Set P consists of 5 items. The standard deviation of the items in Set

Statement (1) All the items in Set P are different even numbers between 1 and 19.
in a set of consecutive even integers which would give the set which is minimum spread out i.e one element equal to mean and rest 4 are at distance of 2 and 4 each 2 respectively .. SD will be \(\sqrt{\frac{2*2^2 + 2*4^2 + 0 }{ 5}}\) =2\(\sqrt{2}\) i.e more than 2

Sufficient

Statement (2) The smallest number in the set is 10.
Irrelevant

(A) is the answer IMO

SD=\( ∑ ∣x−μ∣ ^2/N\)

Where ∑ means "sum of", x is a value in the data set, μ is the mean of the data set, and N is the number of data points in the population.

Logic:
SD will be minimum for items close to mean/close to each other. If items are spread away from the mean then SD is Max. So to say SD is always >2 we need to just prove it's minimum >2.

We need to conclusively prove SD >2 in all the cases of the selected set.

Statement 1: All the items in Set P are different even numbers between 1 and 19.
According to the above logic we will pick set with close items : {2,4,6,8,10} Avg/Mean = 6 and SD = \sqrt{8} >2

Hence using A we can prove SD will be >2.

Hold on !! We need to check another statement as well.

Statement 2: The smallest number in the set is 10
So to find Min SD the set should be {10,11,12,13,14} with SD \sqrt{2}<2

But we need to check Max condition as well, If even Max<2 then we can solve this by using Stmt 2.

To find Max set should be {10,11,12,19,20} with SD clearly SD>2. Hence it is not consistent and Stmt 2 is not useful.

Hence A is a perfect choice.

Note: Although this seems lengthy, once we understand it and with mental calculation, it can be done in <2min



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Set P consists of 5 items. The standard deviation of the items in Set P is S. Is S>2?

Stat1: All the items in Set P are different even numbers between 1 and 19.
\(p_{min}\) ={2,4,6,8,10}, \(mean_{min}\) = 6, \(SD_{min}\) = \(\sqrt{40/5}\) = 2\(\sqrt{2}\) >2. Sufficient

Stat2: The smallest number in the set is 10.
p can have same 10 as numbers or different nos. Not Sufficient

So, I think A.

1) The minimum SD = \(\sqrt{\frac{40}{5}} = 2\sqrt{3}\) > 2. Sufficient!
2) The smallest number = 10
Case1: P = {10, 10, 10, 10, 10}, SD = 0
Case2: P = {10, 12, 14, 16, 18}, SD! = 0 Insufficient!,
Option (C) is a classic GMAT trap!
(A) IMO!