Weighted average and mean concepts – UGC NET Question

Statements A, B and C give the standard description and formula for arithmetic and weighted means. A is correct because the simple mean assigns equal weight to each observation. B and C correctly explain that weighted means use different weights and are computed as the weighted sum divided by the total weight. Statement E is true because if all weights are equal, the weighted mean collapses to the ordinary mean. Statement D is false since a weighted mean, like the arithmetic mean, can lie between observed values and need not coincide with any one observation, so the correct set is A, B, C and E only.

Option A:
Option A is incomplete because it omits C and E and so does not show how weighted means are actually computed or how they relate back to the arithmetic mean when weights are equal. It therefore does not represent all correct statements.

Option B:
Option B is also incomplete as it does not include E, ignoring the simplifying case in which weights are equal, even though it correctly includes A, B and C. The missing link to the arithmetic mean makes it insufficient.

Option C:
Option C is wrong because, although it includes B, C and E, it omits A, which states the key contrast between arithmetic and weighted means, and still does not highlight the equal-weight case as a special instance in context.

Option D:
Option D is correct because it brings together the accurate description of both kinds of means and the proper computation rule, while explicitly excluding D, which incorrectly insists that the mean must be one of the data points. This aligns with the conceptual treatment of averages in NET questions.