The two sets are $\(\{9,9,10,11,11\} \&\{6,8,10,12,14\}\)$


The arithmetic mean (average) of the two sets is $\(=\frac{9+9+10+11+11}{5}=\frac{50}{5}=10 \quad\)$ \& $\(=\frac{6+8+10+12+14}{5}=\frac{50}{5}=10\)$ respectively. $\(\frac{Arithmetic Mean }{ Average }= \frac{Sum of terms }{Number of terms} \)$

The values of the second set i.e. $\(\{6,8,10,12,14\}\)$ are more dispersed $\((Gaps- $-4,-2,0,2,4\}\)$ from the mean when compared with the values of the first set i.e. $\(\{9,9,10,11,11\}\)$ which has a gap of $\(\{-1,-1,0,1,1\}\)$.

Hence the standard deviation of set $\(B\)$ would be higher when compared with set $\(A\)$, so the answer is (B).