Operator A --> 3, 7, 7, 3: Mean = 5
Operator B --> 4, 5, 5, 6: Mean = 5
Operator C --> 8, 2, 5, 5: Mean = 5
Operator D --> 6, 4, 4, 6: Mean = 5
Operator E --> 3, 4, 5, 6: Mean = 5

Deviations from Mean:
Operator A --> 3-5, 7-5, 7-5, 3-5 --> -2, 2, 2, -2
Operator B --> 4-5, 5-5, 5-5, 6-5 --> -1, 0, 0, 1
Operator C --> 8-5, 2-5, 5-5, 5-5 --> 3, -3, 0, 0
Operator D --> 6-5, 4-5, 4-5, 6-5 --> 1, -1, -1, 1
Operator E --> 3-5, 4-5, 5-5, 6-5 --> -2, -1, 0, 1

We can clearly see Operator B is least deviated from Mean value

IMO Option B

mean value for each operator is 5
and the least variation along mean is observed for operator B ;
IMO B

We notice that the sum for each operator is 20; therefore, the mean number of calls for each operator is 5.

Since the standard deviation is a measure of how far data points are from the mean, we should look for the row with the values closest to 5.

Operator B has two calls equal to the mean, and two other calls that are pretty close to the mean. All the other operators either have no calls equal to the mean, or have other calls that are far away from the mean (or both). Thus, Operator B has the smallest standard deviation.

Answer: B
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To find the least we have to find out the least dispersed set , B's numbers are least dispersed

Hence B