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The mean and standard deviation of the grades of a statistics course and an English course are \((\mu=80, \sigma=4.5)\) and \((\mu=85, \sigma=4.0),\) respectively. A student attends both the courses and scores 85 in statistics and 95 in English. Which grade is relatively better? Explain why.

Short Answer

Expert verified
The grade in English (95) is relatively better as it has a higher Z-score (2.5 vs 1.11 for Statistics).

Step by step solution

01

Understanding Z-scores

To compare grades from different distributions, we can use Z-scores, which tell us how many standard deviations a score is from the mean. The formula is \( Z = \frac{X - \mu}{\sigma} \) where \( X \) is the score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
02

Calculate Z-score for Statistics

The student's score in statistics is 85. The mean \( \mu \) for the statistics course is 80 and the standard deviation \( \sigma \) is 4.5. Substitute these into the Z-score formula: \[ Z = \frac{85 - 80}{4.5} = \frac{5}{4.5} = 1.11 \]
03

Calculate Z-score for English

The student's score in English is 95. The mean \( \mu \) for the English course is 85 and the standard deviation \( \sigma \) is 4.0. Substitute these into the Z-score formula:\[ Z = \frac{95 - 85}{4.0} = \frac{10}{4.0} = 2.5 \]
04

Compare Z-scores to Determine Relative Performance

The Z-score for statistics is 1.11, while the Z-score for English is 2.5. The higher the Z-score, the better the performance relative to the class. Since 2.5 is greater than 1.11, the student's performance is relatively better in the English course compared to the statistics course.

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Key Concepts

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

Mean
The concept of the "mean" is fundamental in understanding how individual scores relate to the group as a whole. In statistics, the mean is often referred to as the "average" and is determined by summing all data points in a set and dividing by the total count of those data points. For example, if the grades of a statistics course are given, the mean grade provides a central value that represents how students have performed on average.

To calculate the mean, use the formula:
  • Sum all the scores in the data set.
  • Divide the total sum by the number of scores.
The mean helps to understand whether an individual score, say a student's grade, is below, above, or around the average performance of the class. If the mean of a statistics course is 80, a student's score of 85 indicates a performance above the average.
Standard Deviation
Standard deviation is a key measure in statistics that tells us about the spread of data points in a dataset. It shows how much variation or "spread" exists around the mean.
  • A smaller standard deviation indicates that the data points are close to the mean.
  • A larger standard deviation means the data points spread out over a wider range of values.
To calculate it, you need to understand how far each score is from the mean, square those differences, calculate their average, and then take the square root.

For instance, in the context of the given courses, a standard deviation of 4.5 in the statistics course suggests that the grades are more spread out compared to the English course, which has a standard deviation of 4.0. Hence, understanding standard deviation helps in determining how significant a particular score divergence is in relation to the typical performance.
Relative Performance
"Relative performance" refers to comparing individual scores not only to the mean, but also considering the spread of scores. Z-scores are a useful tool to assess this by showing how far away a particular score is from the mean in terms of standard deviations.

For instance, the student's scores in both statistics and English courses can be converted into Z-scores to gauge relative performance within each class:
  • Statistics: Z-score is 1.11, indicating the score is 1.11 standard deviations above the mean.
  • English: Z-score is 2.5, indicating a score 2.5 standard deviations above the mean.
Since a higher Z-score signifies a better performance relative to the classmates, the student performed relatively better in their English course compared to the statistics course.
Statistics Course
The statistics course uses mathematical principles to analyze and interpret data. By applying concepts such as the mean and standard deviation, students can gain insights into data trends and variations. In this context, a student’s performance is evaluated not just by raw scores but by how these scores compare to class averages and spread.

When a student scores significantly above or below the mean, it can suggest varying levels of understanding of the material. The ability to calculate Z-scores and other statistical measures provides a structured way to assess where a student stands in relation to their peers. This comparative analysis is crucial in fields that heavily rely on numerical data and statistical inference.
English Course
While English courses focus on developing language skills and comprehension, understanding one's performance relative to peers remains important. In this context, grades alone might not provide the full picture of a student's capabilities or their grasp of the subject.

By utilizing statistical tools such as the mean and standard deviation, we can interpret the results meaningfully and understand a student's abilities compared to the class average. For example, a Z-score analysis can reveal a student's position within their peer group, as it reflects how much better or worse the student did compared to the class norm.

Ultimately, applying these statistical concepts in an English course provides clarity on the impact and extent of a student's achievement, much like how language and analysis work hand in hand to enhance understanding and proficiency.

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Most popular questions from this chapter

A professor of statistics wants to prepare a test paper by selecting five questions randomly from an online test bank available for his course. In the test bank, the proportion of questions labeled "HARD" is 0.3 . a. Find the probability that all the questions selected for the test are labeled HARD. b. Find the probability that none of the questions selected for the test is labeled HARD. c. Find the probability that less than half of the questions selected for the test are labeled HARD.

For a normal distribution, use Table A, software, or a calculator to find the probability that an observation is a. at least 1 standard deviation above the mean. b. at least 1 standard deviation below the mean. c. within 1 standard deviation of the mean.

Binomial needs fixed \(n\) For the binomial distribution, the number of trials \(n\) is a fixed number. Let \(X\) denote the number of girls in a randomly selected family in Canada that has three children. Let \(Y\) denote the number of girls in a randomly selected family in Canada (that is, the number of children could be any number). A binomial distribution approximates well the probability distribution for one of \(X\) and \(Y\), but not for the other. a. Explain why. b. Identify the case for which the binomial applies and identify \(n\) and \(p\).

Over roughly the past 100 years, the mean monthly April precipitation in Williamstown, Massachusetts, equaled 3.6 inches with a standard deviation of 1.6 inches. (Source: http://web.williams.edu/ weather/) a. In April \(1983,\) the wettest April on record, the precipitation equaled 8.4 inches. Find its z-score. If the distribution of precipitation were roughly normal, would this be unusually high? Explain. b. Assuming a normal distribution, an April precipitation of 4.5 inches corresponds to what percentile? c. Of the 119 measurements of April precipitation on record (reaching as far back as 1892 ), \(66.5 \%\) fell within one, \(97.5 \%\) within two, and \(99.1 \%\) within three standard deviations of the mean. Do you think that the distribution of April precipitation is approximately normal? Why or why not?

Most phones use lithium-ion (Li-ion) batteries. These batteries have a limited number of charge and discharge cycles, usually falling between 300 and \(500 .\) Beyond this lifespan, a battery gradually diminishes below \(50 \%\) of its original capacity. a. Suppose the distribution of the number of charge and discharge cycles was normal. What values for the mean and the standard deviation are most likely to meet the assumption of normality of this variable? b. Based on the mean and the standard deviation calculated in part a, find the 95 th percentile.

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