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Chapter 2: Problem 86

1\( in the How Extreme Values Affect the Mean and Median applet. This applet loads with a dotplot for the following \)n=5\( observations: 2,5,6,9,11 a… # Refer to Data Set \)\\# 1\( in the How Extreme Values Affect the Mean and Median applet. This applet loads with a dotplot for the following \)n=5\( observations: 2,5,6,9,11 a. What are the mean and median for this data set? b. Use your mouse to change the value \)x=11\( (the moveable green dot) to \)x=13 .\( What are the mean and median for the new data set? c. Use your mouse to move the green dot to \)x=33$. When the largest value is extremely large compared to the other observations, which is larger, the mean or the median? d. What effect does an extremely large value have on the mean? What effect does it have on the median?

Short Answer

Expert verified
Answer: An extreme value in a data set can cause the mean to be significantly larger or smaller than the median, as it gets pulled towards the extreme value. The median, however, remains relatively stable and is less sensitive to the extreme value due to its consideration of only the middle value(s) in the sorted data set. This makes the median a better measure of central tendency when outliers are present.

Step by step solution

01

Part a: Calculate the mean and median of the original data set

First, let's calculate the mean and median for the data set {2, 5, 6, 9, 11}. (1) To calculate the mean, we sum the observations and divide by the number \(n\) of observations: Mean \(= \dfrac{2 + 5 + 6 + 9 + 11}{5} = \dfrac{33}{5} = 6.6\) (2) To find the median, we must first arrange the observations in ascending order (which they already are) and take the middle value (or the average of the middle two values if \(n\) is even). Since there are 5 observations, the median is the third value: Median \(= 6\)
02

Part b: Change the value \(x = 11\) to \(x = 13\) and calculate the new mean and median

Now let's change the last value from 11 to 13 and find the mean and median for the new data set {2, 5, 6, 9, 13}. (1) Calculate the mean: Mean \(= \dfrac{2 + 5 + 6 + 9 + 13}{5} = \dfrac{35}{5} = 7\) (2) Notice that the numbers in the set are already in ascending order, so to find the median, we still look for the third value: Median \(= 6\)
03

Part c: Move the green dot to \(x = 33\) and compare the mean and median

Let's change the last value from 13 to 33 and find the mean and median for the new data set {2, 5, 6, 9, 33}. (1) Calculate the mean: Mean \(= \dfrac{2 + 5 + 6 + 9 + 33}{5} = \dfrac{55}{5} = 11\) (2) The numbers are still in ascending order, so to find the median, we still look for the middle value: Median \(= 6\) Now, compare the mean and median: When the largest value (33) is extremely larger than the other observations, the mean (11) is larger than the median (6).
04

Part d: Effects of an extremely large value on the mean and median

An extremely large value has the following effects on the mean and median: (1) Effect on the mean: When an extremely large value is present, the mean will be pulled towards that value, making it larger than it would be without the extreme value. (2) Effect on the median: The median is less sensitive to extreme values because it considers only the middle value(s) of the data set. In all three cases, the median remains the same (6). The median is more resistant to the effects of extreme values than the mean, and it gives us a better measure of the data's central tendency when there are outliers present.

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Key Concepts

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

Mean
The mean, also known as the average, is a central concept in statistics that provides an overall picture of a dataset. To find the mean, you sum all the observations and divide by the total number of observations. It's a crucial measure because it includes every value in your dataset, making it sensitive to changes and outliers.
In our original dataset {2, 5, 6, 9, 11}, the mean is calculated by adding all the numbers together and dividing by five, resulting in a mean of 6.6. When we altered the largest number from 11 to 13, the mean rose to 7.
This demonstrates the mean's sensitivity to changes: a small increase in one observation affects the mean directly. Therefore, while the mean is informative about the general trend of your data, it can sometimes be misleading if the dataset contains extreme values.
Median
Understanding the median is essential, especially when dealing with skewed datasets. The median represents the middle value of your data when arranged in ascending order. It is particularly useful because it is not swayed by extreme values.
In the case of our dataset {2, 5, 6, 9, 11}, the median is simply the middle number, which is 6. Even when we changed 11 to 13 or even 33, the median remained at 6.
This stability confirms the median's robustness in representing a dataset's central tendency when there are outliers. It serves as an ideal snapshot of the dataset's center, unaffected by possibly misleading extreme values or single large changes.
Outliers
Outliers are values in a dataset that are significantly different from the majority of the other data points. These values stand out because they do not fit the general pattern.
Outliers can skew the mean significantly. For instance, in the original dataset {2, 5, 6, 9, 11}, changing the 11 to a more extreme 33 transformed the mean from 6.6 to 11. However, the median stayed stable at 6.
Thus, while outliers help highlight variations in the data, they can also distort the mean, making the median a more reliable measure of central tendency in such cases.
Extreme Values
Extreme values, or values far removed from the rest of your dataset, can profoundly impact descriptive statistics. While similar to outliers, extreme values tend to deviate more dramatically. They often arise due to data entry errors, experimental errors, or a natural but rare occurrence.
In practical terms, like in the dataset alteration where the value 33 was introduced, the mean was heavily influenced, highlighting how unrepresentative it can become in the presence of an extreme value.
In contrast, the median provides a fairer, more reliable statistic under such conditions because it only looks at the midpoint of the data, offering a measure of central location that is resistant to distortion by extreme changes. By understanding extreme values, you can choose the most appropriate statistical measures for analysis.

