Weighted Mean vs Mean

A weighted mean lets some values count for more than others. It is $\bar{x}_w = \frac{\sum w_i x_i}{\sum w_i}$, said as weighted x-bar, where $w_i$ is the weight attached to the value $x_i$. Set every weight to 1 and the formula collapses to the ordinary mean, so the plain mean is not a different kind of average. It is the special case in which the values genuinely deserve equal say.

A shop sells 120 mugs at a mean price of 8 dollars and 30 teapots at a mean of 28 dollars. The mean price across all 150 items is $\frac{120(8) + 30(28)}{150} = \frac{1800}{150} = 12$ dollars. The weights are the counts, and 12 is simply the total money divided by the total items.

"Average the two means: $(8 + 28)/2 = 18$." That is 6 dollars too high, and the 6 splits unevenly. Promoting the teapots from 20 percent of the items to 50 percent of the say contributes $(0.50 - 0.20)(28 - 12) = 4.80$ of it, while demoting the mugs from 80 percent to 50 percent contributes $(0.50 - 0.80)(8 - 12) = 1.20$. Four fifths of the error comes from over-weighting the small expensive group and only one fifth from under-weighting the large one.

With non-negative weights that are not all zero, the weighted mean always lands between the smallest and largest value being averaged, which is a quick sanity check: 12 sits between 8 and 28. What the formula does not do is combine spread. Averaging two group standard deviations the same way is wrong, and rebuilding a combined one needs the original values or their sum of squares.

The Fall 2026 course does not name the weighted mean, but topic 2.9 contains one. The expected value of a discrete random variable, $\mu_X = \sum x_i P(x_i)$, is a weighted mean whose weights are probabilities and already add to 1, which is why the denominator disappears there.

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.