Question

here is a data set: 1 2 3 3 4 4 4 4 4 5 5 6 7
a. what happens to the mean and standard deviation of the data set when the 7 is changed to a 70?
b. for the data set with the value of 70, why would the median be a better choice for the measure of center than the mean?

Explanation:

Step1: Calculate original mean

The original data set is \(1, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 7\). The sum of the data is \(1 + 2+3 + 3+4 + 4+4 + 4+5 + 5+6 + 7=48\), and there are \(n = 12\) data - points. The original mean \(\bar{x}_1=\frac{48}{12}=4\).

Step2: Calculate new mean

When the \(7\) is changed to \(70\), the new sum is \(48 - 7+70 = 111\). The new mean \(\bar{x}_2=\frac{111}{12}=9.25\). So the mean increases.

Step3: Analyze standard - deviation change

The standard deviation measures the spread of the data. The value of \(7\) is replaced by \(70\), which is a much larger value. This makes the data more spread out. So the standard deviation increases.

Step4: Explain why median is better

The original data set has \(n = 12\) values. The median is the average of the \(6^{th}\) and \(7^{th}\) ordered values, which is \(\frac{4 + 4}{2}=4\). After changing \(7\) to \(70\), the median is still \(\frac{4 + 4}{2}=4\) (since the middle - two values don't change). The mean is affected by the outlier (\(70\)), while the median is not. An outlier can pull the mean in its direction, but the median only depends on the middle values of the ordered data set. So, when there is an outlier, the median is a more robust measure of the center of the data set.

Answer:

a. The mean increases and the standard deviation increases.
b. The median is not affected by the outlier (\(70\)) while the mean is pulled in the direction of the outlier, so the median is a more robust measure of the center.