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

Day - 1

From the information given in the question, we know that cumulative averages on day 1 and day 2 were 3 and 3.1 respectively.

Let the number of buyers in day 1 = $x$.

$\small \dfrac{3x + 170}{x + 50} = 3.1$

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

As per statement 1: Total number of buyers in day 3 = 100.

Also using statements 2, 3, and 4:

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.

Adding all of them we get:

Sum of the number of buyers = $4A + 2B = 100$.

The only possible solution for the equation with B value being the mode is $A = 10$ and $B = 30$.

Average on day 3:

$\small \dfrac{1 \times 10 + 2 \times 10 + 3 \times 20 + 4 \times 30 + 5 \times 30}{100}$

$=3.6$

Cumulative average:

$= \small \dfrac{(150 \times 3) + (50 \times 3.4) + (100 \times 3.6)}{150 + 50 + 100}$

$= 3.266$

We get the above final table.

To find the median of all ratings

we know 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.