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91Ó°ÊÓ

Grading managers. Some companies "grade on a bell curve" to compare the performance of their managers and professional workers. This forces the use of some low performance ratings so that not all workers are listed as "above average." Ford Motor Company's "performance management process" for this year assigned \(10 \%\) A grades, \(80 \%\) B grades, and \(10 \% \mathrm{C}\) grades to the company's managers. Suppose Ford's performance scores really are Normally distributed. This year, managers with scores less than 25 received \(\mathrm{C}\) grades and those with scores above 475 received A grades. What are the mean and standard deviation of the scores?

Short Answer

Expert verified
The mean is approximately 250, and the standard deviation is about 175.78.

Step by step solution

01

Understand the Problem

We have a normal distribution of performance scores. We need to find the mean \( \mu \) and the standard deviation \( \sigma \) of these scores. We know the cutoff for 'C' grades is below 25 and that for 'A' grades is above 475.
02

Assign Z-scores for Grade Cutoffs

The lowest 10% of scores fall below 25, and the highest 10% fall above 475. In a standard normal distribution, a Z-score of -1.28 corresponds to the 10th percentile, and a Z-score of 1.28 corresponds to the 90th percentile.
03

Write the Z-score Equations

Use the Z-score formula, \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is a score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.- For score 25: \( Z = \frac{25 - \mu}{\sigma} = -1.28 \)- For score 475: \( Z = \frac{475 - \mu}{\sigma} = 1.28 \)
04

Solve the Equations

We have two equations: 1. \( 25 - \mu = -1.28 \cdot \sigma \)2. \( 475 - \mu = 1.28 \cdot \sigma \)Subtract the first equation from the second to eliminate \( \mu \) and solve for \( \sigma \):\( (475 - \mu) - (25 - \mu) = 1.28\cdot\sigma - (-1.28\cdot\sigma) \)\( 450 = 2.56\cdot\sigma \)\( \sigma = \frac{450}{2.56} \approx 175.78 \)
05

Solve for Mean \( \mu \)

Substitute \( \sigma \approx 175.78 \) back into either equation to find \( \mu \). Using the first equation:\( 25 - \mu = -1.28 \cdot 175.78 \)\( 25 - \mu = -224.9984 \)\( \mu = 25 + 224.9984 = 249.9984 \approx 250 \) (rounded to whole number)

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Key Concepts

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

Z-score
Understanding Z-scores is fundamental when dealing with normal distributions in statistics. A Z-score, also known as a standard score, tells us how many standard deviations a data point is from the mean of a data set. It's a way to standardize scores on a common scale, making it easier to compare them. For instance, in our example involving Ford Motor Company's performance ratings:
  • A score of 25 is assigned a Z-score of -1.28, indicating it is 1.28 standard deviations below the mean.
  • A score of 475 corresponds to a Z-score of 1.28, showing it's 1.28 standard deviations above the mean.
These particular Z-scores correlate with the 10th and 90th percentiles, which helps in assigning these cutoffs for performance grading. The Z-score formula is given by:\[ Z = \frac{X - \mu}{\sigma} \]where \(X\) is the score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Calculating Z-scores helps us understand where an individual score lies in a standard normal distribution.
Percentiles
Percentiles are a fantastic tool in statistics, particularly when dealing with normal distribution. They divide your dataset into 100 equal parts, giving you a clear picture of where a certain value stands relative to others in the data set. Each percentile represents a specific "rank" of data by showing the percentage of data values below it. In our Ford Motor Company example:
  • The lowest 10% of scores fall below the 10th percentile, resulting in managers receiving a grade C.
  • The highest 10% of scores fall above the 90th percentile, assigning those scores an A grade.
Percentiles allow organizations to categorize or rank data, making them very useful for decision-making and performance evaluations. Knowing the percentile rank is invaluable, as it reveals how a particular score compares to the rest of the data set.
Mean and Standard Deviation
The mean and standard deviation are pivotal concepts in statistics that help describe the central tendency and dispersion in a dataset, respectively. The mean, denoted by \(\mu\), represents the average score of all the data points, giving an overall distribution center.

In our example, solving for the mean gives us \(\mu \approx 250\), meaning that the average score of Ford's performance ratings is around 250.
Standard deviation, symbolized by \(\sigma\), shows how much variation or "spread" exists from the mean. A larger \(\sigma\) means scores are more spread out, while a smaller \(\sigma\) indicates they're clustered closely around the mean.The standard deviation calculated here is \(\sigma \approx 175.78\), showing a fair amount of spread in the performance scores. Understanding both concepts is crucial because they form the basis of calculating Z-scores and subsequently help interpret the bell curve of the normal distribution.
Bell Curve Grading
Bell curve grading, as used in Ford Motor Company's performance evaluation, relies on the normal distribution to assign grades based on where scores fall along the curve. This method ensures a statistical approach to grading, instead of all scores being uniformly high or low.

In a bell curve, most data points cluster near the mean, while fewer and fewer are found as you move away in either direction, creating a shape like a bell. Grades are then distributed:
  • \(10\%\) of managers were given C grades for scores below the 10th percentile (score \(< 25\)).
  • \(80\%\) fall in the B grade range (scores between 25 and 475).
  • \(10\%\) received A grades, scoring above the 90th percentile (score \(> 475\)).
This type of grading system is beneficial for differentiating performance levels, placing workers into categories based on statistical standardization rather than subjective assessment. However, it can sometimes create competition and stress among employees, as it forces some into lower grades regardless of overall performance.

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Most popular questions from this chapter

Body mass index. Your body mass index (BMI) is your weight in kilograms divided by the square of your height in meters. Many online BMI calculators allow you to enter weight in pounds and height in inches. High BMI is a common but controversial indicator of overweight or obesity. A study by the National Center for Health Statistics found that the BMI of American young men (ages 20-29) is approximately Normal with mean \(26.8\) and standard deviation 5.2. 12 (a) People with BMI less than \(18.5\) are often classified as "underweight." What percent of men aged 20-29 are underweight by this criterion? (b) People with BMI over 30 are often classified as "obese." What percent of men aged \(20-29\) are obese by this criterion?

Understanding density curves. Remember that it is areas under a density curve, not the height of the curve, that give proportions in a distribution. To illustrate this, sketch a density curve that has a tall, thin peak at 0 on the horizontal axis but has most of its area close to 1 on the horizontal axis without a high peak at \(1 .\)

To completely specify the shape of a Normal distribution, you must give (a) the mean and the standard deviation. (b) the five-number summary. (c) the median and the quartiles.

The distribution of hours of sleep per week night, among college students, is found to be Normally distributed, with a mean of \(6.5\) hours and a standard deviation of 1 hour. The percentage of college students that sleep at least eight hours per night is about (a) \(95 \%\). (b) \(6.7 \%\). (c) \(2.5 \%\)

The scores of adults on an IQ test are approximately Normal with mean 100 and standard deviation 15. Alysha scores 135 on such a test. She scores higher than what percent of all adults? (a) About \(5 \%\) (b) About \(95 \%\) (c) About \(99 \%\)

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