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A student took two national aptitude tests. The mean and standard deviation were 475 and 100 , respectively, for the first test, and 30 and 8 , respectively, for the second test. The student scored 625 on the first test and 45 on the second test. Use z-scores to determine on which exam the student performed better relative to the other test takers. (Hint: See Example 3.18 )

Short Answer

Expert verified
The student performed better on the test where the z-score is higher.

Step by step solution

01

Calculate Z-score for the first test.

Calculate the Z-score for the first test using the formula \( z = (X - μ) / σ \). In this case, X=625, μ=475 and σ=100. So, the Z-score for the first test would be \( z = (625 - 475) / 100 \).
02

Calculate Z-score for the second test.

Use the same formula to calculate the Z-score for the second test. Here, X=45, μ=30 and σ=8. So, the Z-score for the second test would be \( z = (45 - 30) / 8 \).
03

Compare Z-scores.

The test where the student's score has a higher z-score is the one the student performed better on, compared to others. Compare the calculated z-scores of the two tests.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation
Standard deviation is a way of measuring how spread out numbers are in a set of data. It tells us how much the values in a dataset differ from the mean, or average value. If the standard deviation is small, it means that the data points are close to the mean. Conversely, a large standard deviation indicates that the data points are spread out over a wider range of values.

In the context of standardized test scores, a lower standard deviation means scores were more clustered around the mean. Conversely, a larger standard deviation would indicate a wider variation in the scores. Calculating standard deviation allows us to gauge the typical distance scores deviate from the average score, giving us important statistical insight.
Mean
The mean is another word for the average. To find it, we add up all the numbers in a dataset and then divide by the total number of data points. This gives us a single number that can summarize the central value of the dataset.

In exam contexts, the mean helps us understand the average score that students achieved on a test. By comparing an individual score to the mean, we can get a sense of how a student performed in comparison to their peers. Understanding the mean is crucial when calculating z-scores, as it acts as the point of comparison for measuring how far away a specific score is from the average.
Standardized Test Scores
Standardized test scores are converted into scale scores that allow us to compare scores across different tests or test sections. This conversion is often achieved using z-scores, which tells us exactly how many standard deviations away a particular score is from the mean.

This process of converting scores is vital because it provides a common scale for measuring performance. Standardized test scores, through z-scores, allow us to determine whether a student's score is above or below the average, and by how much. This comparison helps educators and students understand relative performance, regardless of the scale of the original test scores.
  • Calculate a z-score using the formula, where:
    • \( z = \frac{X - \mu}{\sigma} \)
    • \( X \) is the student's score.
    • \( \mu \) is the mean score of all students.
    • \( \sigma \) is the standard deviation.
This provides a clear, standardized view of performance in a diverse educational context.

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Most popular questions from this chapter

Data on a customer satisfaction rating (called the APEAL rating) are given for each brand of car sold in the United States (USA Today, July 17,2010 ). The APEAL rating is a score between 0 and 1,000 , with higher values indicating greater satisfaction. \(\begin{array}{lllllllll}822 & 832 & 845 & 802 & 818 & 789 & 748 & 751 & 794 \\ 792 & 766 & 760 & 805 & 854 & 727 & 761 & 836 & 822 \\\ 820 & 774 & 842 & 769 & 815 & 767 & 763 & 877 & 780 \\ 764 & 755 & 750 & 745 & 797 & 795 & & & \end{array}\) Calculate and interpret the mean and standard deviation for this data set.

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The article "Rethink Diversification to Raise Returns, Cut Risk" (San Luis Obispo Tribune, January 21,2006 ) included the following paragraph: In their research, Mulvey and Reilly compared the results of two hypothetical portfolios and used actual data from 1994 to 2004 to see what returns they would achieve. The first portfolio invested in Treasury bonds, domestic stocks, international stocks, and cash. Its 10 -year average annual return was \(9.85 \%\) and its volatility-measured as the standard deviation of annual returns-was \(9.26 \%\). When Mulvey and Reilly shifted some assets in the portfolio to include funds that invest in real estate, commodities, and options, the 10-year return rose to \(10.55 \%\) while the standard deviation fell to \(7.97 \% .\) In short, the more diversified portfolio had a slightly better return and much less risk. Explain why the standard deviation is a reasonable measure of volatility and why a smaller standard deviation means less risk.

Children going back to school can be expensive for parents-second only to the Christmas holiday season in terms of spending (San Luis Obispo Tribune, August \(18,\) 2005). Parents spend an average of \(\$ 444\) on their children at the beginning of the school year, stocking up on clothes, notebooks, and even iPods. However, not every parent spends the same amount of money. Imagine a data set consisting of the amount spent at the beginning of the school year for each student at a particular elementary school. Would it have a large or a small standard deviation? Explain.

Research by the U.S. Food and Drug Administration (FDA) shows that acrylamide (a possible cancer-causing substance) forms in high-carbohydrate foods cooked at high temperatures, and that acrylamide levels can vary widely even within the same brand of food (Associated Press, December 6,2002 ). FDA scientists analyzed McDonald's french fries purchased at seven different locations and found the following acrylamide levels: \(\begin{array}{lllll}497 & 193 & 328 & 155 & 326\end{array}\) \(\begin{array}{ll}245 & 270\end{array}\) a. Calculate the mean acrylamide level. For each data value, calculate the deviation from the mean. b. Verify that, except for the effect of rounding, the sum of the seven deviations from the mean is equal to 0 forthis data set. (If you rounded the sample mean or the deviations, your sum may not be exactly zero, but it should still be close to zero.) c. Use the deviations from Part (a) to calculate the variance and standard deviation for this data set.

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