Resistant Statistic vs Mean

A statistic is resistant when moving a few observations arbitrarily far barely changes its value. Resistance belongs to the statistic, not to the data: the same numbers can carry a resistant summary and a nonresistant one at once. The median, the quartiles, and the IQR are resistant. The mean $\bar{x}$ (x-bar), the standard deviation $s$, and the range are not.

Take 21, 23, 24, 26, 28, 30, 31, 33, 36. The mean is $252/9 = 28$, the median is the fifth value, also 28, and $s = 4.95$. Change the 36 to 96. The median stays at 28, $Q_1$ at 23.5 and $Q_3$ at 32, so the IQR is still 8.5, while the mean climbs to $312/9 = 34.67$ and $s$ to 23.33, almost five times over. Only the summaries that read magnitudes moved.

"The median is resistant, so it is the better measure of center" does not follow. Resistance is stability, not accuracy: a resistant summary buys it by refusing to look at how far the extreme values actually are. On a roughly symmetric distribution with no outliers, the mean uses more of the information and is what t procedures are built on. It is a reason to switch summaries when the shape calls for it, not a ranking.

Resistant is not a spare word for robust. Resistant describes a statistic: how much its value moves when a few observations are extreme. Robust describes a procedure: whether it still performs as advertised when a condition is only approximately met, the sense in which t procedures tolerate mild non-normality. AP does not keep them apart: topic 1.7 writes "a resistant (or robust) measure of center", so on the exam the two words point at one idea.

Resistance has a limit. A value already beyond a quartile can move further out without touching $Q_1$ or $Q_3$, but one pulled from inside a half does move it: in 1, 2, 3, 4, 5, 6, 7, 8 the third quartile is 6.5, and changing the 5 to 100 raises it to 7.5. Once more than a quarter of the data sits far out on one side, that quartile moves too.

The sample mean is $\bar{x} = \frac{\sum x_i}{n}$ ("x-bar"), the sum of the values divided by how many there are. Compute the same thing for a whole population and it is called $\mu$ ("mu"). The arithmetic is identical; what differs is whether the numbers came from a sample or from everyone. Balance point is the exact version of the intuition: the deviations from the mean always add to zero, for every data set without exception.

For 2, 5, 5, 8, 10 the sum is 30, so $\bar{x} = 30/5 = 6$. The deviations are $-4$, $-1$, $-1$, $2$, and $4$, and they cancel to 0: the pull from below the mean exactly matches the pull from above. Now change the 10 to 40. The sum becomes 60 and the mean becomes 12, while the median has not budged from 5. Every value contributes its full size, so a single value moving 30 units drags the mean $30/5 = 6$ units after it.

Here is the sentence to drop: "the mean is the value half the data fall below." That is the median. In 2, 5, 5, 8, 40 the mean is 12 and four of the five values sit below it, 80 percent of the data. Nothing makes the mean split a data set evenly, and the more skewed the data, the further from an even split it lands.

The everyday version of the same error is averaging averages. A class of 10 students averages 75 and a class of 30 averages 85, so the combined average is 80. It is not. The 30 students count three times as heavily, and the real figure is $\frac{10(75) + 30(85)}{40} = \frac{3300}{40} = 82.5$. Means combine correctly only through their totals, which is exactly what a weighted mean does.

The mean also needs values it makes sense to add. Averaging jersey numbers or zip codes is arithmetic with nothing behind it. And the mean need not be a value the variable can take: 1.75 siblings is an ordinary answer. Summary statistics for one quantitative variable are Unit 1 topic 1.7.