Day 2
| Ratings |
Number of Buyers |
| 1 |
5 |
| 2 |
10 |
| 3 |
5 |
| 4 |
20 |
| 5 |
10 |
Total number of Buyers in Day 2 = 50.
Daily Average on Day 2 =
(1×5 + 2×10 + 3×5 + 4×20 + 5×10) / 50 = 170 / 50 = 3.4
DAY 1:
Cumulative averages on Day 1 and Day 2 were 3 and 3.1 respectively.
Let the number of Buyers in Day 1 = x.
(3x + 170) / (x + 50) = 3.1
Solving, 3x + 170 = 3.1(x + 50);
3x + 170 = 3.1x + 155;
0.1x = 15;
x = 150.
Therefore, the number of Buyers in Day 1 = 150.
DAY 3:
Total number of Buyers in Day 3 = 100.
The numbers of buyers giving each product rating are non-zero multiples of 10.
Number of buyers giving rating 1 = Number of buyers giving rating 2 = Half of number of buyers giving rating 3.
The modes of the product ratings were 4 and 5.
DAY 3
| Ratings |
Number of Buyers |
| 1 |
A |
| 2 |
A |
| 3 |
2A |
| 4 |
B (Mode) |
| 5 |
B (Mode) |
Sum of the number of buyers = 4A + 2B = 100.
The only possible solution for the equation and B value being the mode is A = 10 and B = 30.
DAY 3
| Ratings |
Number of Buyers |
| 1 |
10 |
| 2 |
10 |
| 3 |
20 |
| 4 |
30 |
| 5 |
30 |
Product rating on Day 3 = (1×10 + 2×10 + 3×20 + 4×30 + 5×30) / 100 = 360 / 100 = 3.6
Cumulative rating on Day 3 = (3×150 + 3.4×50 + 3.6×100) / (150 + 50 + 100) = 980 / 300 = 3.266
| DAY |
NUMBER OF PASSENGERS |
DAILY AVERAGE |
CUMULATIVE AVERAGE |
| 1 |
150 |
3 |
3 |
| 2 |
50 |
3.4 |
3.1 |
| 3 |
100 |
3.6 |
3.266 |
The question is "What is the median of all ratings given on Day 3?"
Answer: No of 1s’= 10; No of 2’s= 10; No of 3’s= 20; No of 4’s= 30; No of 5’s= 30.
Therefore, the median will be average of 50th and 51st term which is (4+4)/2 = 4.
Hence, the answer is '4'
