Symmetric Distribution vs Bimodal Distribution

Symmetry is a mirror test. Fold the graph at its center and a symmetric distribution has its two halves land on each other, which means a value a given distance above the center is about as common as one the same distance below. For real data the honest word is approximately symmetric; exact symmetry belongs to models such as the normal curve. When a distribution is symmetric, the mean $\bar{x}$ ("x-bar") and the median sit in the same place.

The six values 3, 5, 7, 7, 9, 11 are symmetric about 7: the distances from 7 are $-4$, $-2$, $0$, $0$, $2$, $4$, each matched by its opposite. The mean is $42/6 = 7$ and the median is $(7 + 7)/2 = 7$. Now take 1, 1, 2, 9, 10, 10. That set is just as perfectly symmetric, about 5.5, with two clumps and a hole where its center is.

The second set kills the sentence "it looks symmetric, so use the empirical rule." Symmetric is not the same as normal. For 1, 1, 2, 9, 10, 10 the mean is 5.5 and the sample standard deviation is about 4.59, so one standard deviation either side spans roughly 0.91 to 10.09 and captures all six values, 100 percent rather than the 68 percent the empirical rule would predict. That rule needs the bell shape, and symmetry alone does not supply it.

The implication runs one way only. Symmetric gets you the mean equal to the median; equal centers do not get you symmetry. For 1, 2, 4, 4, 9 the mean and the median are both 4, and yet the largest value sits 5 above that center while the smallest sits only 3 below. Read symmetry off a graph, then check the centers, not the reverse.

Judge symmetry from a display with equal-width bins, and remember that a dozen values wobble too much for the word to be more than a description. Ask whether the departure from a mirror image is bigger than the bumpiness that sample size would produce anyway. Shape vocabulary is Unit 1 topic 1.6.

Bimodal describes the shape of a graph: two prominent high regions with a real dip between them. Prominent is the operative word, because real data are bumpy and nearly every histogram carries small wiggles that are not peaks. The Fall 2026 course fixes the vocabulary in topic 1.6, where one prominent peak is unimodal, two are bimodal, and frequencies that are all about the same with no prominent peak are approximately uniform.

Twelve quiz scores out of 10: 3, 3, 4, 4, 4, 5, 8, 9, 9, 9, 10, 10. A dotplot shows one clump in the low scores and another in the high ones with nothing between. The mean is $78/12 = 6.5$ and the median is $(5 + 8)/2 = 6.5$ as well, so both land in the empty middle. No student scored 6, 6.5, or 7. A single center is the one number these data most clearly reject.

The peaks do not have to be the same height. Students see a histogram with one tall hump and one shorter one, decide "there is only one mode," and call it unimodal. A second peak qualifies when it is a clear local high separated from the first by a genuine dip, whatever its height. Requiring equal heights would make bimodality almost unobservable, since two peaks in real data are never exactly level.

Bimodality is a judgment made from a graph, and the graph can be tuned. Too many narrow bins turn ordinary sampling noise into a row of spikes; too few merge two real groups into one hump. Look at the shape at more than one bin width before committing to the word. A boxplot is no help at all here, since it is drawn from five numbers that carry nothing about peaks, so a bimodal set and a single-peaked set can produce identical boxplots.

When two peaks do show up, ask which two groups got mixed together, then describe them separately rather than averaging them into a population that does not exist.