Weighted percentages and data cases

What you will learn

A group rate describes successes, defects or ingredient mass relative to that group's own base. To combine groups, return to the underlying amounts. This lesson builds that method, shows when averaging rates is justified, and connects a table, a pie chart and count bars without silently changing the denominator.

You should be comfortable finding a percentage of a whole and distinguishing a percentage-point difference from relative percentage change. Throughout the examples, the given counts and percentages are exact unless a question says otherwise.

1 Recover amounts before combining rates

Suppose group i has a positive base wᵢ and a rate rᵢ written as a decimal. Its numerator is wᵢrᵢ. Therefore:

Overall decimal rate = (w₁r₁ + w₂r₂ + …)/(w₁ + w₂ + …).

Multiply the result by 100 to express it as a percentage. Alternatively, if Pᵢ is the percentage number such as 80 rather than 0.80, the overall percentage number is (w₁P₁ + w₂P₂ + …)/(w₁ + w₂ + …). Do not divide by 100 twice or mix the two conventions.

The weights are the denominators: people for a person-based success rate, units inspected for a defect rate, and total mass for a mass percentage. Do not multiply a success count by group size again. The success count has already accounted for the group size.

2 Check that aggregation means what you think it means

The groups must use the same rate definition and compatible units. Add disjoint groups, or explicitly account for overlap; counting the same person twice is not a combined rate over distinct people. If the observational unit is a trial and the question asks about all trials, repeated trials can be separate observations. Name the unit.

The total denominator must be positive. An empty group's own rate is undefined, not automatically zero. For a genuine part-within-whole rate, the numerator is between zero and its matching denominator, so the percentage lies from 0% to 100%. Relative increases are a different kind of comparison and can exceed 100%.

Use unrounded amounts when they are supplied. Rounding group rates early can change the final answer; a rounded displayed rate may not determine an exact whole-number numerator. The practice here supplies exact data so that ambiguity does not arise.

3 When can you average the rates directly?

Equal denominator weights guarantee the ordinary arithmetic mean for arbitrary group rates. If two equal-size groups have rates r and s, their combined rate is (wr + ws)/(2w) = (r + s)/2.

Equal weights are sufficient, but not necessary in every case. If both groups have the same rate, say 30%, every weighted combination of them also has rate 30%, whatever their positive sizes. With more groups, other coincidences can occur: rates 20%, 50%, 80% with sizes 10, 20, 10 give (2 + 10 + 8)/40 = 50%, also their simple mean. These exceptions do not make unweighted averaging a safe default. Recover amounts or justify the weights.

4 Bounds and missing information

An overall rate with positive weights lies between the smallest and largest group rates. It is strictly between the extremes when the groups include genuinely different rates and all their weights are positive. If every rate is the same, the combined rate equals it. The bound is a check, not a substitute for an exact calculation.

Knowing only two different group rates does not determine the overall rate. For example, 20% and 80% can combine to 50% with equal sizes, but to 65% when the 80% group is three times as large. The missing information is the relative group size; a ratio of sizes is enough even when absolute counts are unknown.

5 Changing composition can change the overall rate

A total can improve because the higher-rate group receives more weight, even if neither group's own rate improves. Compare both the within-group rates and the group sizes before drawing a conclusion. An aggregate change alone is not proof of individual improvement, nor does it by itself establish a cause.

Also separate counts from rates across time. More successful units can accompany a lower success rate when the number of units attempted grows faster. Calculate the two ratios before comparing them; use percentage points for their difference.

6 Read the display and keep its denominator

A table can show counts, rates or both. Read each column heading: “unused as a percentage of its own allocation” has a different denominator from “share of all allocated tickets”. Totals of counts can be added; percentages from different group bases must not be added as though they partitioned one whole.

A pie chart divides one specified whole into disjoint, exhaustive shares. Sector angle = share × 360°. If a sector is 40% of all tickets, its angle is 144° and its count is 0.40 × the total. The number 40 does not mean that 40% of the sector's own tickets were unused.

A count bar starts at zero and its height represents the displayed number of items. In a stacked count bar, segment counts add to the bar's total. A taller unused segment means more unused tickets, not necessarily a higher within-group unused percentage. Divide by the appropriate bar total to get that rate. A 100% stacked bar, in contrast, gives equal-height totals and cannot alone show unequal group sizes.

To transform representations: read the whole and shares from the pie; recover group counts; apply each within-group rate; put those counts in a table or stacked count bars; then check that every group and grand total still match. The worked ticket case below performs each step.

A four-step method

1. Identify a consistent numerator and its matching base for every group.
2. Recover the numerator amounts and put them in common units.
3. Add numerators and bases separately, then divide.
4. Check bounds, total consistency, the display's labels and any change in composition.

