What is the median of the ratings given by R3 to the five workers? (in numerial value)

	U	V	W	X	Y	Total
R1						17
R2						11
R3						19
R4						14
R5						17
Total

Summary Table

	Ullas	Vasu	Waman	Xavier	Yusuf
Mean	$2.2$	$3.8$	$3.4$	$3.6$	$2.6$
Median	$2$	$4$	$4$	$4$	$3$
Modal	$2$	$4$	$5$	$5$	$1 \text{ and } 4$
Range	$3$	$3$	$4$	$4$	$3$

Given that 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.

Total Ratings Given by Restaurants

• $\text{R1} = 3.4 \times 5 = 17$
• $\text{R2} = 2.2 \times 5 = 11$
• $\text{R3} = 3.8 \times 5 = 19$
• $\text{R4} = 2.8 \times 5 = 14$
• $\text{R5} = 3.4 \times 5 = 17$

	U	V	W	X	Y	Total
R1						17
R2						11
R3						19
R4						14
R5						17
Total	11	19	17	18	13

Summary Table

	Ullas	Vasu	Waman	Xavier	Yusuf
Mean	$2.2$	$3.8$	$3.4$	$3.6$	$2.6$
Median	$2$	$4$	$4$	$4$	$3$
Modal	$2$	$4$	$5$	$5$	$1 \text{ and } 4$
Range	$3$	$3$	$4$	$4$	$3$

Also from the table we know mean ratings given by U, V, W, X, Y which is 2.2, 3.8, 3.4, 3.6, 2.6.

Using the same concept as before we can find the sum of ratings given to them:

Total Ratings Given to Each Person

• $U = 2.2 \times 5 = 11$
• $V = 3.8 \times 5 = 19$
• $W = 3.4 \times 5 = 17$
• $X = 3.6 \times 5 = 18$
• $Y = 2.6 \times 5 = 13$

Restaurant Ratings Table (after using clues)

	U	V	W	X	Y	Tot
R1	1		5			$17$
R2			1	5	1	$11$
R3			5	5	1	$19$
R4						$14$
R5		5				$17$
Total	11	19	17	18	13
Med	2	4	4	4	3
Mod	2	4	5	5	1&4
Range	3	3	4	4	3

Now, using clues given in the question:

(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.

We fill the table accordingly.

Restaurant Ratings Table (U deduced)

	U	V	W	X	Y	Tot
R1	1		5			$17$
R2			1	5	1	$11$
R3			5	5	1	$19$
R4						$14$
R5		5				$17$
Total	11	19	17	18	13
Ratings	1,2,2,2,4
Med	2	4	4	4	3
Mod	2	4	5	5	1&4
Range	3	3	4	4	3

Now, consider U.

Given: median = 2, mode = 2 and range = 3.

So ratings are of the form: 1, a, 2, b, 4 (median 2, range 3 → max 4).

Sum condition: $1 + 2 + 4 + a + b = 11 \Rightarrow a + b = 4$.

For mode = 2, take $a = b = 2$. Hence U’s ratings are 1, 2, 2, 2, 4 (in some order).

Restaurant Ratings Table (U, V placed)

	U	V	W	X	Y	Tot
R1	1		5			$17$
R2			1	5	1	$11$
R3			5	5	1	$19$
R4						$14$
R5		5				$17$
Total	11	19	17	18	13
Ratings	1,2,2,2,4	2,4,4,4,5
Med	2	4	4	4	3
Mod	2	4	5	5	1&4
Range	3	3	4	4	3

V: median = 4, mode = 4, range = 3

Form: 2, a, 4, b, 5; sum $= 19 \Rightarrow a+b=8$; mode 4 → $a=b=4$ → 2, 4, 4, 4, 5.

