disper

Let x1, x2, x3, ..., x30 (allin kg) be the weights of the 30 students.

Original mean :
(x1 + x2 + ...... + x30)/30 = 53
x1 + x2 + ...... + x30 = 1590

New mean
= (x1 + x2 + ...... + x30 + 53)/31 kg
= (1590 + 53)/31 kg
= 53 kg
(unchanged)

Original standard deviation :
√{[(x1 - 53)² + (x2 - 53)² + ... + (x30 -53)²]/29} = 6
[(x1 - 53)² + (x2 - 53)² + ... + (x30 -53)²]/29 = 36
(x1 - 53)² + (x2 - 53)² + ... + (x30 - 53)² = 1044

New standard deviation :
= √{[(x1 - 53)² + (x2 - 53)² + ... + (x30 -53)² + (53 - 53)²]/30} kg
= √[(1044 + 0)/30] kg
= 5.90 kg < Original standard deviation

new weight=mean weight of 30 students
=>addition of the new weight does not affect the original mean
=>new mean = original mean = 53kg

new weight = mean
=> deviation of new weight from mean = 0
=> sum of squares of deviations from mean of 30 weights
= sum of squares of deviations from mean of 31 weights

Standard deviation, to a certain extent, is the average of sum of squares of deviations.
With the addition of a new weight, the new standard deviation is smaller.