When a distribution is mound-shaped symmetrical, what is the general relationship among the values of the mean, median, and mode? The mean is less than the median and mode. The mean is greater than the median and mode. The median and mean are approximately equal, but much greater than the mode. The mean, median, and mode are approximately equal.

The mean, median, and mode are approximately equal.

Step-by-step explanation:

The mean, median, and mode are central tendency measures in a distribution. That is, they are measures that correspond to a value that represents, roughly speaking, "the center" of the data distribution.

In the case of a normal distribution, these measures are located at the same point (i.e., mean = median = mode) and the values for this type of distribution are symmetrically distributed above and below the mean (mean = median = mode).

When a distribution is not symmetrical, we say it is skewed. The skewness is a measure of the asymmetry of the distribution. In this case, the mean, median and mode are not the same, and we have different possibilities as the mentioned in the question: the mean is less than the median and the mode (negative skew), or greater than them (positive skew), or approximately equal than the median but much greater than the mode (a variation of a positive skew case).  

In the case of the normal distribution, the skewness is 0 (zero).

Therefore, in the case of a mound-shaped symmetrical distribution, it resembles the normal distribution and, as a result, it has similar characteristics for the mean, the median, and the mode, that is, they are all approximately equal. So, the general relationship among the values for these central tendency measures is that they are all approximately equal for mound-shaped symmetrical distributions, considering they have similar characteristics of the normal distribution, which is also a mound-shaped symmetrical distribution (as well as the t-student distribution).