Match the LIST-I with LIST-II LIST-I LIST-II A. Median of 4, 4, 5, 7, 6, 7, 7,…

Concept: The three measures of central tendency describe the centre of a data set in three different ways — the median is the middle value once the data are arranged in ascending order, the mean is the total of all the values divided by how many values there are, and the mode is the value that occurs with the highest frequency.

For an ungrouped list of n values with n odd, the median is the term standing at position (n + 1)/2 in the sorted list. When a data set is written in terms of an unknown, the condition on its mean becomes a linear equation in that unknown.

Applying the concept to each List-I entry

1. A — Median of 4, 4, 5, 7, 6, 7, 7, 12, 3: arranging the values in ascending order gives 3, 4, 4, 5, 6, 7, 7, 7, 12. Here n = 9, which is odd, so the median is the (9 + 1)/2 = 5th term of the sorted list, which is 6.

2. B — Mean of 7, 5, 9, 8, 15, 3, 8, 9: the total is 7 + 5 + 9 + 8 + 15 + 3 + 8 + 9 = 64 and there are n = 8 values, so the mean is 64 ÷ 8 = 8.

3. C — Mode of 2, 3, 7, 1, 3, 2, 3: counting the frequencies, 1 occurs once, 2 occurs twice, 3 occurs three times and 7 occurs once. The highest frequency belongs to the value 3, so the mode is 3.

4. D — Value of x, if the mean of x, x+2, x+4, x+6, x+8 is 11: the total of the five terms is 5x + 20, so the mean is (5x + 20)/5 = x + 4. Setting x + 4 = 11 gives x = 7.

Cross-check and the resulting matching

Substituting x = 7 into the five-term expression of List-I entry D rebuilds that list as 7, 9, 11, 13, 15, whose total 55 divided by 5 is 11, which confirms the value obtained for x. Note also that the median of the nine-value list is read off the sorted arrangement, not the order in which the values were written down — reading the middle of the unsorted list would give 6 only by coincidence and is not a valid method.

Reading the table row by row, the matching is A-III, B-IV, C-I, D-II.