Comprehension:

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 cumu lative 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.
1. 100 buyers gave product ratings on Day 3.
2. The modes of the product ratings were 4 and 5.
3. The numbers of buyers giving each product rating are non-zero multiples of 10.
4. 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.

What is the median of all ratings given on Day 3?

Correct Answer: 4

To find the median of the ratings given on Day 3, we first analyze the distribution of ratings using the information provided.

First, we look at the chart titled "Distribution of Ratings on Day 2". From the chart, we can extract the number of buyers for each rating on Day 2:
- Rating 5: 10 buyers
- Rating 4: 20 buyers
- Rating 3: 5 buyers
- Rating 2: 10 buyers
- Rating 1: 5 buyers
Total buyers on Day 2 = 10 + 20 + 5 + 10 + 5 = 50 buyers.

Now, let us denote the number of buyers who gave ratings 1, 2, 3, 4, and 5 on Day 3 as:

$x_1,x_2,x_3,x_4,\text{ and }{x}_5$

respectively. From the given conditions, we establish the following equations and constraints:
1. A total of 100 buyers gave ratings on Day 3:

$x_1+x_2+x_3+x_4+x_5=100$

2. The number of buyers for each rating on Day 3 must be a non-zero multiple of 10:

$x_i\in \{10,20,30,40,\dots \}\text{ for }i=1,2,3,4,5$

3. The same number of buyers gave ratings of 1 and 2, which is half the number of buyers who gave a rating of 3:

$x_1=x_2=\frac{1}{2}{x}_3⟹x_3=2x_1=2x_2$

Let us evaluate the possible cases for $x_1$:
Case A: If $x_1=10$, then:
- $x_2=10$
- $x_3=20$
The sum of ratings 1, 2, and 3 is $10+10+20=40$.
This leaves $x_4+x_5=100-40=60$ buyers.
Since the modes of the product ratings are 4 and 5, ratings 4 and 5 must have the same count and it must be strictly greater than any other rating count:
- $x_4=x_5=30$
Since 30 is greater than 20 (the count for rating 3), 4 and 5 are indeed the joint modes. This configuration is valid.

Case B: If $x_1=20$, then:
- $x_2=20$
- $x_3=40$
The sum is $20+20+40=80$, leaving $x_4+x_5=20$.
For 4 and 5 to be modes, their counts must be equal, so $x_4=x_5=10$. However, this would mean the mode is 3 (with count 40), contradicting the condition that the modes are 4 and 5. Thus, Case B is invalid.

Therefore, the distribution of ratings on Day 3 is uniquely determined as:
- Rating 1: 10 buyers (sorted positions 1 to 10)
- Rating 2: 10 buyers (sorted positions 11 to 20)
- Rating 3: 20 buyers (sorted positions 21 to 40)
- Rating 4: 30 buyers (sorted positions 41 to 70)
- Rating 5: 30 buyers (sorted positions 71 to 100)

To find the median of the 100 ratings, we take the average of the 50th and 51st values in the sorted order.
Since positions 41 through 70 all correspond to a rating of 4:
- The 50th rating is 4.
- The 51st rating is 4.
The median is therefore:

$\text{Median}=\frac{4+4}{2}=4$