An online e-commerce firm receives daily integer product ratings from 1 through 5 given by buyers. The daily average is the average of the ratings given on that day. The cumulative average is the average of all ratings given on or before that day.
The rating system began on Day 1, and the cumulative averages were 3 and 3.1 at the end of Day 1 and Day 2, respectively. The distribution of ratings on Day 2 is given in the figure below.
The following information is known about ratings on Day 3.
- 100 buyers gave product ratings on Day 3.
- The modes of the product ratings were 4 and 5.
- The numbers of buyers giving each product rating are non-zero multiples of 10.
- The same number of buyers gave product ratings of 1 and 2, and that number is half the number of buyers who gave a rating of 3.
Q. 1 How many buyers gave ratings on Day 1?
Q. 2 What is the daily average rating of Day 3?
1) 3.2 2) 3.5 3) 3.0 4) 3.6
Q. 3 What is the median of all ratings given on Day 3?
Q. 4 Which of the following is true about the cumulative average ratings of Day 2 and Day 3?
1) The cumulative average of Day 3 decreased from Day 2.
2) The cumulative average of Day 3 increased by more than 8% from Day 2.
3) The cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2.
4) The cumulative average of Day 3 increased by less than 5% from Day 2.
Let’s solve the questions step-by-step using the provided data.
From the bar graph (Day 2 Ratings):
| Rating | Number of Buyers |
|---|---|
| 1 | 5 |
| 2 | 10 |
| 3 | 5 |
| 4 | 20 |
| 5 | 10 |
| Total | 50 |
👉 Total score on Day 2 = (1×5)+(2×10)+(3×5)+(4×20)+(5×10)=5+20+15+80+50=170(1×5) + (2×10) + (3×5) + (4×20) + (5×10) = 5 + 20 + 15 + 80 + 50 = 170
👉 Average of Day 2 = 17050=3.4\frac{170}{50} = 3.4
Given cumulative average at end of Day 1 = 3
Let xx be the number of buyers on Day 1
Let total score on Day 1 = 3x3x
Then cumulative average after Day 2 = 3.1 3x+170x+50=3.1\frac{3x + 170}{x + 50} = 3.1
Multiply both sides: 3x+170=3.1(x+50)3x+170=3.1x+1550.1x=15⇒x=1503x + 170 = 3.1(x + 50) \\ 3x + 170 = 3.1x + 155 \\ 0.1x = 15 \Rightarrow x = 150
✅ Q.1 Answer: 150 buyers gave ratings on Day 1
Day 3 Clues and Constraints:
- 100 buyers.
- Modes: 4 and 5 → highest frequencies occur at ratings 4 and 5 (and both have same high frequency).
- All values are non-zero multiples of 10.
- Let:
- aa: number of 1-star buyers
- aa: number of 2-star buyers
- 2a2a: number of 3-star buyers
- xx: number of 4-star buyers
- xx: number of 5-star buyers
- Total: a+a+2a+x+x=4a+2x=100a + a + 2a + x + x = 4a + 2x = 100
4a+2x=100⇒2a+x=50⇒x=50−2a4a + 2x = 100 \Rightarrow 2a + x = 50 \Rightarrow x = 50 – 2a
Try small non-zero values of aa that make all terms multiples of 10.
Try a=10a = 10
Then:
- 1⋆1\star: 10
- 2⋆2\star: 10
- 3⋆3\star: 20
- x=50−20=30x = 50 – 20 = 30 → 4⋆=304\star = 30, 5⋆=305\star = 30 ✅
Valid distribution.
| Rating | No. of Buyers |
|---|---|
| 1 | 10 |
| 2 | 10 |
| 3 | 20 |
| 4 | 30 |
| 5 | 30 |
| Total | 100 |
👉 Total score = (1×10)+(2×10)+(3×20)+(4×30)+(5×30)(1×10) + (2×10) + (3×20) + (4×30) + (5×30)
= 10+20+60+120+150=36010 + 20 + 60 + 120 + 150 = 360
👉 Daily average (Day 3) = 360100=3.6\frac{360}{100} = 3.6
✅ Q.2 Answer: Option 4) 3.6
Q.3 Median on Day 3
Cumulative frequency:
| Rating | Count | Cumulative |
|---|---|---|
| 1 | 10 | 10 |
| 2 | 10 | 20 |
| 3 | 20 | 40 |
| 4 | 30 | 70 |
| 5 | 30 | 100 |
Median position: 50th and 51st ratings → both fall in 4-star
✅ Q.3 Answer: 4
Q.4 Cumulative Average on Day 3
- Day 1: 150 buyers → score = 150 × 3 = 450
- Day 2: 50 buyers → score = 170
- Day 3: 100 buyers → score = 360
- Total ratings = 300 buyers
- Total score = 450+170+360=980450 + 170 + 360 = 980
Cumulative average = 980300=3.2667\frac{980}{300} = 3.2667
% increase from Day 2 (3.1): 3.2667−3.13.1×100≈5.38%\frac{3.2667 – 3.1}{3.1} \times 100 ≈ 5.38\%
✅ Q.4 Answer: Option 3) Increased by between 5% and 8%
✅ Final Answers:
| Q.No. | Answer |
|---|---|
| Q5 | 150 buyers |
| Q6 | 3.6 (Option 4) |
| Q7 | 4 |
| Q8 | Option 3 |
