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ashwink
If m, s are the average and standard deviation of integers a, b, c, and d, is s > 0?

1. m > a
2. a + b + c + d = 0
\(a,b,c,d\,\,\,\,{\text{ints}}\)

\(m = \frac{{a + b + c + d}}{4}\)

\(s = \sigma \left( {a,b,c,d} \right)\)

\(s\,\,\mathop > \limits^? \,\,0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{?\,\,\,:\,\,\,\,a = b = c = d\,\,\,\,{\text{false}}}\,\)


\(\left( 1 \right)\,\,\,m > a\,\,\,\, \Rightarrow \,\,\,\,a = b = c = d\,\,\,\underline {{\text{false}}} \,\,\,({\text{otherwise}}\,\,m = a)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)



\(\left( 2 \right)\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {a,b,c,d} \right) = \left( {0,0,0,0} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {a,b,c,d} \right) = \left( {1, - 1,0,0} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\, \hfill \\ \\
\end{gathered} \right.\)


The above follows the notations and rationale taught in the GMATH method.
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Statement 1:

If a = b = c = d, the average m will be the same as a.

Since m > a, all the elements in the set cannot be the same, and therefore, s > 0. Sufficient.

Statement 2:

When a = b = c = d = 0, s = 0. NO

When a = -4, b = 0, c = 0, and d = 4, s > 0. YES. Insufficient.

The answer is A.
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standard deviation of any set of numbers is always greater than zero except for the set having all the values same where standard deviation=0
1. if average is greater than any one of the value that certainly means that not all the values in the set are zero hence SUFFICIENT that SD>0
2. a+b+c+d=0 :IF a,b,c,d all equal to zero so SD=0
in all other scenarios SD>0 hence INSUFFICIENT
HENCE A
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