91Ó°ÊÓ

A normal distribution is a common way to represent data in a bell-shaped curve. This curve is symmetrical with a single peak at the mean. The mean is the central value and, in a normal distribution, it is also the median and mode.
This type of distribution is characterized by its mean (\( \mu \)) and standard deviation (\( \sigma \)). The standard deviation defines the spread of the data around the mean. Larger values mean greater variability, while smaller values indicate data clustered closely around the mean.

• The total area under the curve represents probability and equals 1.
• Approximately 68% of data falls within one standard deviation of the mean in a normal distribution.
• About 95% falls within two standard deviations, and roughly 99.7% falls within three standard deviations.

Understanding normal distribution is essential, as it forms the foundation for statistical methods, including the Central Limit Theorem.

The sampling distribution is a probability distribution of a given statistic based on a random sample. In our case, it refers to the distribution of the sample mean. The Central Limit Theorem tells us that, regardless of the shape of the original population distribution, the sampling distribution of the sample mean approaches a normal distribution as the sample size increases.
In the problem at hand, the population is already normally distributed. This means the sample mean from the 9 trees will also be normally distributed.

• The mean of the sampling distribution (\( \mu_\bar{x} \)) is equal to the population mean (\( \mu \)).
• The standard deviation of the sampling distribution, called the standard error (\( \sigma_\bar{x} \)), is given by the formula: \[ \sigma_\bar{x} = \frac{\sigma}{\sqrt{n}} \]Here, \( \sigma \) is the population standard deviation and \( n \) is the sample size.

This tells us about how much the sample mean would vary from the true population mean if we repeatedly took samples of the same size.

Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of the exercise, it represents the average variation of tree basal areas from the mean basal area in the forest.
The standard deviation in the problem was directly given as 4 square inches for the population. When working with sample means, however, we need to calculate the standard error, which is the standard deviation of the sampling distribution.
The standard error is found by dividing the population standard deviation by the square root of the sample size:\[ \sigma_\bar{x} = \frac{\sigma}{\sqrt{n}} \]For the problem, this calculation was:\[ \sigma_\bar{x} = \frac{4}{3} \approx 1.33 \] square inches.
This value helps in estimating how much sample means deviate from the population mean, highlighting the reliability of the sample mean as an estimate of the population mean.

Z-scores are a way to standardize individual values on different scales of a normal distribution so that they can be compared.
They are calculated by taking the difference between a value and the mean, then dividing by the standard deviation. In mathematical terms:\[ Z = \frac{X - \mu}{\sigma} \]For a sampling distribution, the formula becomes:\[ Z = \frac{\bar{x} - \mu}{\sigma_\bar{x}} \]where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, and \( \sigma_\bar{x} \) is the standard error.
In this particular problem:

• Z-scores were calculated to determine the probability of the sample mean falling within a specific range around the population mean.
• This involved converting the limits \( \mu \pm 2 \) into Z-scores: \( Z_1 \approx -1.50 \) and \( Z_2 \approx 1.50 \)

By using the Z-scores in a Z-table, we found the probabilities corresponding to these scores, enabling us to conclude the probability of our sample mean being within that desired range.