Six worked cases

Worked example 1

In one class, 30 students take a test and 80% pass. In another, 70 take it and 60% pass. Find the combined pass percentage.

1. Class 1 contributes 30 × 0.80 = 24 passes. Class 2 contributes 70 × 0.60 = 42.
2. Together there are 24 + 42 = 66 passes among 30 + 70 = 100 candidates. Overall = 66/100 × 100 = 66%.
3. The larger class has the lower rate and receives more weight. An unweighted mean of 80% and 60% would be 70%.

Answer: 66%

Check and trap: 66% lies between 60% and 80% and nearer 60%. Do not multiply the 24 passes by 30 again.

Worked example 2

A factory reports 24 defective units among 600 units in one month and 30 among 1000 the next. Find both monthly rates and the combined rate.

1. First rate = 24/600 × 100 = 4%; second = 30/1000 × 100 = 3%.
2. The defect count rises from 24 to 30, but its rate falls by 1 percentage point because the total inspected increases more.
3. Combined rate = (24 + 30)/(600 + 1000) × 100 = 54/1600 × 100 = 3.375%.

Answer: 4%, 3%; combined 3.375%

Check and trap: 3.375% is between 3% and 4% and closer to the larger batch’s 3%. The simple mean 3.5% assigns equal weight to unequal batches.

Worked example 3

A non-reacting mixture combines 200 g of material containing 30% of an ingredient by mass with 300 g containing 10%. Assume no mass is lost. Find the ingredient percentage.

1. Ingredient mass from the first portion = 200 × 0.30 = 60 g. From the second = 300 × 0.10 = 30 g.
2. Total ingredient = 90 g; total mixture = 200 + 300 = 500 g.
3. Mass percentage = 90/500 × 100 = 18%.

Answer: 18% by mass

Check and trap: The total mass is the weight, not the portion count “two”. The answer is between 10% and 30%. The no-reaction/no-loss assumptions permit adding these masses; no practical mixing is required.

Worked example 4

Two nonempty groups have success rates 60% and 80%. Can their combined success rate be 55%? Explain without choosing group sizes.

1. Let the positive sizes be n and m. Combined decimal rate is (0.60n + 0.80m)/(n + m).
2. Rewrite it as 0.60 + 0.20m/(n + m). Since 0 < m/(n + m) < 1, the result is strictly between 0.60 and 0.80.
3. 55% is below the lower bound, so it is impossible under the complete disjoint-group model.

Answer: No

Check and trap: We can reject 55% without knowing n:m, but cannot identify the exact combined rate without that relative size.

Worked example 5

An event allocates 800 tickets: A gets 40%, B gets 35%, C gets 25%. A leaves 10% of its allocation unused, B 20%, C 5%. How many tickets are used?

One dataset in three forms

Text equivalent: of all 800 allocated tickets, A receives 40%, B 35% and C 25%. The pie angles are 144°, 126° and 90° respectively.

Group	Allocated tickets	Unused as % of own allocation	Unused tickets	Used tickets
A	320	10%	32	288
B	280	20%	56	224
C	200	5%	10	190
Total	800	Do not sum this column	98	702

Text equivalent: A has 288 used plus 32 unused tickets, total 320. B has 224 plus 56, total 280. C has 190 plus 10, total 200. The vertical axis counts tickets from zero in steps of 50; it is not a percent axis. A's taller full bar reflects more allocated tickets. B's unused fraction is 56/280 = 20%, while A's is 32/320 = 10%.

1. Read the whole from the pie: 800 allocated tickets. The shares give A = 0.40 × 800 = 320, B = 280, C = 200. The matching sector angles are 144°, 126° and 90°; their sum is 360°.
2. The unused rates apply within each allocation. Unused counts are 0.10 × 320 = 32, 0.20 × 280 = 56 and 0.05 × 200 = 10.
3. Subtract within each group: used counts are 288, 224 and 190. Put these at the bottom of the stacked bars and the unused counts above them. The three full heights must remain 320, 280 and 200.
4. Total unused = 98 and total used = 702. Overall unused rate = 98/800 × 100 = 12.25%; used rate = 87.75%.

Answer: 702 tickets used

Check and trap: The allocation shares sum to 100%. The within-group unused rates 10%, 20%, 5% do not partition one common whole and must not be added or directly averaged. The count check is 702 + 98 = 800.

Worked example 6

Two lines keep fixed success rates A=90%, B=60%. Last week A handled 100 jobs and B 300; this week A handles 300 and B 100. Find both overall rates and explain the change.

