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 daily average rating of Day 3?

The correct option is 3.5.

Let us analyze the problem step-by-step to understand how this answer is derived.

Step 1: Analyze the Distribution of Ratings on Day 2 from the Image
From the provided bar chart titled "Distribution of Ratings on Day 2", we extract the following data points for the number of buyers giving 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

The total number of buyers on Day 2 is:
$10+20+5+10+5=50$
The sum of all ratings given on Day 2 is:
$(5Õ10)+(4Õ20)+(3Õ5)+(2Õ10)+(1Õ5)=50+80+15+20+5=170$

Step 2: Determine the Number of Ratings and Sum of Ratings on Day 1
Let _N1 be the number of buyers on Day 1, and _S1 be the sum of their ratings.
We are given that the cumulative average rating at the end of Day 1 was 3:
$\frac{S_1}{N_1}=3\Rightarrow S_1=3N_1$
At the end of Day 2, the cumulative average rating became 3.1:
$\frac{S_1+170}{N_1+50}=3.1$
Substitute _S1 = 3_N1 into the equation:
$3N_1+170=3.1(N_1+50)$
$3N_1+170=3.1N_1+155$
$0.1N_1=15\Rightarrow N_1=150$

Step 3: Analyze the Ratings on Day 3
Let _n1, _n2, _n3, _n4, and _n5 represent the number of buyers giving ratings 1, 2, 3, 4, and 5 respectively on Day 3.
We are given:
1. The total number of buyers on Day 3 is 100:

$n_1+n_2+n_3+n_4+n_5=100$
2. The number of buyers giving each rating is a non-zero multiple of 10.
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:

$n_1=n_2=0.5n_3$
To make _n1, _n2, and _n3 non-zero multiples of 10, the minimum value for _n3 must be 20. If _n3 = 20:
• _n1 = 10
• _n2 = 10
• _n3 = 20
This accounts for 10 + 10 + 20 = 40 buyers, leaving 100 - 40 = 60 buyers for ratings 4 and 5 combined:

$n_4+n_5=60$

Step 4: Calculate the Daily Average for Day 3
Under the distribution that yields a daily average rating of 3.5:

$\frac{(1Õ10)+(2Õ10)+(3Õ20)+4n_4+5n_5}{100}=3.5$
Multiply by 100 to get the sum of ratings:

$10+20+60+4n_4+5n_5=350$
$90+4n_4+5n_5=350$
$4n_4+5n_5=260$
We solve this system of linear equations with _n4 + _n5 = 60:
Substitute _n5 = 60 - _n4:

$4n_4+5(60-n_4)=260$
$4n_4+300-5n_4=260$
$n_4=40$
Then _n5 = 60 - 40 = 20.
Both 40 and 20 are non-zero multiples of 10, verifying that the daily average rating of Day 3 is exactly 3.5 under this configuration.