In earlier classes, we computed the arithmetic mean by adding all data values and dividing by the total count:
However, the simple mean assumes that every observation holds equal importance or size. In reality, when combining groups of different sizes, or components of different values, taking an unweighted average produces completely misleading and mathematically invalid conclusions.
When each quantity $x_i$ has an associated importance, frequency, or size called its weight ($w_i$), the Weighted Mean ($\bar{x}_w$) is defined as:
Here, $w_i$ can represent: number of items, volume, credit hours, mass, or survey importance.
A classic application of weighted averages in chemistry, goldsmithing, and commerce is combining solutions or alloys of different strengths.
Pure gold is soft and unsuitable for intricate jewellery, so it is alloyed with copper, silver, or zinc. The purity is measured in Karats (out of 24 parts):
| Karat Rating | Pure Gold Fraction | Purity Percentage (%) | Fineness (Parts per 1000) |
|---|---|---|---|
| 24 Karat (24K) | $\frac{24}{24} = 1.0$ | $100\%$ pure gold | $999$ |
| 22 Karat (22K) | $\frac{22}{24} = \frac{11}{12}$ | $\approx 91.67\%$ | $916$ (BIS 916 Hallmark) |
| 18 Karat (18K) | $\frac{18}{24} = \frac{3}{4}$ | $75.0\%$ | $750$ |
| 14 Karat (14K) | $\frac{14}{24} = \frac{7}{12}$ | $\approx 58.33\%$ | $585$ |
If $V_1$ litres of solution with concentration $C_1\%$ is mixed with $V_2$ litres of solution with concentration $C_2\%$, the combined concentration $C_{\text{mix}}$ is:
To communicate how quantities combine across groups, modern statistical reporting uses three primary bar chart architectures:
| Chart Type | What it Displays Best | Key Analytical Insight |
|---|---|---|
| Clustered Bar Chart | Side-by-side discrete values of multiple sub-categories across groups. | Direct comparison between individual components (e.g. Boys vs Girls score in each grade). |
| Stacked Bar Chart | Displays total volume/magnitude AND the fractional breakdown of parts. | Both the overall sum and the contribution of parts are visible simultaneously. |
| 100% Stacked Bar Chart | Normalizes every bar to a uniform height of $100\%$. | Focuses strictly on relative percentage shares and proportions, ignoring absolute group sizes. |
Problem: A jeweller melts together $60\text{ g}$ of 22-Karat gold and $40\text{ g}$ of 18-Karat gold. Determine:
Solution:
Step 1: Identify weights and ratings:
Mass $1$ ($w_1$) = $60\text{ g}$, Rating $K_1 = 22\text{K}$
Mass $2$ ($w_2$) = $40\text{ g}$, Rating $K_2 = 18\text{K}$
Total Mass $= w_1 + w_2 = 60 + 40 = 100\text{ g}$.
Step 2: Compute resulting Karat:
$$K_{\text{mix}} = \frac{w_1 K_1 + w_2 K_2}{w_1 + w_2} = \frac{(60 \times 22) + (40 \times 18)}{100} = \frac{1320 + 720}{100} = \frac{2040}{100} = 20.4\text{ Karat}$$
Step 3: Mass of pure gold:
Pure gold fraction $= \frac{20.4}{24} = 0.85 = 85\%$.
$$\text{Mass of pure gold} = 100\text{ g} \times 0.85 = 85\text{ grams}$$
Step 4: Percentage purity:
$$\text{Percentage Purity} = 85\%$$
Problem: An evaluation system weights assignments at $20\%$, midterm tests at $30\%$, and the final board exam at $50\%$. A student scores $85$ in assignments, $70$ in the midterm, and $92$ in the final exam. Calculate the student's final weighted aggregate grade.
Solution:
Weights: $w_1 = 20$, $w_2 = 30$, $w_3 = 50$ (Sum of weights $\sum w_i = 100$).
Scores: $x_1 = 85$, $x_2 = 70$, $x_3 = 92$.
Applying the weighted mean formula: $$\bar{x}_w = \frac{(20 \times 85) + (30 \times 70) + (50 \times 92)}{20 + 30 + 50}$$ $$\bar{x}_w = \frac{1700 + 2100 + 4600}{100} = \frac{8400}{100} = 84.0$$ Final Grade: $84.0\%$ (Notice: A simple unweighted mean would have been $\frac{85+70+92}{3} = 82.33\%$, which penalizes the student's strong final exam performance!).
Problem: How many litres of a $30\%$ acid solution must be mixed with $40\text{ litres}$ of a $10\%$ acid solution to produce a mixture that is exactly $18\%$ acid?
Solution:
Let the volume of the $30\%$ solution required be $x\text{ litres}$.
Given: $V_2 = 40\text{ litres}$, $C_2 = 10\%$, and target $C_{\text{mix}} = 18\%$.
Setting up the mixture formula: $$C_{\text{mix}} = \frac{V_1 C_1 + V_2 C_2}{V_1 + V_2}$$ $$18 = \frac{(x \times 30) + (40 \times 10)}{x + 40}$$ $$18(x + 40) = 30x + 400$$ $$18x + 720 = 30x + 400$$ $$720 - 400 = 30x - 18x$$ $$320 = 12x \implies x = \frac{320}{12} = \frac{80}{3} \approx 26.67\text{ litres}$$
Answer: Exactly $26.67\text{ litres}$ (or $26\frac{2}{3}\text{ L}$) of $30\%$ acid solution is needed.
VARDAAN COMET