Question

1. the average size tree in a forest is 15 feet tall with standard deviation of 3. x = height of a randomly selected tree. find:

a. p(x > 20)
b. p(x < 9)
c. p(10 < x < 20)
d. p(15 < x < 21)
e. p(x < 16)

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 15\) (mean) and \(\sigma=3\) (standard deviation).
For \(x = 20\), \(z_1=\frac{20 - 15}{3}=\frac{5}{3}\approx1.67\)
For \(x = 9\), \(z_2=\frac{9 - 15}{3}=\frac{-6}{3}=- 2\)
For \(x = 10\), \(z_3=\frac{10 - 15}{3}=\frac{-5}{3}\approx - 1.67\)
For \(x = 21\), \(z_4=\frac{21 - 15}{3}=2\)
For \(x = 16\), \(z_5=\frac{16 - 15}{3}=\frac{1}{3}\approx0.33\)

Step2: Use the standard normal distribution table

• For \(P(X>20)\):

\(P(X > 20)=P(Z>1.67)\)
Since \(P(Z > z)=1 - P(Z\leq z)\), and from the standard - normal table \(P(Z\leq1.67) = 0.9525\)
\(P(X > 20)=1 - 0.9525=0.0475\)

• For \(P(X < 9)\):

From the standard - normal table, \(P(Z\leq - 2)=0.0228\)
So \(P(X < 9)=P(Z < - 2)=0.0228\)

• For \(P(10 < X < 20)\):

\(P(10 < X < 20)=P(-1.67<Z<1.67)\)
\(P(-1.67 < Z < 1.67)=P(Z < 1.67)-P(Z < - 1.67)\)
From the standard - normal table, \(P(Z < 1.67)=0.9525\) and \(P(Z < - 1.67)=0.0475\)
\(P(-1.67 < Z < 1.67)=0.9525-0.0475 = 0.905\)

• For \(P(15 < X < 21)\):

\(P(15 < X < 21)=P(0<Z<2)\)
\(P(0 < Z < 2)=P(Z < 2)-P(Z < 0)\)
From the standard - normal table, \(P(Z < 2)=0.9772\) and \(P(Z < 0)=0.5\)
\(P(0 < Z < 2)=0.9772 - 0.5=0.4772\)

• For \(P(X < 16)\):

\(P(X < 16)=P(Z < 0.33)\)
From the standard - normal table, \(P(Z < 0.33)=0.6293\)

Answer:

A. \(0.0475\)
B. \(0.0228\)
C. \(0.905\)
D. \(0.4772\)
E. \(0.6293\)