Small-group project: Make a group of 3–4 students. Choose one scenario to design a custom rating scheme by assigning appropriate weights: (a) shopping experience at a cloth store, (b) travel experience in a bus, (c) tourism experience of a nearby tourist spot, (d) clinic/hospital experience
- (i) Discuss and arrive at 4–6 aspects to rate. For each aspect chosen, justify why it matters for an overall experience.
- (ii) Discuss and decide the relative weights and justify.
- (iii) Collect or imagine ratings from at least 10 people on a scale of 1–5 (5 being very good).
- (iv) Compute each individual rating. Compute the overall average across individuals.
- (v) Make a 100% stacked bar chart using your data.
- (vi) Write a short note on the observations and inferences, what your rating system captures well and what it misses.
Step-by-step solution
Idea: A custom rating is a weighted mean of the aspect ratings. The weights say how much each aspect matters; the overall average across people is the mean of the individual weighted ratings.
(i) Discuss and arrive at 4–6 aspects to rate. For each aspect chosen, justify why it matters for an overall experience.
- Cleanliness (prevents infection), waiting time (patients are unwell and time matters), doctor’s care (the main purpose of the visit), staff behaviour (comfort and guidance), cost (affordability).1 mark
(ii) Discuss and decide the relative weights and justify.
- Weights 2 : 3 : 4 : 2 : 1 (total 12). Doctor’s care gets the most (4) because treatment is why people come; waiting time next (3); cleanliness and staff 2 each; cost 1, as a government clinic charges little.1 mark
(iii) Collect or imagine ratings from at least 10 people on a scale of 1–5 (5 being very good).
- Sample ratings of 10 people (with each person’s weighted rating):½ mark
Person Clean (2) Wait (3) Doctor (4) Staff (2) Cost (1) Rating P1 4 3 5 4 3 4812 = 4.00 P2 5 2 4 4 2 4212 = 3.50 P3 3 2 4 3 4 3812 = 3.17 P4 4 4 5 5 3 5312 = 4.42 P5 5 3 5 4 2 4912 = 4.08 P6 2 1 3 3 4 2912 = 2.42 P7 4 3 4 4 3 4412 = 3.67 P8 3 2 5 4 3 4312 = 3.58 P9 5 4 4 5 2 5012 = 4.17 P10 4 2 4 3 3 3912 = 3.25
(iv) Compute each individual rating. Compute the overall average across individuals.
- Individual rating = 2C + 3W + 4D + 2S + 1K12. For P1: 8 + 9 + 20 + 8 + 312 = 4812 = 4.00. All ten: 4.00, 3.50, 3.17, 4.42, 4.08, 2.42, 3.67, 3.58, 4.17, 3.25.1 mark
- Overall average = 48 + 42 + 38 + 53 + 49 + 29 + 44 + 43 + 50 + 3912 × 10 = 435120 ≈ 3.63.½ mark
(v) Make a 100% stacked bar chart using your data.
- For each aspect, find the share of 5★, 4★, 3★, 2★ and 1★ ratings among the 10 people (each person = 10%) and draw one 100% bar per aspect (diagram).½ mark
(vi) Write a short note on the observations and inferences, what your rating system captures well and what it misses.
- Doctor’s care is rated highly (all 3 or more), but waiting time is poor (half gave 2 or less), pulling the overall score down. The system captures what patients value most, since doctor’s care counts 4 times as much as cost. It misses things like distance, medicine availability and how serious the illness was, and 10 people is a small sample.½ mark
Check: Weights add to 12, so each individual rating is between 1 and 5 ✓; the overall average 3.63 lies between the lowest (2.42) and highest (4.42) individual ratings ✓.
Answer to write in the exam
(i)
Cleanliness: hygiene and safety
Waiting time: patients are unwell
Doctor’s care: main purpose of visit
Staff behaviour: comfort and help
Cost: affordability
(ii)
Weights: cleanliness 2, waiting 3, doctor 4, staff 2, cost 1 (sum 12)
Doctor’s care matters most; cost least for a low-fee clinic
(iii)
10 sets of ratings (1–5) collected for the 5 aspects (see table)
(iv)
Rating = 2C + 3W + 4D + 2S + K12
P1 = 4812 = 4.00, …, P10 = 3912 = 3.25
Overall = 435120 = 3.63
∴ Overall average rating ≈ 3.63
(v)
Each person = 10% of a bar
One bar per aspect, split by 5★ to 1★ shares
(vi)
Doctor’s care rated high; waiting time rated low
Weights reflect what patients value most
Misses medicines, distance; sample of 10 is small
Common mistakes that cost marks
- Dividing the weighted sum by the number of aspects (5) instead of the sum of the weights (12).
- Giving weights without reasons; the project asks you to justify them.
- Drawing the 100% chart with bars of different lengths.
How this can come in the exam
A bus service is rated on punctuality, comfort and fare with weights 3 : 2 : 1. Two passengers rate it (4, 3, 5) and (2, 4, 4).
(i) Find each passenger’s rating. (ii) Find the average rating. (iii) Which aspect affects the rating most? (iv) If punctuality improves by 1 point for both passengers, by how much does the average rise?
Show answer
(i) 12 + 6 + 56 = 3.83 and 6 + 8 + 46 = 3 (1 mark). (ii) 23 + 1812 = 4112 ≈ 3.42 (1 mark). (iii) Punctuality (weight 3) (1 mark). (iv) Each rating rises by 36 = 0.5, so the average rises by 0.5 (1 mark).Try one yourself
A cloth store is rated on variety, price and service with weights 2 : 2 : 1. A customer gives 5, 3, 4. Find her rating.
Show answer
10 + 6 + 45 = 205 = 4.
More questions like this
- Whole class project: Each student shares the average age of their family and the number of family members. Discuss among the class and come up with a way to find out the average age of all the families of the class.
- Given some data with corresponding weights, what would happen to the weighted average if all the weights are increased by a constant value, say 1? If required, experiment with some data. What do you observe? Justify your answer using algebra.
- A farm has some cows, sheep, and chickens. Last year the cows made up 60%, the sheep 25%, and the chickens 15%. There was a decrease in the number of all three animals’ population over the year. Choose the possibilities for the change in their respective shares of the population —
(i) % of cows decreased, % of sheep decreased, % of chickens decreased
(ii) % of cows increased, % of sheep increased, % of chickens increased
(iii) % of cows remained the same, % of sheep remained the same, % of chickens remained the same
(iv) % of cows decreased, % of sheep increased, % of chickens remained the same
(v) % of cows increased, % of sheep increased, % of chickens decreased. - In a badminton academy, there is a group of 11 trainees — 8 seniors and 3 juniors. Their heights and average heights (in cm) are given in the table below. Find the average height of the whole group.
Two students calculate the average height of the whole group in two ways:
Method 1: 165.5 + 149.332 = 314.832 = 157.415
Method 2: 165 + 169 + 164 + 167 + 170 + 159 + 164 + 166 + 146 + 149 + 15311 = 177211 = 161.09
Whose calculation gives the correct average height of the whole group? - Why did Method 1 not work?