91Ó°ÊÓ

The concept of a z-score is fundamental in the study of statistics, especially when dealing with the normal distribution. A z-score tells us how many standard deviations a particular data point is from the mean of the distribution. In more straightforward terms, it helps us understand where a specific value sits within a standard normal distribution.

For example, if a z-score is 2.17, like in the exercise, it means that the point is 2.17 standard deviations above the mean. Calculating a z-score requires knowing the mean (0 for a standard normal distribution) and the standard deviation (1 for a standard normal distribution). The formula is:

\[\text{z-score} = \frac{(X - \mu)}{\sigma}\]

Where:

• X is the value for which the z-score is being calculated.
• \mu is the mean of the distribution.
• (\sigma) is the standard deviation.

This concept helps statisticians and researchers understand the probability of occurrence of a given value within a dataset.

The z-table is an essential tool when working with z-scores in a standard normal distribution. It allows us to find probabilities associated with a z-score. However, most z-tables provide probabilities from the left of the mean up to the value of the z-score, denoted as \(P(z \leq a)\).

In the exercise, to find \(P(z \geq 2.17)\), we first look up \(P(z \leq 2.17)\) in the z-table. This value helps us find the probability that the z-score is less than or equal to 2.17.

By using a z-table:

• Locate the row for z = 2.1.
• Move horizontally across to the column representing 0.07.
• The intersecting value gives \(P(z \leq 2.17)\).

In many cases, knowing how to read a z-table allows you to determine probabilities, such as 98.50% for a z-score of 2.17, to be used further in probability calculations.

The normal distribution, often known as the bell curve, is critical in statistics and is used to represent real-valued random variables. It is characterized by its bell-shaped curve, symmetrical about its mean, and is determined entirely by its mean (\(\mu\)) and standard deviation (\(\sigma\)).

The standard normal distribution is a special case where the mean is 0, and the standard deviation is 1. This makes it easier to work with since it provides a foundation for z-score calculations.

Features of a normal distribution include:

• Approximately 68% of data falls within one standard deviation from the mean.
• About 95% falls within two standard deviations.
• Nearly 99.7% is within three standard deviations.

Understanding the normal distribution allows statisticians to make inferences about population parameters based on sample statistics.

Probability calculation in the context of a standard normal distribution involves determining the likelihood of a specific outcome. In the exercise, we calculated the probability that a z-score is greater than or equal to 2.17, symbolized as \(P(z \geq 2.17)\).

To do this, we first find \(P(z \leq 2.17)\) using a z-table, which provides the cumulative probability up to that z-score. The complement rule then helps find \(P(z \geq 2.17)\) by subtracting the found probability from 1:

\[P(z \geq 2.17) = 1 - P(z \leq 2.17)\]

Following these steps:

• Use the z-table to find \(P(z \leq 2.17)\), which is 0.9850.
• Apply the complement rule to calculate \(P(z \geq 2.17) = 1 - 0.9850 = 0.0150\).

This result tells us that there is a 1.50% probability of z being 2.17 or higher, providing a clear understanding of the event's rarity.