Question

from unit 1, lesson 12
here is a data set:
5
10
10
10
15
100
a. after studying the data, the researcher realized that the value 100 was meant to be recorded as 15. what happens to the mean and standard deviation of the data set when the 100 is changed to a 15?
b. for the original data set, with the 100, would the median or the mean be a better choice of measure for the center? explain your reasoning.

Explanation:

Step1: Calculate the original mean

The original data set is \(5,10,10,10,15,100\).
The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 6\) and \(\sum_{i=1}^{6}x_{i}=5 + 10+10+10+15+100=150\).
So the original mean \(\bar{x}_{1}=\frac{150}{6}=25\).
The original data set in ascending order is \(5,10,10,10,15,100\). The median \(M_{1}=\frac{10 + 10}{2}=10\) (since \(n = 6\) (even), median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th values).

Step2: Calculate the new mean

After changing \(100\) to \(15\), the new data set is \(5,10,10,10,15,15\).
\(\sum_{i = 1}^{6}x_{i}=5+10 + 10+10+15+15=65\).
The new mean \(\bar{x}_{2}=\frac{65}{6}\approx10.83\).
The new data set in ascending order is \(5,10,10,10,15,15\). The new median \(M_{2}=\frac{10 + 10}{2}=10\)

Step3: Analyze the effect on standard deviation

The formula for the standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\) (for sample, \(n-1\); for population \(n\), here assume population).
For the original data set:
\((5 - 25)^{2}+(10 - 25)^{2}+(10 - 25)^{2}+(10 - 25)^{2}+(15 - 25)^{2}+(100 - 25)^{2}\)
\(=(- 20)^{2}+(-15)^{2}+(-15)^{2}+(-15)^{2}+(-10)^{2}+75^{2}\)
\(=400+225+225+225+100 + 5625=6800\)
\(s_{1}=\sqrt{\frac{6800}{6}}\approx33.66\)
For the new data set:
\((5-10.83)^{2}+(10 - 10.83)^{2}+(10 - 10.83)^{2}+(10 - 10.83)^{2}+(15 - 10.83)^{2}+(15 - 10.83)^{2}\)
\(=(-5.83)^{2}+(-0.83)^{2}+(-0.83)^{2}+(-0.83)^{2}+4.17^{2}+4.17^{2}\)
\(=33.99+0.69+0.69+0.69+17.39+17.39=70.84\)
\(s_{2}=\sqrt{\frac{70.84}{6}}\approx3.43\)

Step4: Answer part b

The median is a better measure of the center.
The mean is affected by extreme values (the original \(100\)). The median is less sensitive to extreme values. In the original data set, the \(100\) pulled the mean up, while the median was not affected. After changing the extreme value, the mean changed significantly, but the median remained the same.

Answer:

a. The mean decreases from \(25\) to approximately \(10.83\), and the standard deviation decreases from approximately \(33.66\) to approximately \(3.43\).
b. The median is a better measure of the center. The mean is sensitive to extreme values (the original \(100\)), while the median is not. When the extreme value (\(100\)) is changed, the mean changes significantly (\(25\) to \(\frac{65}{6}\approx10.83\)), but the median remains \(10\) (both before and after the change of the extreme value).