Question

the same value, x, is both the mean and the median of a set of data. two values, a and b, are added to the set of data. a is 25 less than x, and b is 100 greater than x. given this information, which of the following is true? a the mean and median are unchanged. b both the mean and median increase. c both the mean and median decrease. d the mean is now greater than the median. e the median is now greater than the mean.

Explanation:

Step1: Analyze the effect on the mean

Let the original sum of data be \(S\) and the number of data points be \(n\). So the original mean \(\bar{x}=\frac{S}{n}=x\). After adding \(a=x - 25\) and \(b=x + 100\), the new sum is \(S'=S+(x - 25)+(x + 100)=S + 2x+75\), and the new number of data points is \(n'=n + 2\). The new mean \(\bar{x}'=\frac{S'}{n'}=\frac{S + 2x+75}{n + 2}\). Since \(\frac{S}{n}=x\), we can rewrite \(\bar{x}'=\frac{nx+2x + 75}{n + 2}=\frac{(n + 2)x+75}{n + 2}=x+\frac{75}{n + 2}\). So the mean increases.

Step2: Analyze the effect on the median

Since \(a=x - 25\) and \(b=x + 100\), when we order the data set (after adding \(a\) and \(b\)), the middle - value (median) will be at least \(x\) (if \(n\) was odd, assume the original data set was ordered. The new data set's median will be the value in the \(\frac{n + 3}{2}\) - th position (if \(n\) was odd) or the average of the \(\frac{n + 2}{2}\) - th and \(\frac{n+4}{2}\) - th positions (if \(n\) was even). In either case, since \(b=x + 100\gt x\) and \(a=x - 25\lt x\), the median will be greater than or equal to \(x\) (in fact, it will be greater than \(x\) because of the addition of \(b\)).

Answer:

B. Both the mean and median increase.