This question asks us to identify the measure of central tendency from which the sum of the absolute deviations of the data points is the smallest. In statistics, "ignoring signs" means we are considering the absolute value of the deviations.
A fundamental property related to the Median is that it uniquely minimizes the sum of the absolute deviations for a given dataset. This means if you calculate the difference between each data point and the Median, take the absolute value of each difference, and then sum them up, you will get the smallest possible sum compared to using any other single value from the dataset or outside of it.
For a dataset denoted as {$x_1, x_2, ..., x_n$}, the value $M$ that minimizes the sum $S = \sum_{i=1}^{n} |x_i - M|$ is the Median of the dataset.
Let's look at why the other options are not correct:
Based on the properties of measures of central tendency, the sum of the absolute deviations (i.e., deviations ignoring signs) of the items from the Median is the least.
Let $X_1, X_2, X_3$ be a random sample of size 3 from an absolutely continuous distribution that is symmetric about 0. For $i=1,2,3$, let $R_i$ denote the rank of $|X_i|$ among $|X_1|, |X_2|$ and $|X_3|$.
If $T^+ = \sum_{i=1, X_i>0}^3 R_i$
is the Willcoxon signed-rank statistic, then which of the following statements are true?,
The mean marks of the following distribution is:
| Marks Obtained | No. of Students |
| 81 | 15 |
| 35 | 4 |
| 73 | 3 |
| 56 | 16 |