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 1455 µm? At least 975 µm? (Round your answers to four decimal places.) less than 1455 µm 0.9965 Correct: Your answer is correct. at least 975 µm 0.5 Incorrect: Your answer is incorrect. (b) What is the probability that the size of a single droplet is between 975 and 1455 µm? (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 µm in size. (d) If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds 1455 µm? (Round your answer to four decimal places.)

Answer 1

X : Droplet size for water

X follows normal distribution with mean 1050 m and standard deviation 150m

Probability that the size of a single droplet is less than 1455 µm = P(X<1455)

Z-score for 1455 = (1455-1050)/150 = 2.7

From standard normal tables, P(Z<2.7) = 0.9965

P(X<1455) = P(Z<2.7) = 0.9965

Probability that the size of a single droplet is less than 1455 µm = 0.9965

Probability that the size of a single droplet At least 975 µm = P(X > 975) = 1-P(X<975)

Z-score for 975 = (975-1050)/150 = -0.5

From Standard normal tables, P(Z<-0.5) = 0.3085

P(X<975) = P(Z<-0.5) = 0.3085

P(X > 975) = 1-P(X<975) = 1 - 0.3085 = 0.6915

Probability that the size of a single droplet At least 975 µm = 0.6915

(b) probability that the size of a single droplet is between 975 and 1455 µm = P(975 < X < 1455) = P(X<1455)-P(X<975)

From (a)

P(X<1455) = 0.9965 ; P(X<975) = 0.3085

P(X<1455)-P(X<975) = 0.9965 - 0.3085 = 0.688

Probability that the size of a single droplet is between 975 and 1455 µm = 0.688

(c) How would you characterize the smallest 2% of all droplets

Let X₂ be the 2% of all droplets therefore . P(X<X₂) = 2/100 =0.02

Z-score for for X₂ be Z₂

Z₂ = (X₂ - 1050)/150 ; X₂ = 1050 + 150Z₂

Also, P(Z<Z₂) = P(X<X₂) = 0.02

From standard normal  tables,

P(Z<-2.05) = 0.0202

Therefore Z₂ = -2.05

X₂ = 1050 + 150Z₂

X₂ = 1050 + 150 x -2.05 = 1050-307.5= 742.5

The smallest 2% of droplets are those smaller than 742.5 µm in size

(d)

If the sizes of five independently selected droplets are measured,

Probability that at least one exceeds 1455 µm = 1 - [Probability that sizes of all five independently selected droplets size does not exceed 1455)

p : Probability that the size of single droplet does not exceeds 1455 = P(X<1455)

From (a), p = P(X<1455) = 0.9965

Probability that sizes of all five independently selected droplets size does not exceed 1455 =

= Probability that first of the five independently selected droplets size does not exceed 1455 x

Probability that second of the five independently selected droplets size does not exceed 1455 x

Probability that third of the five independently selected droplets size does not exceed 1455 x

Probability that fourth of the five independently selected droplets size does not exceed 1455 x

Probability that fifth of the five independently selected droplets size does not exceed 1455

= p⁵ = (0.9655)⁵ = 0.8390

Probability that sizes of all five independently selected droplets size does not exceed 1455 = 0.838998898

From (a) Probability that the size of a single droplet is less than 1455 µm = P(X<1455)= 0.9965

Probability that at least one exceeds 1455 µm = 1 - [Probability that sizes of all five independently selected droplets size does not exceed 1455) = 1 - 0.8390=0.161

If the sizes of five independently selected droplets are measured, Probability that at least one exceeds 1455 µm = 0.161