Problem 13 - Thinking Mathematically

Question

In Exercises 9-16, find the percentage of data items in a normal distribution that lie between \(z=-1.5\) and \(z=1.5\).

Step-by-Step Solution

Answer

In order to obtain a short answer to this question, you need a Z-table. The Z-table provides values that you substitute into the equation \(B - A\) to find the percentage of data within the \(z\) scores.

Step 1: Identify the Given Z Scores

The exercise gives two \(z\) scores: \(z=-1.5\) and \(z=1.5\). Data within these two \(z\) scores is what we are trying to find.

Step 2: Refer to the Z-table

To find the percentage of data between two \(z\) scores, we need to refer to the Z-table. The Z-table provides us the probability that a value is less than a given \(z\) score. So first, we find the values for \(z=-1.5\) and \(z=1.5\) in the Z-table. Let's say these values are \(A\) and \(B\), respectively.

Step 3: Calculate the Percentage

However, to find the percentage of data between \(z=-1.5\) and \(z=1.5\), we need to take \(B - A\). The result will be the desired percentage.

Key Concepts

Understanding Z-Scores

Z-scores are crucial in analyzing a normal distribution. To put simply, a z-score tells us how many standard deviations a data point is from the mean. If you have ever heard someone say "how far off the mark," then you already have a basic understanding of z-scores. For example, a z-score of 1.5 means that the data point is 1.5 standard deviations away from the average or mean.

• Z-scores help compare data from different distributions.
• A z-score close to 0 indicates the data point is nearer to the mean.
• Positive or negative z-scores indicate data points above or below the mean, respectively.

In the context of this exercise, we have two z-scores: -1.5 and 1.5, and we aim to understand the distribution of data lying between these points.

Using the Z-Table

The Z-table is like a treasure map when working with normal distributions. But instead of finding gold, we're finding probabilities. The Z-table gives us the probability that a value is less than a given z-score in a standard normal distribution. Whenever you're asked about data percentages in normal distributions, you often use this table.

• Find the z-score on the first column and corresponding row.
• Follow the row and column to find the associated probability value.
• The table typically only shows the positive z-scores; use symmetry for negative scores.

For instance, to solve any such question like the present exercise, consult the Z-table for z-scores -1.5 and 1.5 to identify their respective probability values. Calculating the difference between them gives us useful insights.

Calculating Probability

Probability is the mathematical way of representing our intuition about how likely an event is. In the setting of normal distributions, it helps us find out how much of the data falls within a particular range defined by z-scores. Here's a concrete approach:

Step-by-Step Probability Calculation

• Locate the first z-score on the Z-table to find its cumulative probability.
• Locate the second z-score and find its cumulative probability as well.
• Subtract the smaller probability from the larger one to find the probability of data between these two z-scores.

This difference gives you the probability or the likelihood of a data point falling between those two z-scores.

Interpreting Data Percentage

Data percentage is the friendliest way to make sense of probabilities, as they give us the result in a more tangible manner. In simpler terms, data percentage allows us to know the chunk of data in the distribution that exists between certain z-scores. When you grasp the probability from the Z-table and subtract as instructed, convert it to a percentage for a more understandable output: - Multiply the probability value by 100. For example, if the difference in probability is 0.87, this translates to 87%. Meaning, 87% of the data falls between your specified z-scores of -1.5 and 1.5 in a standard normal distribution.
Thus, expressing findings as percentages aids in effective communication of statistical results. It makes them relatable and easily interpretable for those less familiar with statistical jargon.