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 1470 µm? At least 975 µm? (Round your answers to four decimal places.)

less than 1470 µm
at least 975 µm

(b) What is the probability that the size of a single droplet is between 975 and 1470 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 1470 µm? (Round your answer to four decimal places.)

_________________________

Answer 1

Let X be the random variable that  size of a single droplet.

Here X ~ N(mu = 1050 µm, sigma = 150 µm)

(a) What is the probability that the size of a single droplet is less than 1470 µm? At least 975 µm?

To find P(X < 1470 )

Convert x = 1470 into z-score.

z-score is defined as,

z = (x - mu) / sigma

z = (1470 - 1050) / 150 = 2.8

Now we have to find P(Z < 2.8)

We can find this probability in excel.

syntax :

=NORMSDIST(z)

where z = 2.8

P(Z < 2.8) = 0.9974

Again we have to find P(X >= 975)

z-score for x = 975 is,

z = (975 - 1050) / 150 = -0.5

Now we have to find P(Z >= -0.5)

Excel syntax :

=1 - NORMSDIST(z)

where z = -0.5

P(Z > = -0.5) = 0.6915

b) What is the probability that the size of a single droplet is between 975 and 1470 µm?

To find P(975 < X < 1470)

z-score for x = 975 and x = 1470 are,

z = -0.5

z = 2.8

Now we have to find P(-0.5 < Z < 2.8)

P(-0.5 < Z < 2.8) = P(Z < 2.8) - P(Z < -0.5)

= 0.9974 - 0.6915

= 0.3060

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

We have to find x when probability = 2% = 0.02

Symbolically we can write as,

P(X < x) = 0.02

The formula for x is,

x = mu + z * sigma

where z is z-score

We can find z in excel.

syntax :

=NORMSINV(probability)

where probability = 0.02

z = -2.05

x = 1050 + (-2.05) * 150 = 741.94