An experimenter publishing in the Annals of Botany investigated whether the stem

Question

An experimenter publishing in the Annals of Botany investigated whether the stem diameters of the dicot sunflower would change depending on whether the plant was left to sway freely in the wind or was artificially supported. Suppose that the unsupported stem diameters at the base of a particular species of sunflower plant have a normal distribution with an average diameter of 35 millimeters (mm) and a standard deviation of 3 mm.

(a) What is the probability that a sunflower plant will have a basal diameter of more than 39 mm? (Round your answer to four decimal places.)

(b) If two sunflower plants are randomly selected, what is the probability that both plants will have a basal diameter of more than 39 mm? (Round your answer to four decimal places.)

(c) Within what limits would you expect the basal diameters to lie, with probability 0.95? (Round your answers to two decimal places.)

(d) What diameter represents the 90th percentile of the distribution of diameters? (Round your answer to two decimal places.) ***in mm

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 35
standard Deviation ( sd )= 3
a.
P(X > 39) = (39-35)/3
= 4/3 = 1.3333
= P ( Z >1.3333) From Standard Normal Table
= 0.0912
b.
mean of the sampling distribution ( x ) = 35
standard Deviation ( sd )= 3/ Sqrt ( 2 ) =2.1213
sample size (n) = 2
P(X > 39) = (39-35)/3/ Sqrt ( 2 )
= 4/2.121= 1.8856
= P ( Z >1.8856) From Standard Normal Table
= 0.0297
c.
LOWER/BELOW
P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.96
P( x-u/s.d < x - 35/3 ) = 0.025
That is, ( x - 35/3 ) = -1.96
--> x = -1.96 * 3 + 35 = 29.12
UPPER/TOP
P ( Z > x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is 1.96
P( x-u / (s.d) > x - 35/3) = 0.025
That is, ( x - 35/3) = 1.96
--> x = 1.96 * 3+35 = 40.88
d.
P ( Z < x ) = 0.9
Value of z to the cumulative probability of 0.9 from normal table is 1.282
P( x-u/s.d < x - 35/3 ) = 0.9
That is, ( x - 35/3 ) = 1.282
--> x = 1.282 * 3 + 35 = 38.85

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