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91Ó°ÊÓ

For each of the following variables, would you use the median or mean for describing the center of the distribution? Why? (Think about the likely shape of the distribution.) a. Salary of employees of a university b. Time spent on a difficult exam c. Scores on a standardized test

Short Answer

Expert verified
Use median for university salaries due to skewness; use mean for exam time and test scores assuming normal distribution.

Step by step solution

01

Analyzing Salary Data

University employee salaries typically include outliers like very high pay for top positions. This means the distribution of salaries will likely be right-skewed. When data is skewed, the median is often a better measure of central tendency because it is not affected by extreme values as the mean is.
02

Understanding Exam Time

The time students spend on a difficult exam is often normally distributed, with most students spending around the average time, but with some taking much more or less time. Here, the mean is appropriate because the data is roughly symmetric.
03

Considering Test Scores

Scores from standardized tests are generally normally distributed, as tests are designed to minimize skewness and provide a wide range of performance. Therefore, the mean is suitable since the distribution is expected to be approximately symmetrical.

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Key Concepts

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

Median
The median is a popular measure of central tendency, especially useful for data sets with outliers. When we talk about the median, picture this: if you list all your data points in order, the median is smack dab in the middle. Imagine you’re looking at university employee salaries. You might find most salaries are moderate, then *bam*, a very high salary figure for a top executive. This extremely high salary pulls the mean (or average) up. However, the median remains unaffected by these extreme values.
So, for a skewed distribution, particularly like the one you’d find with salaries, the median truly shines. It keeps things fair and mellow, showing the center without getting swayed by the extremities. Using the median helps:
  • Give an accurate picture of where the *middle* of your data truly is.
  • Prevent distortion from those way-out-there numbers.
  • Reflect what a "typical" data point looks like in a skewed series.
Mean
The mean is another key player when measuring central tendency. It's what most people think of when they hear "average" – you add up all your numbers and divide by the count of numbers. For situations like the time taken on a difficult exam, where students finish at varying times but still cluster around an average, the mean gives us a true sense of the typical experience. When data is symmetrically distributed with few outliers, the mean and median will generally be close together. This makes the mean ideal for data sets like exam times or standardized test scores, which often tend to have a symmetrical shape. The mean gives us:

  • A general insight into the "center" of the data when extremes aren’t throwing it off.
  • A straightforward calculation useful in symmetrical distributions.
  • A reliable measure when data points are evenly spread.
Distribution Shape
The shape of the distribution is like the DNA of your data set—it tells you a lot about how your data behaves. Think of a distribution as the spread of data points across a graph. The shape could be symmetrical, like a bell curve (normal distribution), or skewed, where more data accumulates on one side. In scenarios with a normal distribution, like standardized test scores designed to display a range of abilities, data points generally spread out evenly on both sides of the mean. This normal, symmetrical shape signifies that using the mean provides a clear picture of central tendency. However, when we examine distributions such as employee salaries, which likely skew right with a tail dipping towards higher values, the median gets into the spotlight as it isn't swayed by those extremes. Understanding the shape helps decide:

  • Which measure, mean or median, gives the best picture of the "center" of your data.
  • How to interpret the data's spread, helping to decode its story.
  • Whether to expect more extreme (outlier) values on one side.

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Most popular questions from this chapter

The Human Development Report 2013, published by the United Nations, showed life expectancies by country. For Western Europe, some values reported were Austria \(81,\) Belgium \(80,\) Denmark \(80,\) Finland 81 , France 83 , Germany \(81,\) Greece \(81,\) Ireland 81 , Italy 83, Netherlands 81 , Norway 81 , Portugal 80 , Spain 82 , Sweden \(82,\) Switzerland \(83 .\) For Africa, some values reported were Botswana \(47,\) Dem. Rep. Congo \(50,\) Angola \(51,\) Zambia \(57,\) Zimbabwe 58 , Malawi 55 , Nigeria \(52,\) Rwanda 63 , Uganda 59 , Kenya 61 , Mali 55 , South Africa 56 , Madagascar 64 , Senegal \(63,\) Sudan \(62,\) Ghana \(61 .\) a. Which group (Western Europe or Africa) of life expectancies do you think has the larger standard deviation? Why? b. Find the standard deviation for each group. Compare them to illustrate that \(s\) is larger for the group that shows more variability from the mean.

A study of 13 children suffering from asthma (Clinical and Experimental Allergy, vol. \(20,\) pp. \(429-432,1990\) ) compared single inhaled doses of formoterol (F) and salbutamol (S). Each child was evaluated using both medications. The outcome measured was the child's peak expiratory flow (PEF) eight hours following treament. Is there a difference in the PEF level for the two medications? The data on PEF follow: $$ \begin{array}{ccc} \hline \text { Child } & \mathbf{F} & \mathbf{S} \\ \hline 1 & 310 & 270 \\ 2 & 385 & 370 \\ 3 & 400 & 310 \\ 4 & 310 & 260 \\ 5 & 410 & 380 \\ 6 & 370 & 300 \\ 7 & 410 & 390 \\ 8 & 320 & 290 \\ 9 & 330 & 365 \\ 10 & 250 & 210 \\ 11 & 380 & 350 \\ 12 & 340 & 260 \\ 13 & 220 & 90 \\ \hline \end{array} $$ a. Construct plots to compare formoterol and salbutamol. Write a short summary comparing the two distributions of the peak expiratory flow. b. Consider the distribution of differences between the PEF levels of the two medications. Find the 13 differences and construct and interpret a plot of the differences. If on the average there is no difference between the PEF level for the two brands, where would you expect the differences to be centered?

For each of the following variables, indicate whether you would expect its histogram to be symmetric, skewed to the right, or skewed to the left. Explain why. a. The price of a certain model of smartwatch in different stores in your district b. The amount of time students use to take an exam in your school c. The grade point average (GPA) in your academic program this year d. The salary of all the employees in a company.

According to the U.S. Census Bureau, Current Population Survey, 2015 Annual Social and Economic Supplement, the mean income for males is $$\$ 47,836.10$$ with a standard deviation of $$\$ 58,353.55$$ and the mean income for females is $$\$ 28,466$$ with a standard deviation of $$\$ 36,961.10 .$$ a. Is it appropriate to use the empirical rule for male incomes? Why? b. Compare the center and variability of the income distributions for females and males. c. Which income is relatively higher-a male's income of $$\$ 55,000$$ or a female's income of $$\$ 45,000 ?$$

Data collected over several years from college students enrolled in a business statistics class regarding their shoe size is numerically summarized by \(\bar{x}=9.91, \mathrm{Q} 1=8,\) median \(=10, \mathrm{Q} 3=11\) a. Interpret the quartiles. b. Would you guess that the distribution is skewed or roughly symmetric? Why?

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