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

Question

Spray drift is a constent 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 ot herbicide formulation on spray atomization. A figure in paper suggested the normal distribution with mean 1050 m and standard deviation 150 pm was reason ble 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 gle draplet is less than 1365 u? At least 900n? (Round your aiswers to four decimal places.) less than 1365 m at least 900 m (b) what is the probability that the size of a single droplet is between 900 and 1365 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.) Th. smallest 2% of droplets are those smaller than jarm in size. (d) If the sizes of five independently selected droplets are measured, what is the probability that at lesst one exceeds 1365 um? (Round your answer to four decimal places.)

Explanation / Answer

Solution:-
a) = 1050 = 150
standardize x to z = (x - ) /
P(x < 1365) = P( z < (1365-1050) / 150)
= P(z < 2.1)
= 0.9821 (From Normal probability table)
-----------------------------------

P(x > 975) = P( z > (900-1050) / 150)
= P(z > -1) = 0.8413 (From Normal probability table)

b) = 1050 = 150
standardize x to z = (x - ) /
P( 900 < x < 1365) = P[( 900 - 1050) / 150 < Z < ( 1365 - 1050) / 150]
P( -1 < Z < 2.1) = 0.9821 - 0.1587
= 0.8234 (From Normal probability table)

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

  The smallest 2% of droplets are those smaller than 742.50 µm in size.

d)The probability of one droplet exceeding 1440 µm is :
= 1050 = 150
standardize x to z = (x - ) /
P(x > 1365) = P( z > (1365-1050) / 150)
= P(z > 2.1)
= 0.0179 (From Normal probability table)

Use the binomial probability with n = 5, p = 0.0179, x=1,2,3,4,5
P( x >=1) = 1-P(x=0)
P(x=0) = 5C0 (0.0179)^0 (1-0.0179)^(5-0)
P(x=0) = (1) (1) (0.9821)^5
= 0.9136
1-P(x=0) = 1-0.9136 = 0.0864

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