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Suppose Professor Alpha and Professor Omega each teach Introductory Biology. You need to decide which professor to take the class from and have just completed your Introductory Statistics course. Records obtained from past students indicate that students in Professor Alpha's class have a mean score of \(80 \%\) with a standard deviation of \(5 \%,\) while past students in Professor Omega's class have a mean score of \(80 \%\) with a standard deviation of \(10 \% .\) Decide which instructor to take for Introductory Biology using a statistical argument.

Short Answer

Expert verified
Take Professor Alpha's class due to its lower standard deviation and more consistent scores.

Step by step solution

01

Understand the Problem

Evaluate the performance of students in Professor Alpha's and Professor Omega's classes based on the given statistical measures - mean and standard deviation.
02

Recognize Key Statistics

Identify the mean and standard deviation for each professor: Professor Alpha's class has a mean score of 80% and a standard deviation of 5%, while Professor Omega's class also has a mean score of 80% but with a higher standard deviation of 10%.
03

Interpret the Standard Deviation

Compare the standard deviations. A lower standard deviation indicates that the scores are more consistent and closer to the mean, while a higher standard deviation indicates more variability in the scores.
04

Make a Decision Based on Consistency

Since both professors have the same mean score, but Professor Alpha's class has a lower standard deviation of 5% compared to Professor Omega's 10%, you should choose the class with the more consistent scores – i.e., Professor Alpha's class.

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Key Concepts

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

Mean Score
The mean score is a fundamental concept used to determine the average performance of students in a class. It's calculated by summing all the individual scores and dividing by the total number of scores. In this exercise, both Professor Alpha and Professor Omega have a mean score of 80%. This signifies that, on average, students in both classes achieve similar results. The mean score helps gauge the overall performance but doesn't inform about the distribution or consistency of these scores. When comparing two classes with identical mean scores, other statistical measures like standard deviation become crucial for more informed decision-making.
Standard Deviation
The standard deviation is a measure of the spread or dispersion of a set of scores around the mean. It indicates how much individual scores vary from the average score. In the provided exercise, Professor Alpha's class has a standard deviation of 5%, while Professor Omega's class has a higher standard deviation of 10%. The formula for calculating standard deviation is given by:
\[ \text{Standard Deviation} = \frac{1}{N}\frac{\text{sum of squared deviations}}{mean} \]
In simpler terms, a smaller standard deviation suggests that most students' scores are closer to the average (mean score), whereas a larger standard deviation implies greater variation in the scores, with more students scoring significantly higher or lower than the mean.
Consistency in Performance
Consistency in performance is crucial when deciding which professor's class to take. Consistency is represented by how closely individual scores cluster around the mean score, which is directly influenced by the standard deviation. Lower standard deviation means higher consistency. In this scenario, Professor Alpha’s class shows more consistent performance with a standard deviation of 5%, compared to Professor Omega’s class with 10%. High consistency implies that you are likely to predict your performance more accurately. Professor Alpha’s class thus provides a stable environment where the majority of the student scores are close to the mean, making it a preferable choice for students aiming for predictable outcomes.

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Use the five test scores of 65,70,71 \(75,\) and 95 to answer the following questions: (a) Find the sample mean. (b) Find the median. (c) Which measure of central tendency better describes the typical test score? (d) Suppose the professor decides to curve the exam by adding 4 points to each test score. Compute the sample mean based on the adjusted scores. (e) Compare the unadjusted test score mean with the curved test score mean. What effect did adding 4 to each score have on the mean?

Explain the meaning of the following percentiles. Source: Advance Data from Vital and Health Statistics (a) The 15 th percentile of the head circumference of males 3 to 5 months of age is \(41.0 \mathrm{~cm}\) (b) The 90 th percentile of the waist circumference of females 2 years of age is \(52.7 \mathrm{~cm}\) (c) Anthropometry involves the measurement of the human body. One goal of these measurements is to assess how body measurements may be changing over time. The following table represents the standing height of males aged 20 years or older for various age groups. Based on the percentile measurements of the different age groups, what might you conclude? $$ \begin{array}{llllll} & {\text { Percentile }} \\\\\hline { } \text { Age } & \text { 10th } & \text { 25th } & \text { 50th } & \text { 75th } & \text { 90th } \\ \hline 20-29 & 166.8 & 171.5 & 176.7 & 181.4 & 186.8 \\ \hline 30-39 & 166.9 & 171.3 & 176.0 & 181.9 & 186.2 \\ \hline 40-49 & 167.9 & 172.1 & 176.9 & 182.1 & 186.0 \\ \hline 50-59 & 166.0 & 170.8 & 176.0 & 181.2 & 185.4 \\ \hline 60-69 & 165.3 & 170.1 & 175.1 & 179.5 & 183.7 \\ \hline 70-79 & 163.2 & 167.5 & 172.9 & 178.1 & 181.7 \\ \hline 80 \text { or older } & 161.7 & 166.1 & 170.5 & 175.3 & 179.4\\\ \hline \end{array} $$

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