Answer:
The mean, median, and mode are approximately equal.
Step-by-step explanation:
The mean, median, and mode are <em>central tendency measures</em> 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 <em>normal distribution</em>, 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 <em>distribution is not symmetrical</em>, we say it is <em>skewed</em>. The skewness is a measure of the <em>asymmetry</em> of the distribution. In this case, <em>the mean, median and mode are not the same</em>, and we have different possibilities as the mentioned in the question: the mean is less than the median and the mode (<em>negative skew</em>), or greater than them (<em>positive skew</em>), or approximately equal than the median but much greater than the mode (a variation of a <em>positive skew</em> case).
In the case of the normal distribution, the skewness is 0 (zero).
Therefore, in the case of a <em>mound-shaped symmetrical distribution</em>, it resembles the <em>normal distribution</em> and, as a result, it has similar characteristics for the mean, the median, and the mode, that is, <em>they are all approximately equal</em>. So, <em>the </em><em>general</em><em> relationship among the values for these central tendency measures is that they are all approximately equal for mound-shaped symmetrical distributions, </em>considering they have similar characteristics of the <em>normal distribution</em>, which is also a mound-shaped symmetrical distribution (as well as the t-student distribution).
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