Weighted arithmetic mean (${\bar{x}}_w$)

The weighted arithmetic mean combines values according to how much each contributes to the quantity being summarized. For a finite list of real numbers $x_i$ and nonnegative weights $w_i$ with $W={\sum }_i{w}_i>0$, defineZero weights ignore an entry; at least one weight must be positive. Multiplying every weight by the same positive number leaves the result unchanged. Writing $p_i=w_i/W$ makes the weighted mean a convex combination, since $p_i\ge 0$ and ${\sum }_i{p}_i=1$. It therefore lies between the smallest and largest input values. ^[ref]

For example, give the values $90$ and $60$ weights $2$ and $8$:The larger weight pulls the result closer to $60$. The balancing identityexplains a geometric interpretation: if the weights were masses placed at positions $x_i$, the weighted mean would be their center of mass. Here $8(66-60)=2(90-66)=48$, so the two sides balance even though their distances from the mean differ.

Weights need an interpretation. They can represent numbers of observations, masses, or probabilities in a finite probability distribution. These meanings lead to the same formula, while answering different questions. For instance, combining means of groups uses group sizes as weights to recover the mean over individual observations. For a function $f$ and nonnegative weight function $w$, the analogous integral is $\int wf/\int w$, provided $wf$ is integrable and the total weight $\int w$ is positive and finite.

Suppose disjoint nonempty groups have sizes $n_j$ and arithmetic means ${\bar{x}}_j$. The arithmetic mean of all observations isEach group contributes the sum $n_j{\bar{x}}_j$; add those totals before dividing by the overall count. This is exactly a weighted mean of group means with weights equal to group sizes. ^[ref]

Take a synthetic example with two observations equal to $90$ and eight equal to $60$. The two group means are $90$ and $60$, and the combined mean is $66$. Giving the two groups equal weights instead produces $(90+60)/2=75$, which summarizes the two group means with equal influence. To summarize all ten individuals with equal influence, use their group counts.

The group means and counts determine the combined arithmetic mean, but do not generally determine the combined variance. Variation within the groups is additional information.

For a finite list of real numbers $x_i$ and nonnegative weights of positive total $W$, the weighted mean uniquely minimizes the weighted sum of squared deviationsExpanding around ${\bar{x}}_w$ and using ${\sum }_i{w}_i(x_i-{\bar{x}}_w)=0$ givesEvery displacement from the weighted mean increases the objective. This is the simplest one-parameter weighted least squares problem: represent all entries by one common value.

For the weighted values $90$ and $60$ with weights $2$ and $8$,Thus $S(66)=1440$, while $S(75)=2250$. The minimum is defined by this squared-error criterion; a different loss can select a different summary, such as a median under unweighted absolute deviations.

For a nonempty finite list of real numbers $x_1,\dots ,x_n$, the arithmetic mean isIt is the weighted mean with equal weights, giving each observation equal influence. For $(2,4,9)$, the sum is $15$ and the arithmetic mean is $5$. Replacing every value by $5$ retains the same total. The residuals $-3,-1,4$ sum to zero, so the mean is also the balancing point of these equally weighted observations.

The arithmetic mean minimizes the sum of squared deviations by the weighted mean minimizes squared deviations identity with all weights one. A large input can consequently have a substantial effect: replacing $9$ by $90$ changes the mean to $32$, while the median remains $4$. The choice of summary depends on what aspect of the data matters.

The harmonic mean of a nonempty finite list of positive real numbers is the multiplicative inverse of the arithmetic mean of their inverses:For two inputs, $H=2x_1{x}_2/(x_1+x_2)$. Replacing each input by $H$ preserves the sum of inverses, making it useful when a fixed amount divided by a rate determines a cost or duration. This positive-input definition avoids division by zero; negative inputs need not retain the usual averaging or ordering properties. ^[ref].

Travel two legs of equal length $d$ at positive speeds $v_1,v_2$, with no stops. Their travel times are $d/v_1$ and $d/v_2$, so the average speed isFor two $60\mathrm{km}$ legs at $30$ and $60km/h$, the times are two hours and one hour. The average speed is $120/3=40km/h$. The same speeds maintained for one hour each instead cover $90\mathrm{km}$ in two hours, giving the arithmetic mean $45km/h$. A slower equal-distance leg occupies more time, so the data being averaged must include what is held equal. This is total distance traveled divided by elapsed time; a return journey can have zero average velocity while keeping a positive average speed. ^[ref].

The arithmetic-geometric mean inequality applied to the inverses gives $H\le G$; together with $G\le A$ it yields the comparison in geometric mean. For two inputs the arithmetic–harmonic gap has the exact formFor $30$ and $60$ the gap is five. With one positive input held fixed and another approaching zero, the harmonic mean approaches zero too: a very slow fixed-distance leg dominates the travel time. Unequal leg lengths instead require a weighted harmonic mean.

For positive real numbers $x_i$ and nonnegative weights with ${\sum }_i{w}_i=1$, the weighted harmonic mean isEqual weights give the harmonic mean, while a zero weight excludes that input. For travel at positive constant speeds $v_i$ over positive leg lengths $d_i$, the weights are fractions of total distance:For $30\mathrm{km}$ at $30km/h$ and $90\mathrm{km}$ at $60km/h$, the weights are $1/4,3/4$, elapsed time is $5/2$ hours, and the average speed is $48km/h$. Weighting the speeds by their time fractions instead gives the same result as a weighted arithmetic mean: $(2/5)30+(3/5)60=48$. The two weight systems describe the same journey in different coordinates and must not be interchanged. ^[ref].