solve part c and d Spray drift is a constant concern for pesticide applicators a
ID: 2946892 • Letter: S
Question
solve part c and d
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 contro treatment ) sprayed through a 760 ml min nozze. (a) what is the probability that the size of a single droplet is less than 1350 ?m? At least 1000 ?m? (Round your answers to four decimal places.) less than 1350 ?m 0.9722 at least 1000 ?m 0.6305 (b) what is the probability that the size of a single droplet is between 1000 and 1350 ?m? (Round your answer to four decimal places.) 0.6077 (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 923.76 X ?m în size. (d) If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds 1350 ?m? (Round your answer to four decimal places.) 0.0228XExplanation / Answer
Data given is:
Mean, m = 1050
Standard Deviation, S = 150
(c)
For the smallest 2%, that is, for the p-value of 0.02, we have:
z = -2.054
Now, (X-m)/S = z
So,
X = z*S + m = -2.054*150 + 1050 = 741.9
So, the smalles 2% are those smaller than 741.9 um in size.
(d)
P(at least one exceeds size 1350) = 1 - P(all have size below 1350)
As calculated in part (a), P(a droplet has size less than 1350) = 0.9722
So,
P(all have size below 1350) = 0.9722^5 = 0.8685
So,
P(at least one exceeds size 1350) = 1 - 0.8685 = 0.1315
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