We are given that sets $\(\mathrm{S} \& \mathrm{~T}\)$ satisfy the condition that all the numbers in the set $S$ are in the set $\(T\)$ and the average of the numbers in set $\(S\)$ is equal to the average of the numbers in set $\(T\)$.

Since the values of the two sets S \& T are not known, a unique comparison cannot be formed between the standard deviations of both the sets.

For example if we consider set $\(\mathrm{T}=\{0,1,2,3,4\} \&\)$ set $\(\mathrm{S}=\{1,2,3\}\)$, the average (arithmetic mean of both the sets is same i.e. 2 each.

(Note: - when numbers form arithmetic sequence average comes same as median)

Now, three elements $\(1,2 \& 3\)$ of the two sets $\(T \& S\)$ are the same and the remaining elements of set $\(\mathrm{T}-0 \& 4\)$ will add to the sum and will result in higher standard deviation for set T .

But if we consider the two sets as $\(\mathrm{T}=\{1,1,1,1,1\} \& \mathrm{~S}=\{1,1,1\}\)$, both the sets have the same arithmetic mean 1 and also as the sets have equal numbers, the standard deviations of both the sets would come zero.


Note: - Standard deviation is never negative and is equal to zero only when all the terms of the set are equal whereas in rest of the cases the standard deviation is positive.

Hence the answer is (D).