Problem 29 - Algebra 2

Question

Critical Thinking Jake and Elena took the same standardized test, but with different groups of students. They both received a score of \(87 .\) In Jake's group, the mean was 80 and the standard deviation was \(6 .\) In Elena's group, the mean was 76 and the standard deviation was \(4 .\) Did either student score in the top 10\(\%\) of his or her group? Explain.

Step-by-Step Solution

Answer

Elena scored in the top 10% of her group, but Jake did not.

Step 1: Calculate Jake's Z-score

To calculate Jake's z-score, subtract the mean of his group from his score and divide by the standard deviation of his group. The formula is given by: Z = (X - µ) / σ. Substituting the given values, Jake's z-score= (87-80) / 6 = 1.17.

Step 2: Calculate Elena's Z-score

In a similar fashion, calculate Elena's z-score by the same method of subtracting her group's mean from her score, then divide by her group's standard deviation. Substituting the given values, Elena's z-score = (87-76) / 4 = 2.75.

Step 3: Compare Z-scores with Threshold

In a standard normal distribution (z-distribution), a z-score above 1.28 corresponds to being in the 90th percentile. Comparing Jake's and Elena's Z-scores with 1.28, we note that Elena's Z-score is higher than 1.28 but Jake's Z-score is not. Therefore, Elena scored in the top 10% of her group, but Jake did not.

Key Concepts

Z-score Calculation

To understand how well someone performed on a standardized test in comparison to others, we can calculate their z-score. A z-score measures how many standard deviations a score is away from the mean. The formula for calculating a z-score is: \[ Z = \frac{(X - \mu)}{\sigma} \]

• X is the raw score (in this case, 87 for both Jake and Elena).
• \(\mu\) is the mean score of the group.
• \(\sigma\) is the standard deviation of the group.

For Jake, we substitute his group's mean and standard deviation into the formula: \( Z = \frac{(87 - 80)}{6} = 1.17 \). Similarly, Elena's z-score is calculated using her group's mean and standard deviation: \( Z = \frac{(87 - 76)}{4} = 2.75 \). Z-scores are essential because they allow scores from different distributions to be compared directly. After calculation, we can interpret these z-scores to understand their relative standing within the group.

Mean and Standard Deviation

The mean and standard deviation are key statistical concepts that help us understand the distribution of test scores. The mean is the average score and provides a central value for the data. It's calculated by adding all the scores together and dividing by the number of scores. For Jake's group, the mean score is 80, while for Elena's group, the mean score is 76. These means represent the typical scores in their respective groups. The standard deviation measures how spread out the scores are around the mean. A smaller standard deviation means that the scores are closer to the mean, while a larger standard deviation indicates more variation among the scores.

• Jake's group has a standard deviation of 6.
• Elena's group has a standard deviation of 4.

In this context, Jake's scores vary more from the mean compared to Elena's scores, which makes Elena's score of 87 more remarkable, as it is farther from her group mean relative to the spread of her group's scores.

Percentiles in Normal Distribution

Percentiles help us determine the relative standings of scores within a group by indicating the percentage of scores that fall below a particular score. In a standard normal distribution (z-distribution), percentiles match with specific z-scores. The 90th percentile is a common benchmark. If a score is in the top 10% of its distribution, it is above the 90th percentile. Generally, this percentile corresponds to a z-score of about 1.28. So, to figure out if a score is in the top 10%, we compare the z-score to 1.28.

• If a z-score is above 1.28, it places that score in the top 10%.
• If it is below 1.28, it does not reach the top 10%.

For our exercise, Elena's z-score of 2.75 exceeds 1.28, meaning she is in the top 10% of her group. On the other hand, Jake's z-score of 1.17 does not reach this threshold, indicating he isn't in the top 10%. Understanding these percentiles is crucial for interpreting performance in standardized tests.