1. Last week: successes = 100 × 0.90 + 300 × 0.60 = 270 of 400, so 67.5%.
2. This week: successes = 300 × 0.90 + 100 × 0.60 = 330 of 400, so 82.5%.
3. The overall rate increases by 82.5 − 67.5 = 15 percentage points. A now handles a larger share, but both line-specific rates are unchanged.

Answer: 67.5% then 82.5%; +15 percentage points

Check and trap: Both overall rates stay between 60% and 90%. The data support a composition explanation; they do not show either line becoming more effective.

Ten fresh practice questions

Try the questions before the solutions. Select all correct options only where instructed. For numeric answers, enter the requested percentage number. All chart facts are also given in the text alternatives.

Practice question 1

Two disjoint inspection teams check different units using the same pass criterion. Team A checks 40 units with 60% passing; B checks 160 with 85% passing. Enter the overall pass percentage.

Practice question 2

Three disjoint batches contain 50, 150 and 300 items. Their acceptance rates under one common criterion are 40%, 60% and 90%. Which combined acceptance percentage is correct?

A. 63⅓%

B. 76%

C. 90%

D. 190%

Practice question 3

For nonempty disjoint groups with the same definition of success, select all true statements.

A. Equal group sizes guarantee that the simple average of group rates equals the combined rate

B. If every group has a 75% rate, the combined rate is 75% even when their sizes differ

C. Unequal group sizes always make the simple average different from the combined rate

D. Success counts should be multiplied by group sizes again before adding

Practice question 4

A table records morning: 200 attempts, 150 successes; evening: 400 attempts, 260 successes. Which statement compares the two periods correctly?

Period	Attempts	Successes
Morning	200	150
Evening	400	260

A. Successes rose and the success rate rose

B. Successes fell but the success rate rose

C. Successes rose by 110 while the success rate fell by 10 percentage points

D. The success rate fell by 10% relative to its old rate

Practice question 5

Two non-overlapping periods record 36 rejected units out of 450, then 18 rejected out of 150. What is the combined rejection percentage? Enter the percentage number.

Period	Inspected units	Rejected units
1	450	36
2	150	18

Practice question 6

Two nonempty disjoint groups have rates 20% and 80%, but neither their sizes nor a size ratio is given. What can be concluded about their combined rate?

A. It must be 50%

B. It is strictly between 20% and 80%, but its exact value is not determined

C. It is at least 80%

D. Even knowing the size ratio would not be enough

Practice question 7

Two lines keep rates A = 60% and B = 80%. In period 1 they handle 75 and 25 jobs; in period 2 they handle 25 and 75 jobs. All jobs are disjoint and counted by the same criterion. Select all true statements.

A. The overall rate is 65% in period 1 and 75% in period 2

B. Both lines improved their own success rates

C. The overall increase is 10 percentage points

D. More weight moved to the higher-rate line, which explains the aggregate change under the stated data

Practice question 8

Use the practice allocation pie and the unused-rate table below. How many tickets allocated to T were used?

Text equivalent: Allocation pie: the whole is 1000 tickets. R receives 20%, S 30%, T 50%. The separate table gives unused rates within each own allocation: R 15%, S 10%, T 4%.

Group	Unused as % of its own allocation
R	15%
S	10%
T	4%

A. 20

B. 480

C. 500

D. 960

Practice question 9

Use the equipment count bars below. Across all three stores, what percentage of units are available rather than on loan? Enter the percentage number.

Text equivalent: Stacked count bars from zero. Store X: 84 on loan and 36 available, total 120. Y: 108 on loan and 72 available, total 180. Z: 108 on loan and 92 available, total 200. Categories are disjoint and exhaustive within each store.

Practice question 10

A table shows Team P: 150 allocated passes, 120 used; Team Q: 450 allocated passes, 180 used. A new pie chart is to show shares of all used passes. Which P and Q percentages belong in that pie?

Team	Allocated passes	Used passes
P	150	120
Q	450	180

A. 25%, 75%

B. 80%, 40%

C. 40%, 60%

D. 20%, 30%

Practice solutions

Solution 1

Answer: 80

Passes = 40 × 0.60 + 160 × 0.85 = 24 + 136 = 160. Total inspected = 200, so 160/200 × 100 = 80%. The simple mean 72.5% ignores B’s fourfold weight.

Solution 2

Answer: B

Accepted counts are 20, 90 and 270; total 380 of 500. The rate is 76%. The largest batch, at 90%, has the greatest weight.

• A: A averages the three rates without their bases.
• B: B adds accepted counts and divides by all items.
• C: C uses only the largest batch rate.
• D: D adds percentages that apply to different groups.

Solution 3

Answer: A, B

Equal sizes give equal denominator weights. Identical rates also remain unchanged by weighting. Thus equal weights are a sufficient condition, not a necessary condition in every individual case. Counts already incorporate their group sizes.

