Question

consider a set (a) of numbers ({a_1,a_2,a_3,a_4}), and a set (b) of numbers ({b_1,b_2,b_3,b_4}). the numbers in each set are distinct and ordered such that (a_n < a_{n + 1}) for set (a) and (b_n < b_{n+1}) for set (b). suppose the values in set (a) are identical to the values in set (b), with the exception of the number with the least value in each set. the number with the least value in set (a) is greater than the number with the least value in set (b). the value of which of the following measures must be greater for set (a) than for set (b)?
mode only
mean and median only
mean, median, and mode
mean only
median only
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Explanation:

Step1: Analyze mode

Since the sets have distinct numbers, there is no mode (as no number repeats). So mode is not greater for set A.

Step2: Analyze median

Set \(A=\{a_1,a_2,a_3,a_4\}\) and \(B = \{b_1,b_2,b_3,b_4\}\) with \(a_n < a_{n + 1}\) and \(b_n < b_{n+1}\). The median of a set of 4 - ordered numbers is the average of the 2nd and 3rd numbers. Since the non - least numbers are the same in both sets, the medians of \(A\) and \(B\) are equal.

Step3: Analyze mean

Let \(a_1>b_1\) and \(a_2 = b_2\), \(a_3 = b_3\), \(a_4 = b_4\). The mean of set \(A\) is \(\bar{a}=\frac{a_1 + a_2+a_3 + a_4}{4}\) and the mean of set \(B\) is \(\bar{b}=\frac{b_1 + b_2+b_3 + b_4}{4}\). Since \(a_1>b_1\) and \(a_2 = b_2\), \(a_3 = b_3\), \(a_4 = b_4\), we have \(\bar{a}>\bar{b}\).

Answer:

Mean only