Understanding Amber's Self-Control Score in the Context of Percentile Rankings

Amber's raw score on the Self-Control Scale was 12. The population scores are normally distributed with a mean (μ) of 22 and a standard deviation (σ) of 4. Using these parameters, we can determine Amber's percentile rank. This information is crucial for understanding how her score compares to the broader population.

Calculating Percentile Rank Using the Unit Normal Table

To find the percentile rank of Amber's score, we can use the Unit Normal Table.

Using a TI-84 Calculator, we follow these steps:

Go to the catalog and find the function for normalcdf. Fill in the values as follows: normalcdf(Lo, High, μ, σ). For the lower bound, you can use a very low number, such as -9000. Enter the upper bound as 12, the mean (μ) as 22, and the standard deviation (σ) as 4. The input should be: normalcdf(-9000, 12, 22, 4)

Upon executing the command, you should get a result of approximately 0.0062, indicating that Amber's score is at the 0.62 percentile.

Conclusion: Amber scored better than only 0.62% of the population. This doesn't seem very high, but considering the standard deviation of 4, it places her score significantly below the mean. The ±3σ ranges, which span from 10 to 30, further emphasize the lower percentile. This indicates that Amber's score is in the lower end of the distribution, well below average.

Diving Deeper into Z-Scores

To further understand Amber's score, we can calculate the Z-score. The Z-score represents how many standard deviations a particular score is from the mean. It is calculated using the following formula:

Z-score (X - μ) / σ

For Amber's score:

Z-score (12 - 22) / 4 -10 / 4 -2.5

This Z-score indicates that Amber's score is 2.5 standard deviations below the mean. In other words, Amber's score is 2.5 standard deviations below the average score. Describing her score as well below average, Amber may have self-control issues, which could be a cause for concern.

Conclusion and Key Takeaways

Amber's raw score of 12 on the Self-Control Scale, when placed in a normally distributed population with a mean of 22 and a standard deviation of 4, gives her a percentile rank of 0.62. This means she scored better than only 0.62% of the population. Her Z-score of -2.5 further confirms this, as it places her significantly below the mean. Understanding these concepts is essential for educational, psychological, and other professional contexts where performance and behavior are analyzed.