Respuesta :
Answer:
Explained below.
Step-by-step explanation:
Let X represents the tree heights.
It is provided that X follows a normal distribution with mean μ = 112 inches and standard deviation σ = 10 inches.
(a)
Compute the 22nd percentile of the tree heights as follows:
[tex]P(X<x)=0.22\\\\P(Z<z)=0.22[/tex]
The z-score for the corresponding probability is, z = -0.77.
Compute the value of x as follows:
[tex]z=\frac{x-\mu}{\sigma}\\\\-0.77=\frac{x-112}{10}\\\\x=112-(0.77\times 10)\\\\x=104.3[/tex]
Thus, the 22nd percentile of the tree heights is 104.3 inches.
(b)
Compute the 80th percentile of the tree heights as follows:
[tex]P(X<x)=0.80\\\\P(Z<z)=0.80[/tex]
The z-score for the corresponding probability is, z = 0.84.
Compute the value of x as follows:
[tex]z=\frac{x-\mu}{\sigma}\\\\0.84=\frac{x-112}{10}\\\\x=112+(0.84\times 10)\\\\x=120.4[/tex]
Thus, the 80th percentile of the tree heights is 120.4 inches.
(c)
Compute the first quartile of the tree heights as follows:
[tex]P(X<x)=0.25\\\\P(Z<z)=0.25[/tex]
The z-score for the corresponding probability is, z = -0.67.
Compute the value of x as follows:
[tex]z=\frac{x-\mu}{\sigma}\\\\-0.67=\frac{x-112}{10}\\\\x=112-(0.67\times 10)\\\\x=113.3[/tex]
Thus, the first quartile of the tree heights is 113.3 inches.
(d)
The tallest 1% of the trees, i.e. P (X > x) = 0.01.
That is, P (X < x) = 0.99.
⇒ P (Z < z) = 0.99
The corresponding z-value is, z = 2.33.
Compute the value of x as follows:
[tex]z=\frac{x-\mu}{\sigma}\\\\0.99=\frac{x-112}{10}\\\\x=112+(0.99\times 10)\\\\x=121.9[/tex]
Thus, the tallest 1% of the trees are more than 121.9 inches tall.
