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Chapter 6: Problem 35

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 \(\mathrm{B}\). 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
New mean is 75, standard deviation remains 12. Adding 15 is not fair. B grade limits are 66.24 to 75.36. Percentile-based curving is fairer.

Step by step solution

01

Analyzing the impact of adding 15 to each grade

When 15 is added to each grade, the mean increases by 15, but the standard deviation remains unchanged. The new mean will be the original mean plus 15.
02

Calculating the new mean and standard deviation

The original mean is 60 and the standard deviation is 12. Adding 15 to each grade changes the mean by adding 15, thus: New Mean = 60 + 15 = 75 New Standard Deviation = 12 (remains the same)
03

Evaluating fairness of adding 15 to each grade

Adding a constant value to each score shifts all grades equally, not accounting for the distribution of scores relative to one another. Hence, it may not be considered fair as it doesn't adjust for any differences in performance among students.
04

Converting percentile limits into grades for part (c)

For grades of B: - Bottom 70% 鈫� Corresponds to a z-score such that 70% of data lies below it. Using a z-table, the z-score for 70% is approximately 0.52. - Top 10% 鈫� Corresponds to a z-score such that 90% of data lies below it. Using a z-table, the z-score for 90% is approximately 1.28.Convert z-scores to actual scores using the formula: Score = Mean + (z-score 脳 Standard Deviation)
05

Calculating numerical limits for a grade of B

Using the given mean (60) and standard deviation (12): - Lower limit for B: (60 + 0.52 脳 12) = 60 + 6.24 = 66.24 - Upper limit for B: (60 + 1.28 脳 12) = 60 + 15.36 = 75.36Thus, the numerical limits for a grade of B are approximately 66.24 to 75.36.
06

Comparing fairness of grading methods

Using a percentile-based curving method, as in part (c), adjusts grades according to the distribution and relative performance, thus providing a more balanced representation of student performance. Hence, it is more fair compared to simply adding 15 to each score.

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Key Concepts

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

normal distribution
The concept of normal distribution is fundamental in statistics, especially in the context of grading systems. A normal distribution is a symmetric, bell-shaped distribution where most of the data points cluster around the mean. The properties of the normal distribution include:
  • Mean (average)
  • Median (middle value)
  • Mode (most frequent value)
  • Standard deviation (measure of spread)
These properties are used to make inferences about the population and to transform raw scores into a standard format using z-scores. In grading, assuming a normal distribution allows for standard adjustments and understanding of how students perform relative to one another.
z-score
A z-score is a measure that describes a value's position relative to the mean of a group of values. It is expressed in terms of standard deviations from the mean. The formula for calculating a z-score is:

\[ z = \frac{(X - \mu)}{\sigma} \]

where
  • \( X \text{ is the raw score}\text{, } \mu \text{ is the mean},\text{ } \sigma \text{ is the standard deviation}\text{.} \)
A z-score tells you how many standard deviations a score is from the mean. For example, a z-score of 1.28 means the score is 1.28 standard deviations above the mean. This is useful in grading to determine how well a student performed compared to the entire class.
percentile rank
Percentile rank is a statistical measure indicating the percentage of scores in a distribution that a given score is above. For example, if a score is in the 70th percentile, it is higher than 70% of the scores in the distribution. Percentiles are useful in grading systems because they provide a clear way to rank and compare students' performance.
To find the percentile rank, you first convert the raw score to a z-score and then use a z-table to find the corresponding percentile. In the provided exercise, we use percentiles to determine the range for a grade of B, based on the bottom 70% and top 10% of the scores.
mean and standard deviation
The mean and standard deviation are critical concepts in understanding distributions like the normal distribution. The mean is the average of all scores, found by summing all the values and then dividing by the number of values. The standard deviation measures how spread out the values are around the mean.
In the given exercise, we know the mean score is 60 and the standard deviation is 12. Adding 15 to each grade changes the mean to 75 but does not affect the standard deviation. This tells us that the relative distance between the scores remains the same even after the adjustment. When curving grades, recognizing these properties ensures adjustments that fairly reflect students' performance.

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Most popular questions from this chapter

Use these parameters (based on Data Set 1 "Body Data" in Appendix B): \- Men's heights are normally distributed with mean \(68.6 \mathrm{in.}\) and standard deviation \(2.8 \mathrm{in} .\) \- Women's heights are normally distributed with mean 63.7 in. and standard deviation \(2.9 \mathrm{in}\). The Gulfstream 100 is an executive jet that seats six, and it has a doorway height of \(51.6 \mathrm{in}\). a. What percentage of adult men can fit through the door without bending? b. Does the door design with a height of \(51.6\) in. appear to be adequate? Why didn't the engineers design a larger door? c. What doorway height would allow \(40 \%\) of men to fit without bending?

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ Find the probability that a male has a back-to-knee length between \(22.0\) in. and \(24.0\) in.

The Ethan Allen tour boat capsized and sank in Lake George, New York, and 20 of the 47 passengers drowned. Based on a 1960 assumption of a mean weight of \(140 \mathrm{lb}\) for passengers, the boat was rated to carry 50 passengers. After the boat sank, New York State changed the assumed mean weight from \(140 \mathrm{lb}\) to \(174 \mathrm{lb}\). a. Given that the boat was rated for 50 passengers with an assumed mean of \(140 \mathrm{lb}\), the boat had a passenger load limit of \(7000 \mathrm{lb}\). Assume that the boat is loaded with 50 male passengers, and assume that weights of men are normally distributed with a mean of \(189 \mathrm{lb}\) and a standard deviation of \(39 \mathrm{lb}\) (based on Data Set 1 "Body Data" in Appendix \(\mathrm{B}\) ). Find the probability that the boat is overloaded because the 50 male passengers have a mean weight greater than \(140 \mathrm{lb}\). b. The boat was later rated to carry only 14 passengers, and the load limit was changed to \(2436 \mathrm{lb}\). If 14 passengers are all males, find the probability that the boat is overloaded because their mean weight is greater than \(174 \mathrm{lb}\) (so that their total weight is greater than the maximum capacity of \(2436 \mathrm{lb}\) ). Do the new ratings appear to be safe when the boat is loaded with \(14 \mathrm{male}\) passengers?

When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 929 peas, with 705 of them having red flowers. If we assume, as Mendel did, that under these circumstances, there is a \(3 / 4\) probability that a pea will have a red flower, we would expect that \(696.75\) (or about 697 ) of the peas would have red flowers, so the result of 705 peas with red flowers is more than expected. a. If Mendel's assumed probability is correct, find the probability of getting 705 or more peas with red flowers. b. Is 705 peas with red flowers significantly high? c. What do these results suggest about Mendel's assumption that \(3 / 4\) of peas will have red flowers?

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ For males, find \(P_{90}\), which is the length separating the bottom \(90 \%\) from the top \(10 \%\).
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