number of beetles in 100 traps i than 0.5. (c) Do you think it is appropriate in this situation to use the central limit theorem? Explain your answer.Explanation / Answer
Mean number of beetles trapped = 0.3
Standard deviation of beetles trapped = 0.8
(a) Here mean number of beetles in 100 traps = = 0.3
standard deviaition of beetles trapped s = /sqrt(n) = 0.8/sqrt(100) = 0.8/10 = 0.08
(b) As we use central limit theorm
the mean number of beeteles = 0.3 and standard deviation = 0.08
if x is the average number of beetles in 100 trap
Pr(x > 0.5) = NORM (x > 0.5 ; 0.3 ; 0.08) = 1 - NORM (x < 0.5 ; 0.3 ; 0.08)
Z = (0.5 - 0.3)/ 0.08 = 2.5
Pr(x > 0.5) = NORM (x > 0.5 ; 0.3 ; 0.08) = 1 - NORM (x < 0.5 ; 0.3 ; 0.08) = 1- Pr(Z < 2.5) = 1 - 0.9938 = 0.0062
(c) Yes, as the number of traps or SRS (n) is greater than 30 then it is appropriate in this situation to use the central limit theorem.