Question

here is a data set: (lesson 1 - 12) 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? the mean decreases from 25 to approximately 10.83 and the standard deviation decreases. 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 original mean

Let the data - set be \(5, 10, 10, 10, 15, 100\). The number of data - points \(n = 6\). The original mean \(\bar{x}_1=\frac{5 + 10+10+10+15+100}{6}=\frac{150}{6}=25\).

Step2: Calculate new mean

When the value \(100\) is changed to \(15\), the new sum of data - points is \(5 + 10+10+10+15+15=65\). The new mean \(\bar{x}_2=\frac{65}{6}\approx10.83\). So the mean decreases.

Step3: Analyze effect on standard deviation

The standard deviation measures the spread of the data. The value \(100\) is an out - lier. When it is changed to \(15\), the data becomes more clustered around the center, so the standard deviation decreases.

Step4: Analyze median and mean for central tendency

The original data - set \(5, 10, 10, 10, 15, 100\) has \(n = 6\) data - points. The median is the average of the 3rd and 4th ordered values, so the median \(M_1=\frac{10 + 10}{2}=10\). The new data - set \(5, 10, 10, 10, 15, 15\) also has \(n = 6\) data - points, and the median \(M_2=\frac{10+10}{2}=10\). The median is not affected by extreme values (out - liers), while the mean is. So, for this data set with an out - lier, the median is a better measure of the center.

Answer:

a. The mean decreases from \(25\) to approximately \(10.83\) and the standard deviation decreases.
b. The median is a better measure of the center because the data set has an out - lier (the value \(100\) originally), and the median is not affected by extreme values while the mean is.