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
OA : A
source:4gmat.com
mean and SD
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- sumgb
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Interesting question.
a,b,c,d are int.
m = avg
s = std dev
Is s > 0?
std dev = sq rt [ (sum of (mean - term)^2) / no of terms ]
which means std dev is always positive and std dev can be 0 if and only if all terms in the set are zero. std dev is greater than 0 even if avg of the set is 0.
so essentially we need to find out if at least 1 term from a, b ,c or d is greater than 0.
stmnt 1 : m > a which means that avg of a,b,c and d is greater than a which implies that there is at least 1 term among a, b, c and d which is not zero. hence std dev is definitely greater than 0.
Suff.
stmnt 2: a + b + c + d = 0
case I : a=b=c=d=0 so m =0 and s = 0
case II: a=-b and c =-d so m = 0 but s > 0
you can not determine if s is definitely greater than 0 hence insuff.
ANswer A
Hope this helps.
a,b,c,d are int.
m = avg
s = std dev
Is s > 0?
std dev = sq rt [ (sum of (mean - term)^2) / no of terms ]
which means std dev is always positive and std dev can be 0 if and only if all terms in the set are zero. std dev is greater than 0 even if avg of the set is 0.
so essentially we need to find out if at least 1 term from a, b ,c or d is greater than 0.
stmnt 1 : m > a which means that avg of a,b,c and d is greater than a which implies that there is at least 1 term among a, b, c and d which is not zero. hence std dev is definitely greater than 0.
Suff.
stmnt 2: a + b + c + d = 0
case I : a=b=c=d=0 so m =0 and s = 0
case II: a=-b and c =-d so m = 0 but s > 0
you can not determine if s is definitely greater than 0 hence insuff.
ANswer A
Hope this helps.