Please help me with parts a-d and box answers to each part. Thank you! Spray dri
ID: 3201466 • Letter: P
Question
Please help me with parts a-d and box answers to each part.
Thank you!
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 mu m and standard deviation 150 mu m was a reasonable model for droplet size for water (the "control treatment") sprayed through a 760 ml/min nozzle. What is the probability that the size of a single droplet is less than 1500 mu m? At least 950 mu m? (Round your answers to four decimal places.) less than 1500 mu m at least 950 mu m What is the probability that the size of a single droplet is between 950 and 1500 mu m? (Round your answer to four decimal places.) How would you characterize the smallest 2% of all droplets? () The smallest 2% of droplets are those smaller than mu m in size. If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds 1500 mu m? (Round your answer to four decimal places.)Explanation / Answer
a) = 1050
= 150
standardize x to z = (x - ) /
P(x < 1440) = P( z < (1440-1050) / 150)
= P(z < 2.6) = 0.9953
(From Normal probability table)
-----------------------------------
= 1050
= 150
standardize x to z = (x - ) /
P(x > 975) = P( z > (975-1050) / 150)
= P(z > -0.5) = 0.6915
(From Normal probability table)
b)
= 1050
= 150
standardize x to z = (x - ) /
P( 975 < x < 1440) = P[( 975 - 1050) / 150 < Z < ( 1440 - 1050) / 150]
P( -0.5 < Z < 2.6) = 0.9953 - 0.3085 =0.6868
(From Normal probability table)
c)
From the normal distribution table, P( z < -2.05) =0.02
z = (x - ) /
-2.05 = (x-1050)/150
solve for x
x = 1050 + (150)(-2.05) = 742.5 µm in size
d)
The probability of one droplet exceeding 1440 µm is :
= 1050
= 150
standardize x to z = (x - ) /
P(x > 1440) = P( z > (1440-1050) / 150)
= P(z > 2.6) = 0.0047
(From Normal probability table)
Use the binomial probability with n=5, p=0.0047, x=1,2,3,4,5
P( x >=1) = 1-P(x=0)
P(x=0) = 5C0 (0.0047)^0 (1-0.0047)^(5-0)
P(x=0) = (1) (1) (0.9953)^5 = 0.9767
1-P(x=0) = 1-0.9767 = 0.0233