Question

The first Stats exam had a mean of 65 and a standard deviation of 10 points; the second had a mean of 80 and a standard deviation of 5 points. Derrick scored an 80 on both tests. Julie scored a 70 on the first test and a 90 on the second. They both totaled 160 points on the two exams, but Julie claims that her total is better. Explain.

Short answer

Julie's scores are relatively better as her Z-scores are higher.

Step-by-step solution

Calculate Derrick's Z-scores

The Z-score measures how many standard deviations a student's score is away from the mean. For Derrick's first test, the Z-score is calculated as: \( Z = \frac{(X - \mu)}{\sigma} = \frac{(80 - 65)}{10} = 1.5 \). For his second test, \( Z = \frac{(80 - 80)}{5} = 0 \).

Calculate Julie's Z-scores

Similarly, find Julie's Z-score for each test. For the first test: \( Z = \frac{(70 - 65)}{10} = 0.5 \). For the second test: \( Z = \frac{(90 - 80)}{5} = 2 \).

Compare Z-scores

Derrick's Z-scores are 1.5 and 0, respectively, for the first and second tests. Julie's scores are 0.5 and 2. Julie's second score is particularly strong, standing out more in terms of standard deviations from the mean compared to Derrick's scores, indicating a better relative performance.

Key concepts

Z-score

The Z-score, also known as a standard score, indicates how many standard deviations a data point is from the mean. Understanding Z-scores helps interpret individual statistical values relative to the group. This concept is especially helpful in comparing results across different sets of data that may have different means and standard deviations.

To calculate a Z-score, use the formula:

• \( Z = \frac{(X - \mu)}{\sigma} \)

Here, \( X \) represents the individual data point, \( \mu \) is the mean of the dataset, and \( \sigma \) represents the standard deviation.

In our example, Derrick and Julie each wrote two exams. By computing the Z-scores, we can understand how each of their scores relates to the average class performance. For instance, when Derrick scored 80 on the first test, his Z-score of 1.5 indicates he did much better than the average student (who scored around 65).

Z-scores give us a clearer picture of performance. They show how scores stack up against statistical norms, despite raw scores appearing similar.

Standard Deviation

Standard deviation measures the amount of variation or dispersion in a set of values. It is a crucial concept for understanding how spread out the data points are from the mean.

In simple terms:

• Low standard deviation means data points are close to the mean.
• High standard deviation implies data points are spread out over a wider range of values.

In the textbook example, the first test had a standard deviation of 10, while the second had a standard deviation of 5. This difference indicates that scores on the first exam were more scattered than on the second.

Understanding the standard deviation helps us interpret the Z-scores better. A higher standard deviation means a specific Z-score represents a wider gap from the mean. Conversely, with a lower standard deviation, as seen in the second test, scores are closer to the mean, making Julie's high score all the more impressive.

Mean

The mean, often called the average, is a measure of central tendency that indicates the sum of all data points divided by the number of points. The mean is helpful for understanding the overall tendency of a data set.

The formula for calculating the mean \( \mu \) is:

• \( \mu = \frac{\sum X}{N} \)

Where \( \sum X \) is the sum of all data points, and \( N \) is the number of data points.

In our scenario, the first test's mean was 65, and the second test's mean was 80. These means offer a basis for calculating Z-scores and understanding how each individual's scores compare to their peers.

While raw scores give immediate results, understanding averages (or means) sets the stage for deeper analysis. This is because the mean provides a benchmark, allowing students like Derrick and Julie to see how they perform relative to the average of their class.