Restaurant Ratings Table (U, V, W placed)

	U	V	W	X	Y	Tot
R1	1		5			$17$
R2			1	5	1	$11$
R3			5	5	1	$19$
R4						$14$
R5		5				$17$
Total	11	19	17	18	13
Ratings	1,2,2,2,4	2,4,4,4,5	1,2,4,5,5
Med	2	4	4	4	3
Mod	2	4	5	5	1&4
Range	3	3	4	4	3

W: median = 4, mode = 5, range = 4

Form: 1, a, 4, 5, 5; sum $= 17 \Rightarrow a=2$ → 1, 2, 4, 5, 5.

Restaurant Ratings Table (U, V, W, X placed)

	U	V	W	X	Y	Tot
R1	1		5			$17$
R2			1	5	1	$11$
R3			5	5	1	$19$
R4						$14$
R5		5				$17$
Total	11	19	17	18	13
Ratings	1,2,2,2,4	2,4,4,4,5	1,2,4,5,5	1,3,4,5,5
Med	2	4	4	4	3
Mod	2	4	5	5	1&4
Range	3	3	4	4	3

X: median = 4, mode = 5, range = 4

Form: 1, a, 4, 5, 5; sum $= 18 \Rightarrow a=3$ → 1, 3, 4, 5, 5.

Restaurant Ratings Table (all distributions fixed)

	U	V	W	X	Y	Tot
R1	1		5			$17$
R2			1	5	1	$11$
R3			5	5	1	$19$
R4						$14$
R5		5				$17$
Total	11	19	17	18	13
Ratings	1,2,2,2,4	2,4,4,4,5	1,2,4,5,5	1,3,4,5,5	1,1,3,4,4
Med	2	4	4	4	3
Mod	2	4	5	5	1&4
Range	3	3	4	4	3

Y: median = 3, mode = 1 & 4, range = 3 → 1, 1, 3, 4, 4.

Now, consider column R3:

The two missing entries should add up to $19 - 1 - 5 - 5 = 8$ (only possibility: 4 + 4).

Thus we can place U = 4 in R3’s column and one 4 in V’s column for R3.

Therefore, in row R2, the missing entry should be

$11 - 2 - 1 - 5 - 1 = 2$.

Restaurant Ratings Table (penultimate fill)

	U	V	W	X	Y	Tot
R1	1	4	5			$17$
R2	2	2	1	5	1	$11$
R3	4	4	5	5	1	$19$
R4	2	4	2/4			$14$
R5	2	5	4/2			$17$
Total	11	19	17	18	13
Ratings	1,2,2,2,4	2,4,4,4,5	1,2,4,5,5	1,3,4,5,5	1,1,3,4,4
Med	2	4	4	4	3
Mod	2	4	5	5	1&4
Range	3	3	4	4	3

Consider row R1:

Missing elements must add to $17 - 5 - 4 - 1 = 7$ → (3 + 4) or (4 + 3). — (1)

Consider R5: the three missing elements should add to $17 - 2 - 5 = 10$ → 2 + 4 + 4 or 4 + 3 + 3.

But (1) requires a 3, so the consistent placement leads to resolving R1 and then R4/R5.

Completed Restaurant Ratings Table

	U	V	W	X	Y	Tot
R1	1	4	5	3	4	$17$
R2	2	2	1	5	1	$11$
R3	4	4	5	5	1	$19$
R4	2	4	4	1	3	$14$
R5	2	5	2	4	4	$17$
Total	11	19	17	18	13
Med	2	4	4	4	3
Mod	2	4	5	5	1&4
Range	3	3	4	4	3

So we can fill row R1 as 3 and 4, and the remaining values in R4 and R5 to match all given totals and summary statistics.

Restaurant Ratings Table

	U	V	W	X	Y	Tot
R1	1	4	5	3	4	$1$
R2	2	2	1	5	1	$1$
R3	4	4	5	5	1	$1$
R4	2	4	4	1	3	$1$
R5	2	5	2	4	4	$1$
Total	11	19	17	18	13
Med	2	4	4	4	3
Mod	2	4	5	5	1&4
Range	3	3	4	4	3

=> Ratings give by R3 are 1, 4, 4, 5, 5 => Median = 4