Spray drift is a constant concern for pesticide applicators and agricultural pro

Question

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 150um was a reasonable de ora opie size a e, e contro treatment sprayed through a 760 ml/min nozzle. (a) what is the probability that the size of a single droplet is less than 1395 um? At least 900 m? (Round your answers to four decimal places.) less than 1395 m at least 900 m (b) what is the probability that the size of a single droplet is between 900 and 1395 m? (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 L335 mRun places.) yo answer to urdea a You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

a)
= 1050
= 150
standardize x to z = (x - ) /
P(x < 1395) = P( z < (1395-1050) / 150)
= P(z < 2.3) = 0.9893
(From Normal probability table)
-----------------------------------
= 1050
= 150
standardize x to z = (x - ) /
P(x > 900) = P( z > (900-1050) / 150)
= P(z > -1) = 10.8413=0.1587
(From Normal probability table)

b)
= 1050
= 150
standardize x to z = (x - ) /
P( 900 < x < 1395) = P[( 900 - 1050) / 150 < Z < ( 1395 - 1050) / 150]
P( -1 < Z < 2.3) = 0.9893 - 0.1587 =0.8306
(From Normal probability table)

c)
From the normal distribution table, P( z < -2.05) =0.02
z = (x - ) /
-2.05 = (x-1050)/150
solve for x
x = 1050 + (150)(-2.05) = 742.5 µm in size

d)
The probability of one droplet exceeding 1395 µm is :
= 1050
= 150
standardize x to z = (x - ) /
P(x > 1395) = P( z > (1395-1050) / 150)
= P(z > 2.3) = 0.0107
(From Normal probability table)

Use the binomial probability with n=5, p=0.0047, x=1,2,3,4,5
P( x >=1) = 1-P(x=0)
P(x=0) = 5C0 (0.0047)^0 (1-0.0047)^(5-0)
P(x=0) = (1) (1) (0.9953)^5 = 0.9767
1-P(x=0) = 1-0.9767 = 0.0233

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