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vipulgoyal
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Note that mean of all the sets are 3.
Now, standard deviation is nothing but the average measure of how far away are the elements from the mean.
Now, elements of option A are uniformly distributed.
Elements of options E and D are more away from 3 than the elements of A, so their standard deviation must be greater than A.
Similarly elements of options B and C are closer to 3 than the elements of A, so their standard deviation must be less than A.
So, option A must have the third highest standard deviation.
Answer : A
Here is solution with standard deviation calculation.
A. {1, 2, 3, 4, 5} # mean = 3 # stdev = √[(2² + 1² + 0² + 1² + 2²)/5] = √2
B. {2, 3, 3, 3, 4} # mean = 3 # stdev = √[(1² + 0² + 0² + 0² + 1²)/5] = √(2/5) = √(0.4)
C. {2, 2, 2, 4, 5} # mean = 3 # stdev = √[(1² + 1² + 1² + 1² + 2²)/5] = √(8/5) = √(1.6)
D. {0, 2, 3, 4, 6} # mean = 3 # stdev = √[(3² + 1² + 0² + 1² + 3²)/5] = √4
E. {-1, 1, 3, 5, 7} # mean = 3 # stdev = √[(4² + 2² + 0² + 2² + 4²)/5] = √10













