Question

A forester measured 20 of the trees in a large woods that is up for sale. He found that their mean diameter was 191 inches and their standard deviation 22.4 inches. Suppose that these trees provide an accurate description of the whole forest and that the diameter of the tree follows a normal distribution. Find the following:

a) What percentage of trees would be above 191 inches in diameter?

b) What percentage of trees would be between 180 and 190 inches?

c) What is the interquartile range (IQR)?

d) What size diameter would you say represents the top 20% of the trees?

Answer 1

Solution:
Given that,
μ =191 Inches, σ=22.4 inches
a) P(X>191)=1-P(X<=191)
=1- p{[(x- μ)/σ]<=[(191 - 191)/22.4]}
= 1- P(z< 0)
=1- 0.5( from Standard Normal table)
=0.5
P(X>191)=0.5
In percentage,
P(X>191)=50%
b)P(180<X< 190)

= p{[(180 - 191)/22.4]<[(X- μ)/σ]<[(190 - 191)/22.4]}
=P(-0.49<Z< -0.04)
= p(Z< -0.04) - p(Z< -0.49)
= 0.4840 - 0.3121 ( from. Standard Normal table)
=0.1719

P(180<X< 190)= 0.1719
In percentage,
P(180<X< 190)= 17.19%

C)let ' a ' be the 1st quartile.
P( X< a)= 0.25
p{[(X- μ)/σ]<[(a - 191)/22.4]}=0.25
P( Z< z)= 0.25 .... where z=(a - 191)/22.4
By using the standard normal table , we get
P( Z< -0.67)= 0.25
z= -0.67
Now using the z-score formula, we can find the value of ' a ' as follows,
z=(a- μ)/σ
z=(a - 191)/22.4
(-0.67)×22.4=( a - 191)
a= 191+[(-0.67)×22.4]
=191- 15.008
a= 175.992
P( X< 175.992)= 0.25
Now,let ' a ' be the 3rd quartile.
P( X< a)= 0.75
p{[(X- μ)/σ]<[(a - 191)/22.4]}=0.75
P( Z< z)= 0.75 .... where z=(a - 191)/22.4
By using the standard normal table , we get
P( Z< 0.67)= 0.75
z= 0.67
Now using the z-score formula, we can find the value of ' a ' as follows,
z=(a- μ)/σ
z=(a - 191)/22.4
(0.67)×22.4=( a - 191)
a= 191+[(0.67)×22.4]
=191+ 15.008
a= 206.008
P( X< 206.008)= 0.75
Interqurtile range(IQR) = Q3-Q1
= 206.008 −175.992
=30.016
d) Let' a 'be the diameter size that represents the top 20% of the trees.
P( X> a)= 0.20
1-P( X<=a)= 0.20
1-p{[(X- μ)/σ]<[(a - 191)/22.4]}=0.20
1-P( Z< z)= 0.20 .... where z=(a - 191)/22.4
P( Z< z)= 0.80
By using the standard normal table , we get
P( Z< 84)= 0.80
z= 0.84
Now using the z-score formula, we can find the value of ' a ' as follows,
z=(a- μ)/σ
z=(a - 191)/22.4
(0.84)×22.4=( a - 191)
a= 191+[(0.84)×22.4]
=191+18.816
a= 209.816
P( X< 209.816)= 0.20
209.816 inches be size of diameter would you say represents the top 20% of the trees