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 pm and standard deviation 150 pm was a reasonable model for droplet size for water (the "control treatment") sprayed through a 760 mlmin nozzle. (a) What is the probability that the size of a single droplet is less than 1395 pm? At least 900 pm? (Round your answers to four decimal places.) less than 1395 pm at least 900 pm (b) What is the probability that the size of a single droplet is between 900 and 1395 pm? (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 pm in size. (d) If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds 1395 pm? (Round your answer to four decimal places.)

Explanation / Answer

Solution:-

Solution:-
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) = 0.8413
(From Normal probability table)

b) P( 900 < x < 1395) = P[( 900 - 1050) / 150 < Z < ( 1395 - 1050) / 150]
P( -1 < Z < 2.3) = 0.9893 - 0.1587 = 0.8306

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=0.0107, x=1,2,3,4,5
P( x >=1) = 1-P(x=0)
P(x=0) = 5C0 (0.0107)^0 (1-0.0107)^(5-0)
P(x=0) = 0.9476
1-P(x=0) = 1-0.9476 = 0.0524

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