please answer these questions. Spray drift is a constant concern for pesticide a

Question

please answer these questions.

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. 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.

b. If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds 1440 µm? (Round your answer to four decimal places.)

Explanation / Answer

Solution

Part (a) is based on Normal Distribution and part (b) is based on Binomial Distribution. Essential theory part is first given below.

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

If Y ~ B(n, p). i.e., Y has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then

probability mass function (pmf) of Y is given by

p(y) = P(Y = y) = (nCy)(py)(1 - p)n – y, y = 0, 1, 2, ……. , n ………………………………..(3)

Now, to work out solution,

Let X = Droplet size (µm) for water and Y = Number of droplets of size exceeding 1440 µm in a random sample of 5. Then, X ~ N(µ, 2) and Y~ B(n, p), where

µ = 1050, = 150 and n = 5 [all given]. p = probability of a droplet of size exceeding 1440 µm will have to be worked out.

Part (a)

Let s be the size of the smallest 2% of droplets. Then, we should have P(X s) = 0.02.

[vide (2) under Back-up Theory], P(X s) = 0.02 is the same as P[Z {(s - 1050)/150}] = 0.02

[using Excel Function], {(s - 1050)/150} = - 2.05375 => s = 741.9377 = 741.94 µm ANSWER

Part (b)

To find p = probability of a droplet of size exceeding 1440 µm = P(X > 1440)

= P[Z > {(1440 - 1050)/150}] = P(Z > 2.6) = 0.0047 [using Excel Function].

Now, we want probability that at least one exceeds 1440 µm, which is equal to

P(Y 1) = 1- P(Y = 0) = 1 - (5C0)(0.00470)(1 – 0.0047)5 – 0 = 1 – 0.99535 = 1 – 0.97672

= 0.02328 = 0.0233 ANSWER

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