Chapter 7.2, Problem 10E

To determine

(a)

To find:

Find the probability of walking down the street and finding on oak tree with a diameter of more than 5 feet.

Answer to Problem 10E

Solution:

The probability of walking down the street and finding on oak tree with a diameter of more than 5 feet is 0.0038.

Explanation of Solution

Given:

$\mu =4000and\sigma =0.375$

Given a population of size $N$ with mean $\left(\mu \right)$ and standard deviation $\left(\sigma \right)$, samples of a fixed discrete number of trials $\left(n\right)$ can be chosen each having a mean called the sample mean $\left(\overline{x}\right)$.

On dealing with the collection of these means for samples chosen from the population, the concept of sampling distribution pops and the statistic dealt with is the sample means, and that the distribution contains all possible samples for the chosen sample size.

The continuity criterion can be extended and here it is known as Central Limit Theorem which states that sampling distribution is approximately normal with:

a. mean ${\mu }_{\overline{x}}=\mu$, and

b. Standard deviation ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

Moreover, the standard score is given by:

$z=\frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}$.

Thereafter, the required probability is accordingly obtained by taking into account the specifics in the question.

Calculation:

The value of the standard score (z-score)

$\begin{array}{rll}z& =& \frac{x-\mu }{\sigma }\\ & =& \frac{5-4}{0.375}\\ & =& \frac{1}{0.375}\\ & =& 2.67\end{array}$

Now probability of x more than 5 feet is calculated as

$\begin{array}{rll}P\left(x>5\right)& =& P\left(z>2.67\right)\\ & =& 1-P\left(z<2.67\right)\\ & =& 1-0.9962\\ & =& 0.0038\end{array}$

So, the probability of walking down the street and finding on oak tree with a diameter of more than 5 feet is 0.0038.

To determine

(b)

To find:

Find the probability of sampling a set of 87 oak trees and finding their mean to be more than 4.1 feet in diameter.

Answer to Problem 10E

Solution:

The probability of sampling a set of 87 oak trees and finding their mean to be more than 4.1 feet in diameter is 0.0064.

Explanation of Solution

Calculation:

The value of the standard score (z-score)

$\begin{array}{rll}z& =& \frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\\ & =& \frac{4.1-4}{\left(\frac{0.375}{\sqrt{87}}\right)}\\ & =& \frac{0.1}{0.0402}\\ & =& 2.49\end{array}$

The probability using z table is calculated as:

$\begin{array}{rll}P\left(\overline{x}>4.1\right)& =& P\left(z>2.49\right)\\ & =& 1-P\left(z<2.49\right)\\ & =& 1-0.9936\\ & =& 0.0064\end{array}$

So the probability of sampling a set of 87 oak trees and finding their mean to be more than 4.1 feet in diameter is 0.0064.

To determine

(c)

To find:

Find the probability of sampling a set of 87 oak trees and finding their mean to be less than 3.92 feet in diameter.

Answer to Problem 10E

Solution:

The probability of sampling a set of 87 oak trees and finding their mean to be less than 3.92 feet in diameter is 0.0233.

Explanation of Solution

Calculation:

The value of the standard score (z-score)

$\begin{array}{rll}z& =& \frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\\ & =& \frac{3.92-4}{\left(\frac{0.375}{\sqrt{87}}\right)}\\ & =& \frac{-0.08}{0.0402}\\ & =& -1.99\end{array}$

The probability using z table is calculated as:

$\begin{array}{rll}P\left(\overline{x}<3.92\right)& =& P\left(z<-1.99\right)\\ & =& 0.0233\end{array}$

The probability of sampling a set of 87 oak trees and finding their mean to be less than 3.92 feet in diameter is 0.0233.

To determine

(d)

To find:

Find the probability of sampling a set of 87 oak trees and finding their mean to differ from the population mean by less than 0.1 feet in diameter.

Answer to Problem 10E

Solution:

The probability of sampling a set of 87 oak trees and finding their mean to differ from the population mean by less than 0.1 feet in diameter is 0.9876.

Explanation of Solution

Calculation:

The probability of sampling a set of 87 oak trees and finding their mean to differ from the population mean by less than 0.1 feet in diameter is calculated as:

$P\left(\left|\overline{x}-\mu \right|\right)<0.1$

$P\left(\left|\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\right|\right)<\frac{0.1}{\frac{0.375}{\sqrt{87}}}$

$P\left(\left|z\right|\right)<\frac{0.1}{0.0402}$

$P\left(z\right)<2.49;P\left(z\right)>-2.49$

The probability using z table is calculated as:

$\begin{array}{rll}P\left(\left|z\right|<2.49\right)& =& P\left(-2.49<z<2.49\right)\\ & =& P\left(z<2.49\right)-P\left(z<-2.49\right)\\ & =& 0.9936-0.0064\\ & =& 0.9872\end{array}$