/*! 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 129 Find the \(z\) -score for test s... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the \(z\) -score for test scores of 92 and 63 on a test that has a mean of 72 and a standard deviation of 12.

Short Answer

Expert verified
The z-scores for test scores 92 and 63 are 1.67 and -0.75 respectively.

Step by step solution

01

Identify the Provided Values

First identify the given values in the problem: The test scores are 92 and 63. The mean for the test is 72 and the standard deviation is 12.
02

Apply the Z-Score Formula for First Score

The formula for z-score is: \(z = \frac{(X - \mu)}{\sigma}\) where \(X\) is the value whose z-score we want to find, \(\mu\) is the mean and \(\sigma\) is the standard deviation. Apply the formula with 92 as \(X\): \(z_1 = \frac{(92 - 72)}{12}\)
03

Calculate the First Z-Score

Now, simplify the expression above to calculate first z-score: \(z_1 = 1.67\) (rounded to two decimal places)
04

Apply the Z-Score Formula for Second Score

Apply the z-score formula again, this time with 63 as \(X\): \(z_2 = \frac{(63 - 72)}{12}\)
05

Calculate the Second Z-Score

Simplify the expression to calculate the second z-score: \(z_2 = -0.75\) (rounded to two decimal places)

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Key Concepts

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

Understanding Mean and Standard Deviation
Mean and standard deviation are fundamental concepts in statistics that help summarize and describe data. The mean is essentially the average of a set of numbers, providing a central value for the data. To find the mean, you simply add up all the numbers in the dataset and divide by the number of numbers. For example, if your test scores are 60, 70, and 80, the mean would be calculated as (60 + 70 + 80) / 3 = 70. The mean helps us understand the overall trend or the typical score in the data set.

On the other hand, the standard deviation provides insight into the variability or spread of the data around the mean. It tells us how much the scores differ from the mean. A smaller standard deviation indicates that the values are close to the mean, whereas a larger standard deviation suggests more spread out data. For test scores, if the standard deviation is 5, most scores are expected to fall within 5 points of the mean, either above or below.

Together, mean and standard deviation are powerful tools that provide a comprehensive view of the data. In the context of the z-score, they are crucial for determining how far a specific test score is from the average in terms of standard deviations.
Decoding Test Scores
Test scores are a simple numerical representation of a student's performance. These scores are often analyzed in statistics to determine how well an individual or group has performed relative to others. Generally, you can look at raw test scores to see if someone passed or failed, but they don't provide much information about how someone's performance compares to their peers. This is where statistical tools like mean, standard deviation, and z-scores come into play.

When comparing test scores, it's important to understand the context in which they were obtained. Knowing the mean score and the standard deviation is essential because it allows us to calculate a z-score, facilitating comparison. The z-score tells us exactly how many standard deviations a particular score is from the mean. For instance, on a test with a mean score of 72 and a standard deviation of 12, a score of 92 would be considered well above average, as shown by its z-score of 1.67. Conversely, a score of 63 would be below average, with a z-score of -0.75.

By using statistical measures like z-scores, we can assess test scores more accurately and understand their significance in the broader context of the entire dataset. This approach makes it easier to identify outliers, trends, and overall performance patterns.
Normal Distribution in Test Scores Analysis
Normal distribution is a vital concept in statistics, especially when dealing with test scores. It is often referred to as a "bell curve" due to its shape when graphed. In a normal distribution, most data points are concentrated around the mean, and the frequency of data points decreases as you move away from the mean. This distribution pattern is common in many natural phenomena, including test scores.

When a dataset follows a normal distribution, it becomes easier to apply statistical measures and make assumptions about the data. For instance, in an educational setting, if test scores are normally distributed, we can expect about 68% of scores to fall within one standard deviation of the mean, about 95% within two standard deviations, and about 99.7% within three standard deviations.

This kind of distribution enables us to better understand and interpret z-scores. For example, a z-score of 1.67, as calculated for a test score of 92, indicates that this score is 1.67 standard deviations above the mean. This means the score is higher than most of the scores in the dataset, as expected in a normal distribution. Understanding normal distribution thus provides a clear framework for analyzing test scores and making informed conclusions about a student's performance relative to the whole group.

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Most popular questions from this chapter

What kinds of financial transactions do you do online? Are you worried about your security? According to Consumer Internet Barometer, the source of a March \(25,2009,\) USA Today Snapshot titled "Security of online accounts," the following transactions and percent of people concerned about their online security were reported. $$\begin{array}{lc} \text { What } & \text { Percent } \\ \hline \text { Bankirg } & 72 \\ \text { Paying bills } & 70 \\ \text { Buying stocks, bords } & 62 \\ \text { Fling toxes } & 62 \\ \hline \end{array}$$ Prepare two bar graphs to depict the percentage data. Scale the vertical axis on the first graph from 50 to \(80 .\) Scale the second graph from 0 to \(100 .\) What is your conclusion concerning how the percentages of the four responses stack up based on the two bar graphs, and what would you recommend, if anything, to improve the presentations?

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Shown below are the heights (in inches) of the basketball players who were the first-round picks by National Basketball Association professional teams for 2009. $$\begin{array}{llllllllll} \hline 82 & 86 & 76 & 77 & 75 & 72 & 75 & 81 & 78 & 74 \\ 77 & 77 & 81 & 81 & 82 & 80 & 76 & 72 & 74 & 74 \\ 73 & 82 & 80 & 84 & 74 & 81 & 80 & 77 & 74 & 78 \\ \hline \end{array}$$ a. Construct a dotplot of the heights of these players. b. Use the dotplot to uncover the shortest and the tallest players. c. What is the most common height and how many players share that height? d. What feature of the dotplot illustrates the most common height?

Comment on the statement: "The mean loss for customers at First State Bank (which was not insured) was \(\$ 150 .\) The standard deviation of the losses was \(-\$ 125 . "\)

a. What proportion of a normal distribution is greater than the mean? b. What proportion is within 1 standard deviation of the mean? c. What proportion is greater than a value that is 1 standard deviation below the mean?

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