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A professor gives a test and the scores are normally distributed with a mean of 60 and a standard deviation of \(12 .\) She plans to curve the scores. a. If she curves by adding 15 to each grade, what is the new mean and standard deviation? b. Is it fair to curve by adding 15 to each grade? Why or why not? c. If the grades are curved so that grades of \(\mathrm{B}\) are given to scores above the bottom \(70 \%\) and below the top \(10 \%,\) find the numerical limits for a grade of B. d. Which method of curving the grades is fairer: adding 15 to each original score or using a scheme like the one given in part (c)? Explain.

Short Answer

Expert verified
The new mean is 75, and the standard deviation is 12. Scores from 54.76 to 75.36 get a B using z-scores; this method is fairer.

Step by step solution

01

Determine the new mean and standard deviation after adding 15 to each grade

When a constant value is added to all values in a distribution, the mean increases by that constant, but the standard deviation remains the same. Here, the mean increases by 15.
02

Calculate New Mean

Original mean = 60. New mean = 60 + 15 = 75.
03

Assess Standard Deviation

The standard deviation remains the same, at 12.
04

Consider the fairness of adding 15 to each grade

Adding 15 to each grade does not change the relative standing of the scores but merely shifts them all upward. Therefore, it treats all students equally by uniform shift, not altering the spread.
05

Interpret grades based on statistical limits

Given a normal distribution, we use z-scores to find the numerical limits for determining grades. We find the z-scores that correspond to the bottom 70% and top 10%.
06

Calculate Z-scores

For bottom 70%, use z = -0.52 (since bottom 70% includes up to 70th percentile). For top 10%, use z = 1.28 (since top 10% starts from 90th percentile).
07

Determine Numerical Limits for Grade B

We use the formula: X = mean + z * standard deviation.
08

Calculate Boundary Scores

Lower boundary: X = 60 + (-0.52 * 12) ≈ 54.76 Upper boundary: X = 60 + (1.28 * 12) ≈ 75.36 Therefore, scores between 54.76 and 75.36 receive a Grade B.
09

Compare fairness of different curving methods

Adding a fixed number (15) preserves the distribution shape but may not adjust sufficiently by student's distinctions and efforts. The z-score method considers the distribution shape to determine relative standing more fairly.

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Key Concepts

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

Mean and Standard Deviation
In statistics, the **mean** is the average of the data set. To find the mean, you sum up all the observations and divide by the number of observations. For example, if the test scores are normally distributed with a mean of 60, this means that the average score is 60. The **standard deviation** measures the spread of the scores around the mean. It tells us how much the scores typically vary from the average. A standard deviation of 12 means most scores fall within 12 points above or below the mean of 60. When you add a constant value to each score, like adding 15 to each grade, the mean increases by that value, but the standard deviation remains the same. This is because the spread of the scores around the mean does not change.
Grading Curve
A **grading curve** is used to adjust the distribution of grades to fit a desired distribution. This can be helpful when the raw scores do not spread in the way the instructor prefers. One simple way to curve grades is by adding a fixed number of points to each student's score. This method shifts the mean upward while keeping the standard deviation unchanged. For instance, in the given example, adding 15 points to each grade pushes the mean from 60 to 75, but the standard deviation stays at 12. However, this approach is simplistic and may not address all fairness concerns. It doesn't account for the performance relative to peers.
Z-scores
A **z-score** transforms a score into how many standard deviations it is away from the mean. The formula for calculating a z-score is: \( z = \frac{(X - \text{mean})}{\text{standard deviation}} \). Z-scores are useful in finding the relative position of a score within a distribution. For example, in our scenario, if we want to give a grade of 'B' to the scores that fall between the bottom 70% and the top 10%, we first determine the z-scores corresponding to these percentiles. For the bottom 70%, the z-score is approximately -0.52, and for the top 10%, it is around 1.28. Using the mean of 60 and standard deviation of 12, we can find the boundary scores: for the lower boundary, \( X = 60 + (-0.52 \times 12) \), which is approximately 54.76. For the upper boundary, \( X = 60 + (1.28 \times 12) \), which is about 75.36.
Statistical Fairness
Fairness in grading is crucial to ensure that performance assessments are equitable. When considering statistical fairness, it is important to consider both absolute and relative performance measures. Adding a fixed number of points, such as 15, to each score treats all students uniformly and does not change their relative standing. However, this might not adequately differentiate between students' performances. A method leveraging z-scores respects the distribution shape and differentiates based on relative position, which can be seen as fairer since it considers students' standing in relation to the overall performance. It ensures that the awarding of grades takes into account not just an absolute shift but how well students perform relative to their peers.

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