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" investigated the effects of herbicide formulation on spray atomization. A figure in a paper suggested the normal distribution with mean 1050 mu m and standard deviation 150 mu m was a reasonable model for droplet size for water (the "control treatment") sprayed through a 760 ml/min nozzle. What is the probability that the size of a single droplet is less than 1425 mu m? At least 950 mu m? (Round your answers to four decimal places.) less than 1425 mu m at least 950 mu m What is the probability that the size of a single droplet is between 950 and 1425 mu m? (Round your answer to four decimal places.) 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 mu m in size. If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds 1425 mu m? (Round your answer to four decimal places.)

Explanation / Answer

a)
= 1050
= 150
standardize x to z = (x - ) /
P(x < 1425) = P( z < (1425-1050) / 150)
= P(z < 2.5) = 0.9938
(From Normal probability table)
-----------------------------------
= 1050
= 150
standardize x to z = (x - ) /
P(x > 950) = P( z > (950-1050) / 150)
= P(z > -0.67) = 1 - 0.2514

(From Normal probability table)

= 0.7486

b)
= 1050
= 150
standardize x to z = (x - ) /
P( 950 < x < 1425) = P[( 950 - 1050) / 150 < Z < ( 1425 - 1050) / 150]
P( -0.67 < Z < 2.5) = 0.9938 - 0.2514 =0.7424
(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)

= 1050 + (-307.5) = 742.5 µm in size

d)
The probability of one droplet exceeding 1425 µm is :
= 1050
= 150
standardize x to z = (x - ) /
P(x > 1425) = P( z > (1425-1050) / 150)
= P(z > 2.5) = 1 – 0.9938
(From Normal probability table)

=0.0062

Use the binomial probability with n=5, p=0.0062, x=1,2,3,4,5
P( x >=1) = 1 - P(x=0)
P(x=0) = 5C0 * (0.0062)^0 * (1-0.0062)^(5-0)
P(x=0) = (1) * (1) * (0.9938)^5 = 0.969382 ~ 0.9694
= 1 - P(x=0) = 1 - 0.9694

= 0.030618 ~ 0.0306

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