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Essay errors Typographical and spelling errors can be either "nonword errors" or "word errors." A nonword error is not a real word, as when "the" is typed as "teh." A word error is a real word, but not the right word, as when "lose" is typed as "loose." When students are asked to write a 250 -word essay (without spell-checking), the number of nonword errors \(X\) in a randomly selected essay has the following probability distribution: $$ \begin{array}{lccccc} \hline \text { Value: } & 0 & 1 & 2 & 3 & 4 \\ \text { Probability: } & 0.1 & 0.2 & 0.3 & 0.3 & 0.1 \\ \hline \end{array} $$ $$ \mu_{X}=2.1 \quad \sigma_{X}=1.136 $$ The number of word errors \(Y\) has this probability distribution: $$ \begin{array}{lcccc} \hline \text { Value: } & 0 & 1 & 2 & 3 \\ \text { Probability: } & 0.4 & 0.3 & 0.2 & 0.1 \\ \hline & & & & \\ & \mu_{Y}=1.0 & \sigma_{Y} & =1.0 & & \end{array} $$ Assume that \(X\) and \(Y\) are independent. An English professor deducts 3 points from a student's essay score for each nonword error and 2 points for each word error. Find the mean and standard deviation of the total score deductions for a randomly selected essay. Show your work.

Step by step solution

01

Define the Total Deduction Formula

The total score deduction, denoted as \(D\), from the essay is calculated based on the nonword errors \(X\) and word errors \(Y\). The formula for the total deduction is \(D = 3X + 2Y\).

02

Calculate the Expected Value of X and Y

From the problem, the expected number of nonword errors \(\mu_X = 2.1\) and the expected number of word errors \(\mu_Y = 1.0\) are already given.

03

Calculate the Expected Value of D

Using the linearity of expectation: \(E(D) = E(3X + 2Y) = 3E(X) + 2E(Y) = 3 \times 2.1 + 2 \times 1.0\). Calculate this to get \(E(D) = 6.3 + 2.0 = 8.3\).

04

Calculate the Variances of X and Y

The variance is the square of the standard deviation. Thus, \(\sigma^2_X = 1.136^2 = 1.291\) and \(\sigma^2_Y = 1.0^2 = 1.0\).

05

Calculate the Variance of D

Since \(X\) and \(Y\) are independent, the variance of their linear combination is: \(Var(D) = Var(3X + 2Y) = 3^2Var(X) + 2^2Var(Y) = 9\times1.291 + 4\times 1.0\). Calculate this to obtain \(Var(D) = 11.619 + 4 = 15.619\).

06

Calculate the Standard Deviation of D

The standard deviation of \(D\) is the square root of its variance: \(\sigma_D = \sqrt{15.619} \approx 3.95\). Thus, the standard deviation of the total deductions is approximately 3.95 points.

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Key Concepts

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

Expectations and Variance

In the context of probability distributions, expectations and variance are crucial measures. They help us understand the average outcome and the variability around this average. The expectation, often called the expected value, is a measure of the center of the distribution. It's like an average value that gives us a sense of what to expect from a random variable. In our exercise, for nonword errors, this expectation is given as \( \mu_X = 2.1 \), and for word errors, \( \mu_Y = 1.0 \). These values mean that, on average, we can expect about 2.1 nonword errors and 1.0 word errors per essay.

Variance, on the other hand, tells us about the spread or dispersion of the values. It quantifies how much the numbers deviate from the expected value. The formula for variance is the square of the standard deviation. For nonword errors, the variance \( \sigma^2_X \) is calculated as \( 1.136^2 = 1.291 \). For word errors, it's straightforward with \( \sigma^2_Y = 1.0^2 = 1.0 \). By understanding these values, we get a clearer picture of the consistency of errors in student essays, helping us anticipate potential score deductions.

Independent Random Variables

When two random variables are independent, the occurrence of one does not affect the probability of the occurrence of the other. In our exercise, the variables \( X \) for nonword errors and \( Y \) for word errors are considered independent. This assumption simplifies the calculations of combined properties like means and variances.

For independent variables, the expectation of their sum can simply be calculated by summing their individual expectations. Therefore, the expectation for the total score deduction \( D = 3X + 2Y \) is calculated as \( E(D) = 3E(X) + 2E(Y) = 8.3 \).

Moreover, when dealing with variance for independent variables, we can add their variances to find the variance of their linear combination. That's why the variance of \( D \), denoted as \( Var(D) \), is calculated using \( Var(D) = 3^2Var(X) + 2^2Var(Y) = 15.619 \). This is possible only because \( X \) and \( Y \) are independent.

Nonword and Word Errors

Nonword and word errors are two types of common mistakes found in essays. Nonword errors are misspellings that don't form real words, such as typing "teh" instead of "the". Word errors, however, occur when a wrong yet real word is used, like writing "loose" when one means "lose". Understanding these errors is important for analyzing the impact they have on assessments.

By observing the respective probability distributions provided in the exercise, we can discern patterns and frequency of these error types. For instance, nonword errors have a probability spread from 0 to 4 errors, while word errors range from 0 to 3 errors. These distributions help in determining the expected value and variance, which are fundamental to evaluating their probable count in essays.

These errors' frequency influences the expected total deductions, thereby directly affecting the grading of essays. By implementing statistical measures, educators can pinpoint common error patterns, allowing for focused feedback and tailored teaching strategies to improve students' writing skills.

Score Deductions

Score deductions in essays can be directly influenced by the frequency of errors such as nonword and word errors. In this scenario, an English professor deducts specific points: 3 points per nonword error and 2 points per word error. The total score deduction for an essay, represented by \( D \), is derived by the formula \( D = 3X + 2Y \).

To find the average deduction a student might face, we calculated \( E(D) \) as 8.3. This average reflects potential penalty points due to spelling mistakes, thus highlighting the importance for students to minimize errors. The variance \( Var(D) = 15.619 \), and its standard deviation, approximately 3.95, give us insight into the variability of these deductions.

Understanding these statistics is valuable for students as it emphasizes the significance of error-proof writing, which can heavily and unpredictably impact their final score. Additionally, for educators, this data underscores the value of providing ample spelling and grammar training to mitigate these avoidable deductions.