• A: A follows by cancelling the common weight.
• B: B follows by factoring 0.75 out of the total numerator.
• C: C is disproved by B.
• D: D applies the weights twice and changes the meaning of the numerator.

Solution 4

Answer: C

The rates are 150/200 = 75% and 260/400 = 65%. Counts rise by 110 while the rate drops 10 percentage points. Its relative fall is 10/75 × 100 = 13⅓%, not 10%.

• A: A mistakes a count increase for a rate increase.
• B: B reverses both directions.
• C: C compares the matching rates and counts correctly.
• D: D confuses percentage points with relative percent.

Solution 5

Answer: 9

The period rates are 8% and 12%. Add rejected units and inspected units: (36 + 18)/(450 + 150) × 100 = 54/600 × 100 = 9%. It is nearer 8% because the first period has three times the units.

Solution 6

Answer: B

Positive weights with distinct rates give a strict interior bound. Equal weights give 50%; weights 1:3 give 65%, so a unique answer is not determined. A size ratio would fix the weights up to a common factor, which cancels.

• A: A silently assumes equal group sizes.
• B: B states the valid bound and the missing information.
• C: C violates the upper bound.
• D: D ignores cancellation of a common scale factor.

Solution 7

Answer: A, C, D

Period 1 has 45 + 20 = 65 successes of 100. Period 2 has 15 + 60 = 75 of 100. The group rates remain exactly 60% and 80%; only their weights change.

• A: A uses both period-specific weights correctly.
• B: B contradicts the fixed rates given.
• C: C is the difference between 75% and 65%.
• D: D identifies the supported composition mechanism without claiming individual improvement.

Solution 8

Answer: B

The pie shows T receives 50% of all 1000 tickets, so its allocation is 500. The table says 4% of T’s own allocation is unused: 0.04 × 500 = 20. Used = 500 − 20 = 480.

• A: A is the unused count in T.
• B: B uses the two nested bases correctly.
• C: C is T’s allocation before removing unused tickets.
• D: D applies 96% to all 1000 instead of to T’s allocation.

Solution 9

Answer: 40

Read the available segments: 36 + 72 + 92 = 200. Full bar totals are 120, 180 and 200, giving 500 units. Overall available = 200/500 × 100 = 40%. Averaging the three within-store rates 30%, 40% and 46% would weight unequal stores equally.

Solution 10

Answer: C

The new whole is all used passes: 120 + 180 = 300. P’s share is 120/300 = 40% and Q’s is 180/300 = 60%. The denominator changes with the stated chart purpose.

• A: A shows shares of all allocated passes, a different whole.
• B: B shows each team’s own usage rate; the two rates are not a partition.
• C: C uses the common whole of 300 used passes and sums to 100%.
• D: D divides each used count by all 600 allocated passes and omits the unused category.

Use mistakes to choose the next step

Questions 1–3: rebuild the weighted numerator and revisit when a simple mean works. Questions 4–5: distinguish counts from rates and keep the period denominators. Question 6: name the missing weights rather than guessing. Question 7: compare group rates separately from composition. Questions 8–10: write the whole represented by each chart or table before calculating.

Every item gets one vote

Imagine one token for every inspected unit: green for an accepted unit and grey for a rejected unit. Combining batches means putting all their tokens together. A batch of 300 contributes more tokens than a batch of 50, so their percentage labels do not deserve equal weight merely because there are two labels.

This picture explains count-based weighting. For a mass percentage, replace one token per item with equal small units of mass; do not weight by the number of containers. The analogy does not resolve overlapping records, different success definitions or missing data. It also does not claim that a higher aggregate proves every group improved: moving more tokens from a higher-rate group can change the pooled proportion while each group's rule stays fixed.

Quick reference

• Match each numerator with its own positive denominator
• Overall rate = total numerator/total matching denominator
• Weight decimal rates by their denominators, then divide by the total weight
• Equal weights guarantee the simple mean; identical rates and other coincidences show equal weights are not always necessary
• With positive weights, the pooled rate lies within the component extremes; distinct rates give a strict interior result
• Unknown relative group sizes generally mean an unknown exact pooled rate
• Compare within-group rates and composition separately
• A pie divides one named whole; a stacked count bar adds segment counts from a zero baseline
• Allocation share, within-group usage rate and share of all used items can have different denominators
• Do not add overlapping groups or incompatible units/definitions without an explicit model

Concept references: OpenStax Introductory Statistics 2e, section 2.5, frequency-weighted mean; section 1.2, categorical counts, percentages, pies and bars. The pooled-rate identity is derived here from the actual numerator and denominator.