My NotesAsk Your -125 points DevoreSa9 4.E.036. Spray drift is a constant concer

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My NotesAsk Your -125 points DevoreSa9 4.E.036. Spray drift is a constant concern for pesticide applicators and agricultural producers. The inverse relationship between droplet size and drift potential is well known. The paper "Effects of 2,4-D Formulation and Quinclorac on Spray Droplet Size and Deposition"t investigated the effects of herbicide formulation on spray atomization. A figure in a paper suggested the normal distribution with mean 1050 m and standard deviation 150 m was a reasonable model for droplet size for water (the "control treatment") sprayed through a 760 ml/min nozzle (a) What is the probability that the size of a single droplet is less than 1365 um? At least 1000 um? (Round your answers to four decimal places.) less than 1365 um at least 1000 um (b) What is the probability that the size of a single droplet is between 1000 and 1365 um? (Round your answer to four decimal places.) (c) How would you characterize the smallest 2% of all droplets? (Round your answer to two decimal places.) The smallest 2% of droplets are those smaller than m in size (d) If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds 1365 um? (Round your answer to four decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question Submit Answer Save ProgressPractice Another Version

Explanation / Answer

Solution:- Given that mean = 1050 , standard deviation = 150

Formula: Z = (X-)/

a) P(X < 1365) = P(Z < (1365 - 1050)/150)
= P(Z < 2.1)
= 0.9821

P(X >= 1000) = p(Z >= (1000 - 1050)/150)
= P(Z >= -0.3333)
= 0.6293

b) P(1000 < X < 1365) = P(-0.3333 < Z < 2.1)
= 0.6114

c) P(Z < (x - 1050)/150) = 0.02
(x - 1050)/150 = -2.0537
x = 1050 - (2.0537*150)
x = 741.95

d)For n = 5
=> P(X >= 1365) = P(Z >= (1365-1050)/(150/sqrt(5))
= P(Z >= 4.6957)
= 0
  

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