Question

A forester studying the effects of fertilization on certain pine forests in the Southeast is interested in estimating the average basal area of pine trees. In studying basal areas of similar trees for many years, he has discovered that these measurements (in square inches) are normally distributed, with standard deviation approximately 4 square inches. If the forester samples n = 9 trees, find the probability that the sample mean will be within 2 square inches of the population mean.

Answer 1

Answer:

$P(-2 \leq \bar X -\mu \leq 2)$

If we divide both sides by $\frac{\sigma}{\sqrt{n}}$ we got:

$P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})$

And we can use the normal distribution table or excel to find the probabilites and we got:

$P(-1.5 \leq Z \leq 1.5)= P(Z$Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the area of a population, and for this case we know the distribution for X is given by:

$X \sim N(M,4)$  

Where $\mu=M$ and $\sigma=4$

We select a a sample of n =4 and since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want to find this probability:

$P(-2 \leq \bar X -\mu \leq 2)$

If we divide both sides by $\frac{\sigma}{\sqrt{n}}$ we got:

$P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})$

And we can use the normal distribution table or excel to find the probabilites and we got:

$P(-1.5 \leq Z \leq 1.5)= P(Z$