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is the standard deviation of number w,x,y,z greater than 2 (1) w =3 (2) the average of four numbers is 8
analysing statement 1
value of x, y,z are not defined hence this statement is insufficent
analysing statement 2
we are given that average of 4 numbers is 8 but again we do not know the individual values for the variables w,x,y,z
therefore statement 2 is insufficent
together both statements put together we get
w=3 average of four numbers is 8, but we still do not know the values of x,y and z
so we cant arrive at the standard deviation of these 4 numbers
IMO E should be the answer (which needless to say is different from what the OA is)
am i missing something in this question
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ANS C .
1 and 2 alone not sufficient
if you combine both the statement: W=3 , now choose other three numbers closely clustered such that avg of W, X, Y,Z is 8 . Such data set is {3 , 9 , 9 ,10} , SD of which is > 2 , hence SD of any other combination will also be > 2 .
standard deviation SD tell us how far / close numbers are from the mean.
if SD is low, means numbers are closer to mean.
Combinig the two statements, we have w = 3 and mean = 8. Lets consider an example set where the numbers are closest to the mean, including w = 3. This set will be - 3,8,8,13. now we can calculate SD for this set. It'll be sqrt((5^2+0^2+0^2+5^2)/4) = sqrt(12.5) which is more than 2.
standard deviation SD tell us how far / close numbers are from the mean.
if SD is low, means numbers are closer to mean.
Combinig the two statements, we have w = 3 and mean = 8. Lets consider an example set where the numbers are closest to the mean, including w = 3. This set will be - 3,8,8,13. now we can calculate SD for this set. It'll be sqrt((5^2+0^2+0^2+5^2)/4) = sqrt(12.5) which is more than 2.
Any other combination, SD will be more than this.
Answer is C.
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Ok guys got the point thanks for the help
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