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Most popular questions from this chapter

The miles per gallon (mpg) for each of 20 medium-sized cars selected from a production line during the month of March follow. $$ \begin{array}{llll} 23.1 & 21.3 & 23.6 & 23.7 \\ 20.2 & 24.4 & 25.3 & 27.0 \\ 24.7 & 22.7 & 26.2 & 23.2 \\ 25.9 & 24.7 & 24.4 & 24.2 \\ 24.9 & 22.2 & 22.9 & 24.6 \end{array} $$ a. What are the maximum and minimum miles per gallon? What is the range? b. Construct a relative frequency histogram for these data. How would you describe the shape of the distribution? c. Find the mean and the standard deviation. d. Arrange the data from smallest to largest. Find the \(z\) -scores for the largest and smallest observations. Would you consider them to be outliers? Why or why not? e. What is the median? f. Find the lower and upper quartiles.

Suppose that some measurements occur more than once and that the data \(x_{1}, x_{2}, \ldots, x_{k}\) are arranged in a frequency table as shown here: $$ \begin{array}{cc} \text { Observations } & \text { Frequency } f_{i} \\ \hline x_{1} & f_{1} \\ x_{2} & f_{2} \\ \cdot & \cdot \\ \cdot & \cdot \\ \cdot & \cdot \\ x_{k} & f_{k} \end{array} $$ The formulas for the mean and variance for grouped data are \(\bar{x}=\frac{\sum x_{i} f_{i}}{n}\) $$ \text { where } n=\Sigma f_{i} $$ and $$ s^{2}=\frac{\sum x_{i}^{2} f_{i}-\frac{\left(\sum x_{i} f_{i}\right)^{2}}{n}}{n-1} $$ Notice that if each value occurs once, these formulas reduce to those given in the text. Although these formulas for grouped data are primarily of value when you have a large number of measurements, demonstrate their use for the sample \(1,0,0,1,3,1,3,2,3,0,\) 0,1,1,3,2 a. Calculate \(\bar{x}\) and \(s^{2}\) directly, using the formulas for ungrouped data. b. The frequency table for the \(n=15\) measurements is as follows: $$ \begin{array}{ll} x & f \\ \hline 0 & 4 \\ 1 & 5 \\ 2 & 2 \\ 3 & 4 \end{array} $$ Calculate \(\bar{x}\) and \(s^{2}\) using the formulas for grouped data. Compare with your answers to part a.

Is your breathing rate normal? Actually, there is no standard breathing rate for humans. It can vary from as low as 4 breaths per minute to as high as 70 or 75 for a person engaged in strenuous exercise. Suppose that the resting breathing rates for college-age students have a relative frequency distribution that is mound-shaped, with a mean equal to 12 and a standard deviation of 2.3 breaths per minute. What fraction of all students would have breathing rates in the following intervals? a. 9.7 to 14.3 breaths per minute b. 7.4 to 16.6 breaths per minute c. More than 18.9 or less than 5.1 breaths per minute

Here are a few facts reported as Snapshots in USA Today. \- The median hourly pay for salespeople in the building supply industry is \(\$ 10.41 .^{15}\) \- Sixty-nine percent of U.S. workers ages 16 and older work at least 40 hours per week. \({ }^{16}\) \- Seventy-five percent of all Associate Professors of Mathematics in the U.S. earn \(\$ 91,823\) or less. \(^{17}\) Identify the variable \(x\) being measured, and any percentiles you can determine from this information.

An analytical chemist wanted to use electrolysis to determine the number of moles of cupric ions in a given volume of solution. The solution was partitioned into \(n=30\) portions of .2 milliliter each, and each of the portions was tested. The average number of moles of cupric ions for the \(n=30\) portions was found to be .17 mole; the standard deviation was .01 mole. a. Describe the distribution of the measurements for the \(n=30\) portions of the solution using Tchebysheff's Theorem. b. Describe the distribution of the measurements for the \(n=30\) portions of the solution using the Empirical Rule. (Do you expect the Empirical Rule to be suitable for describing these data?) c. Suppose the chemist had used only \(n=4\) portions of the solution for the experiment and obtained the readings \(.15, .19, .17,\) and \(.15 .\) Would the Empirical Rule be suitable for describing the \(n=4\) measurements? Why?
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