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. A researcher is interested in the effects of herbicide formulation on spray atomization. The researcher believes a normal distribution with mean 1050 µm and standard deviation 150 µm is 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 1380 µm? Greater than 1000 µm? (Round your answers to four decimal places.)

less than 1380 µm
greater than 1000 µm

(b) What is the probability that the size of a single droplet is between 1000 and 1380 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 1380 µm? (Round your answer to four decimal places.)

Answer 1

Here we have = 1050, = 150

a) Here we need to find

p ( x < 1380 )

= p ( z < 2.2 )

= 0.9861 ---------- ( using standard normal probability table )

-----------------------------------------------------------------------------------------------------------------

p ( x > 1000 )

= p ( z > -0.33 )

= 1 - p ( z - 0.33 )

= 1 - 0.3707

= 0.6293 ------------------ ( using standard normal probability table )

b)

Here we need to find

p ( 1000 < x < 1380 )

= p ( -0.33 < z < 2.2 )

= p ( z< 2.2 ) - p ( z < -0.33 )

= 0.9861 - 0.3707

= 0.6154

c) Smallest 2% of droplets :

p ( Z < z ) = 0.02

z = -2.05 ------------( looking in the probability section of standdard nomal probability table for 0.02 )

x = 742.50

The smallest 2% of droplets are those smaller than _742.50__ µm in size.

d) Here n = 5

First we find probability that droplets exceeds 1380 µm.

p ( x > 1380 ) = 1 - ( x 1380 ) = 1 - p ( x < 1380 )

From ( a) we have p ( x < 1380 ) = 0.9861

So p ( x > 1380 ) = 1 - 0.9861 = 0.0139

Let us denote this probability by P = 0.0139

Now let random variable y = number of droplets selected from five

y ~ B( n , p )

So we need to find

p ( y > 1 ) = 1 - p ( y = 0 )

= 1 - {⁵C₀ * 0.0139⁰ * ( 1 -0.0139 ) ^5-0 }

= 1 - { 1 * 1 * ( 0.9861) ⁵ }

= 1 - 0.9324

= 0.0676