What is the probability that haul time will be at least 10 min? Will excced 10 m
ID: 3126026 • Letter: W
Question
What is the probability that haul time will be at least 10 min? Will excced 10 min? What is the probability that haul time will exceed 12 min? What is the probability that haul time will exceed 15 min? What value c is such that 98% of all haul times are in the interval from 8.46 - c to 8.46 + c? If four haul times are independently selected, what is the probability that at least one of them exceeds 10 min? 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" (Weed Technology, 2005: 1030-1036) investigated the effects of herbicide formulation on spray atomization. A figure in the 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 1000 mu m? What is the probability that the size of a single droplet is between 1000 and 1500 mu m? How would you characterize the smallest 2 % of all droplets? lf the sizes of five independently selected droplets are measured, what is the probability that exactly two of them exceed 1500 mu m?Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 1500
u = mean = 1050
s = standard deviation = 150
Thus,
z = (x - u) / s = 3
Thus, using a table/technology, the left tailed area of this is
P(z < 3 ) = 0.998650102 [ANSWER]
**************
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 1000
u = mean = 1050
s = standard deviation = 150
Thus,
z = (x - u) / s = -0.333333333
Thus, using a table/technology, the right tailed area of this is
P(z > -0.333333333 ) = 0.63055866 [ANSWER]
***************
b)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 1000
x2 = upper bound = 1500
u = mean = 1050
s = standard deviation = 150
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.333333333
z2 = upper z score = (x2 - u) / s = 3
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.36944134
P(z < z2) = 0.998650102
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.629208762 [ANSWER]
***************
c)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.02
Then, using table or technology,
z = -2.053748911
As x = u + z * s,
where
u = mean = 1050
z = the critical z score = -2.053748911
s = standard deviation = 150
Then
x = critical value = 741.9376634
Hence, they are AT MOST 741.938. [ANSWER]
******************
d)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 1500
u = mean = 1050
s = standard deviation = 150
Thus,
z = (x - u) / s = 3
Thus, using a table/technology, the right tailed area of this is
P(z > 3 ) = 0.001349898
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 5
p = the probability of a success = 0.001349898
x = the number of successes = 2
Thus, the probability is
P ( 2 ) = 1.81486*10^-5 [ANSWER]
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