Question

Five restaurants, coded R1, R2, R3, R4 and R5 gave integer ratings to five gig workers – Ullas, Vasu, Waman, Xavier and Yusuf, on a scale of 1 to 5. The means of the ratings given by R1, R2, R3, R4 and R5 were 3.4, 2.2, 3.8, 2.8 and 3.4 respectively. 
The summary statistics of these ratings for the five workers is given below.

	Ullas	Vasu	Waman	Xavier	Yusuf
Mean rating	2.2	3.8	3.4	3.6	2.6
Median rating	2	4	4	4	3
Model rating	2	4	5	5	1 and 4
Range of rating	3	3	4	4	3

* Range of ratings is defined as the difference between the maximum and minimum ratings awarded to a worker.
The following is partial information about ratings of 1 and 5 awarded by the restaurants to the workers.
(a) R1 awarded a rating of 5 to Waman, as did R2 to Xavier, R3 to Waman and Xavier, and R5 to Vasu. 
(b) R1 awarded a rating of 1 to Ullas, as did R2 to Waman and Yusuf, and R3 to Yusuf.
How many individual ratings cannot be determined from the above information? [This question was asked as TITA]

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is A

Given the means of the ratings given by R1, R2, R3, R4, and R5 as 3.4, 2.2, 3.8, 2.8, and 3.4 respectively, we can calculate the total sum of ratings given by each rater as follows:

• R1: 5×3.4=17
• R2: 5×2.2=11
• R3: 5×3.8=19
• R4: 5×2.8=14
• R5: 5×3.4=17

Similarly, the sum of ratings received by U, V, W, X, and Y are:

• U: 5×2.2=11
• V: 5×3.8=19
• W: 5×3.4=17
• X: 5×3.6=18
• Y: 5×2.6=13

Given this information, we can capture the absolute data in the form of a table. Let's represent this partial information as follows:

	U	V	W	X	Y	Sum
R1	a	b	c	d	e	17
R2	f	g	h	i	j	11
R3	k	l	m	n	o	19
R4	p	q	r	s	t	14
R5	u	v	w	x	y	17
Sum	11	19	17	18	13

Where the variables 𝑎,𝑏,𝑐,…,𝑦a,b,c,…,y represent the individual ratings given by each rater to each item. The sums at the end of each row and column represent the total ratings given by each rater and the total ratings received by each item, respectively.
Consider U: Given median = 2, mode = 2, and range = 3:

• His ratings should be of the form 1, a, 2, b, 4, where a and b are unknown.
• The total sum of his ratings is 11 (from previous calculations).
• For mode = 2, both a and b should be 2.
• Therefore, U's ratings are 1, 2, 2, 2, 4.

Consider V: Given median = 4, mode = 4, and range = 3:

• His ratings should be of the form 2, a, 4, b, 5, where a and b are unknown.
• The total sum of his ratings is 19 (from previous calculations).
• For mode = 4, both a and b should be 4.
• Therefore, V's ratings are 2, 4, 4, 4, 5.

Consider W: Given median = 4, mode = 5, and range = 4:

• His ratings should be of the form 1, a, 4, 5, 5, where a is unknown.
• The total sum of his ratings is 17 (from previous calculations).
• Solving, we find that a = 2.
• Therefore, W's ratings are 1, 2, 4, 5, 5.

Consider X: Given median = 4, mode = 5, and range = 4:

• His ratings should be of the form 1, a, 4, 5, 5, where a is unknown.
• The total sum of his ratings is 18 (from previous calculations).
• Solving, we find that a = 3.
• Therefore, X's ratings are 1, 3, 4, 5, 5.
   

Consider Y: Given median = 3, mode = 1 and 4, and range = 3:

• His ratings should be of the form 1, a, 3, b, 4, where a and b are unknown.
• The total sum of his ratings is 13 (from previous calculations).
• We need to solve for a and b.
• Considering the mode, both a and b should be either 1 or 4.
• However, considering the range, the difference between the highest and lowest ratings should be 3.
• Therefore, Y's ratings are 1, 1, 3, 4, 4.

Considering column R3, the two missing entries should add up to 8. The only possibility is 4 + 4. Therefore, we can fill in 4 for row "U" and 4 for row "V."

Consider column R1, where the missing elements should add up to 17−5−4−1=717−5−4−1=7. The possible combinations are 3 + 4 or 4 + 3.

Now, consider column R5, where the missing elements should add up to 10. We cannot have 4 + 3 + 3 as it contradicts the possible combinations for column R1. Therefore, we must have 2 + 4 + 4.

We can fill column R1 as 3 + 4 and the remaining in column R4. With this, we can complete the table.
 

	R1	R2	R3	R4	R5	Total
U	1	2	4	2	2	11
V	4	2	4	4	5	19
W	5	1	5	4	2	17
X	3	5	5	1	4	18
Y	4	1	1	3	4	13
Total	17	11	19	14	17

All ratings can be determined uniquely that is 0.