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 µm and standard deviation 150 µm was a reasonable 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 single droplet is less than 1410 µm? At least 925 µm? (Round your answers to four decimal places.)

less than 1410 µm___
at least 925 µm ____

What is the probability that the size of a single droplet is between 925 and 1410 µm? (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__ µm in size.

If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds 1410 µm? (Round your answer to four decimal places.)___

Answer 1

µ = 1050

sd = 150

a)

                                 

                                  = P(Z < 2.4)

                                  = 0.9918

                           

                            = P(Z > -0.83)

                            = 1 - P(Z < -0.83)

                            = 1 - 0.2033

                            = 0.7967

b)

                                                

                                                 = P(-0.83 < Z < 2.4)

                                                 = P(Z < 2.4) - P(Z < -0.83)

                                                 = 0.9918 - 0.2033

                                                 = 0.7885

c)

or, x = 1050 - 2.05 * 150

or, x = 742.5

The smallest 2% of droplets are those smaller than_742.5_ µm in size.

d) P(X > 1410) = 1 - P(X < 1410) = 1 - 0.9918 = 0.0082

n = 5

P(at least one exceeds 1410) = 1 - P(none exceeds 1410) = 1 - (1- 0.0082)⁵ = 0.0403