/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 A student took two national apti... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A student took two national aptitude tests. The national average 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.

Short Answer

Expert verified
The student performed better on the second test relative to other test takers.

Step by step solution

01

Calculation of \(z\) score for the first test

The formula for a \(z\) score is \( Z = (X - \mu) / \sigma \), where \( Z \) is the \(z\)-score, \( X \) is the observed value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Substituting the values for the first test, we have \( Z = (625 - 475) / 100 = 1.5 \). The \(z\)-score for the first test is 1.5.
02

Calculation of \(z\) score for the second test

Using the same formula, for the second test we plug in \( X = 45 \), \( \mu = 30 \), and \( \sigma = 8 \). Thus, \( Z = (45 - 30) / 8 = 1.875 \). The \(z\)-score for the second test is 1.875.
03

Comparing the \(z\) scores

The \(z\)-score for the second test is higher than the \(z\)-score for the first test. Therefore, relative to other test takers, the student performed better on the second test.

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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 vital statistical concept that helps us understand how data is dispersed around the mean. In essence, it tells us how spread out the numbers in a set are. If the standard deviation is low, it means the numbers are closer to the mean. Conversely, a high standard deviation indicates a wide range of values.

The formula to calculate standard deviation involves several steps:
  • First, find the mean of the data set.
  • Next, subtract the mean from each number to find the deviation for each value.
  • Then, square each deviation to eliminate negative values.
  • Find the average of these squared deviations — this is the variance.
  • Finally, take the square root of the variance to obtain the standard deviation.
In our example, the standard deviation for the first test is 100, suggesting scores widely vary around the mean of 475. The second test has a standard deviation of 8, indicating scores are more tightly clustered around the mean of 30.
Mean in Statistics
The mean is one of the three main measures of central tendency, alongside the median and mode. It is the average of a set of numbers and provides a simple summary statistic.

To calculate the mean:
  • Add up all the values in your data set.
  • Divide the total by the number of values.
In the context of our exercise, the mean gives us an idea of the typical score for each test. For instance, on the first national aptitude test, the mean score is 475. This means that, on average, test takers score around this value. Similarly, the second test has a mean score of 30.

The mean can help put individual scores into perspective, especially when combined with measures like the standard deviation. By understanding the mean, we can assess if a score is above or below average.
Comparing Performance with Z-Scores
Z-scores play a crucial role in comparing individual performance against a population. A z-score, also called a standard score, indicates how many standard deviations an element is from the mean. It provides a way to compare scores from different distributions.

Here's how z-scores are calculated:
  • Subtract the mean from the individual score to find the deviation.
  • Divide this deviation by the standard deviation of the data set.
In our scenario, the student's z-scores help us understand their performance relative to others. The first test z-score is 1.5, meaning the student's score is 1.5 standard deviations above the mean. The z-score for the second test is 1.875, indicating an even better performance relative to others since it is further above the mean in terms of standard deviations.

By comparing these z-scores, we learn the student performed better on the second test. Z-scores are a powerful tool for such comparisons, allowing for objective performance analysis across different data sets.

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Most popular questions from this chapter

The paper "Answer Changing on Multiple-Choice Tests" (Journal of Experimental Education \([1980]: 18-21)\) reported that for a group of 162 college students, the average number of responses changed from the correct answer to an incorrect answer on a test containing 80 multiplechoice items was \(1.4\). The corresponding standard deviation was reported to be \(1.5 .\) Based on this mean and standard deviation, what can you tell about the shape of the distribution of the variable number of answers changed from right to wrong? What can you say about the number of students who changed at least six answers from correct to incorrect?

The amount of aluminum contamination (in parts per million) in plastic was determined for a sample of 26 plastic specimens, resulting in the following data ("The Log Normal Distribution for Modeling Quality Data When the Mean Is Near Zero," Journal of Quality Technology \([1990]: 105-110):\) \(\begin{array}{rrrrrrrrr}30 & 30 & 60 & 63 & 70 & 79 & 87 & 90 & 101 \\ 102 & 115 & 118 & 119 & 119 & 120 & 125 & 140 & 145 \\ 172 & 182 & 183 & 191 & 222 & 244 & 291 & 511 & \end{array}\) Construct a boxplot that shows outliers, and comment on the interesting features of this plot.

Suppose that your statistics professor returned your first midterm exam with only a \(z\) score written on it. She also told you that a histogram of the scores was approximately normal. How would you interpret each of the following \(z\) scores? a. \(2.2\) d. \(1.0\) b. \(0.4\) e. 0 c. \(1.8\)

The average reading speed of students completing a speed-reading course is 450 words per minute (wpm). If the standard deviation is \(70 \mathrm{wpm}\), find the \(z\) score associated with each of the following reading speeds. a. \(320 \mathrm{wpm}\) c. \(420 \mathrm{wpm}\) b. 475 wpm d. 610 wpm

The percentage of juice lost after thawing for 19 different strawberry varieties appeared in the article "Evaluation of Strawberry Cultivars with Different Degrees of Resistance to Red Scale" (Fruit Varieties Journal [1991]: \(12-17):\) \(\begin{array}{llllllllllll}46 & 51 & 44 & 50 & 33 & 46 & 60 & 41 & 55 & 46 & 53 & 53 \\ 42 & 44 & 50 & 54 & 46 & 41 & 48 & & & & & & \end{array}\) a. Are there any observations that are mild outliers? Extreme outliers? b. Construct a boxplot, and comment on the important features of the plot.

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