In the realm of statistics, a Z-score, also known as a standard score, provides a way to understand how far away a particular data point is from the mean of its dataset. The Z-score is measured in terms of standard deviations.
The concept of standardizing data has roots in the early 20th century with the development of statistical methods. The Z-score became a fundamental tool in statistical analysis, allowing researchers to compare data from different distributions. It was popularized alongside the rise of statistical hypothesis testing and quality control.
Follow these steps to calculate Z-scores from raw data:
Let's look at some examples to illustrate the calculation of Z-scores.
Suppose you have the following test scores: 70, 80, 90, 60, 85. Calculate the Z-scores for each test score.
| Test Score (X) | Z-score (Z) |
|---|---|
| 70 | $\frac{70 - 77}{10.25} \approx -0.68$ |
| 80 | $\frac{80 - 77}{10.25} \approx 0.29$ |
| 90 | $\frac{90 - 77}{10.25} \approx 1.27$ |
| 60 | $\frac{60 - 77}{10.25} \approx -1.66$ |
| 85 | $\frac{85 - 77}{10.25} \approx 0.78$ |
Consider a dataset of plant heights (in cm): 12, 15, 18, 20, 14.
| Plant Height (X) | Z-score (Z) |
|---|---|
| 12 | $\frac{12 - 15.8}{2.95} \approx -1.29$ |
| 15 | $\frac{15 - 15.8}{2.95} \approx -0.27$ |
| 18 | $\frac{18 - 15.8}{2.95} \approx 0.75$ |
| 20 | $\frac{20 - 15.8}{2.95} \approx 1.42$ |
| 14 | $\frac{14 - 15.8}{2.95} \approx -0.61$ |
Calculate the Z-scores for the following data points, given a mean ($\mu$) of 50 and a standard deviation ($\sigma$) of 5:
Solutions:
Calculating Z-scores is a fundamental skill in statistics. By understanding how to standardize data, you can effectively compare and analyze data points across different distributions. This guide provides a solid foundation for mastering Z-scores in Algebra 2.
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