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.
Question: 1
How many individual ratings cannot be determined from the above information?
Given that the average ratings given by R1, R2, R3, R4, and R5 were 3.4, 2.2, 3.8, 2.8, and 3.4 respectively, the sum of the ratings given by each reviewer is calculated as 5 times their respective means:
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 the ratings received by U, V, W, X, and Y is also calculated as 5 times their respective means:
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 Capturing the absolute data provided in the partial information (a) and (b) and representing it in a table, we get:
R1
R2
R3
R4
R5
Total
Missing Entries
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
-
So, we can clearly get that All ratings can be determined uniquely is 0
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=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
R2 gave ratings of 1, 1, 2, 2, 5 He gave 4 to 0 workers So, The correct answer is 0
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
From the table, we can see that R1 gave a rating of 3 to Xavier.
So the correct answer is 3.
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0
Question: 4
What is the median of the ratings given by R3 to the five workers?
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
Ratings give by R3 are 1, 4, 4, 5, 5 => Median = 4.
So, the correct answer is 4.
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Question: 5
Which among the following restaurants gave its median rating to exactly one of the workers?
Given that the average ratings given by R1, R2, R3, R4, and R5 were 3.4, 2.2, 3.8, 2.8, and 3.4 respectively, the sum of the ratings given by each reviewer is calculated as 5 times their respective means:
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 the ratings received by U, V, W, X, and Y is also calculated as 5 times their respective means:
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 Capturing the absolute data provided in the partial information (a) and (b) and representing it in a table, we get:
