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BillyZ
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Very nice one.
It shows that SD measure the actually the "Dispersion" around the average value of the items.

(1)- Not sufficient
(2)- Sufficient. Can be done by trying different value of n.
Below is a formula approach
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BillyZ
Are the standard deviations of set A and set B the same?

1) n = 6
2) Set A = {n, 2, 2, 2} and set B = {n, n, n, 2}

Tough question. Took some time to wrap my head around this.

Statement 1 gives us very little info, so insufficient.

Statement 2 requires some visualizing of standard deviations. Let's take a value for n as 10, just as an example.

For Set A, if n is 10, the mean of the set is 4. For Set B, mean is 8.

Standard deviation measures the distance from the mean. For Set A, these distances end up being:
DatapointDistance from Mean
22
22
22
106

For Set B, these distances end up being:
DatapointDistance from Mean
102
102
102
26

Standard deviation cares about the distance from the mean for its calculation, not the datapoint value itself. Since the distances from the mean are the same here, it'll result in the same value for standard deviation. The answer to "Are the standard deviations of Set A and Set B the same?" is a definitive yes for Statement 2. Our answer is B.
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