02 // Visual Math & Finance Solversx̄_w = (w₁x₁ + w₂x₂ + w₃x₃) / (w₁ + w₂ + w₃)
Visual Weighted Average & Center-of-Mass Fulcrum Beam
The weighted arithmetic mean $\bar{x}_w = \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i}$ is mathematically identical to the physical center of mass where net rotational torque $\sum w_i(x_i - \bar{x}_w) = 0$ balances a rigid lever.
x̄_w = (w₁x₁ + w₂x₂ + w₃x₃) / (w₁ + w₂ + w₃)Inputs & Parameters (6)
pt
0pt±1100pt
%
1%±1100%
pt
0pt±1100pt
%
1%±1100%
pt
0pt±1100pt
%
1%±1100%
Weighted Mean (Fulcrum)
82.55
∑(wᵢxᵢ) = 8,255 / ∑w = 100
Unweighted Simple Mean
80.33
Weighting shift Δ = +2.22
Normalized Shares
25 / 35 / 40%
Relative mass distribution on beam
Balanced Torque (∑τ = 0)
±458
Equal counter-clockwise & clockwise moments
20-Row Comparison Table
Compare across:
| # | Value 1 (x₁) (pt) | Weighted Mean (x̄_w) | Simple Mean | Weight Shift Δ |
|---|---|---|---|---|
| 01 | 0 pt | 66.3 | 58.67 | +7.63 |
| 02 | 5 pt | 67.55 | 60.33 | +7.22 |
| 03 | 11 pt | 69.05 | 62.33 | +6.72 |
| 04 | 16 pt | 70.3 | 64 | +6.3 |
| 05 | 21 pt | 71.55 | 65.67 | +5.88 |
| 06 | 26 pt | 72.8 | 67.33 | +5.47 |
| 07 | 32 pt | 74.3 | 69.33 | +4.97 |
| 08 | 37 pt | 75.55 | 71 | +4.55 |
| 09 | 42 pt | 76.8 | 72.67 | +4.13 |
| 10 | 47 pt | 78.05 | 74.33 | +3.72 |
| 11 | 53 pt | 79.55 | 76.33 | +3.22 |
| 12 | 58 pt | 80.8 | 78 | +2.8 |
| 13 | 63 pt | 82.05 | 79.67 | +2.38 |
| 14 | 68 pt | 83.3 | 81.33 | +1.97 |
| 15 | 74 pt | 84.8 | 83.33 | +1.47 |
| 16 | 79 pt | 86.05 | 85 | +1.05 |
| 17 | 84 pt | 87.3 | 86.67 | +0.63 |
| 18 | 89 pt | 88.55 | 88.33 | +0.22 |
| 19 | 95 pt | 90.05 | 90.33 | -0.28 |
| 20 | 100 pt | 91.3 | 92 | -0.7 |
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Key Analytical Takeaways
- 01Increasing a category's weight $w_i$ pulls the triangular fulcrum toward $x_i$ just like adding mass to one side of a seesaw.
- 02When all weights are equal ($w_1 = w_2 = w_3$), the weighted fulcrum collapses to the simple arithmetic mean $(x_1 + x_2 + x_3)/3$.
- 03Notice how the clockwise torque on the right of the fulcrum exactly cancels the counter-clockwise torque on the left.
Frequently Asked Questions
Do the weights in a weighted average have to add up to 100%?+
No. The formula automatically normalizes each weight by dividing by the sum of all weights (w₁ + w₂ + w₃), so raw credit hours, share counts, or percentages all work directly.
Why is the weighted mean equivalent to the center of gravity?+
In statics, a horizontal beam balances at point x̄ when total torque Σ w_i(x_i − x̄) = 0. Solving for x̄ gives Σ(w_i·x_i) / Σ(w_i).
What happens if the weights do not sum to 100% (or 1.0)?+
The weighted average formula automatically normalizes arbitrary positive weights by dividing ∑(wᵢ·xᵢ) by the total weight sum ∑wᵢ, so credit hours (3, 4, 3) work without manual scaling.