Both statements are telling us that the numbers 1 and 19 are being removed from the set to yield the following set :
{3, 5, 7, 9, 11, 13, 15, 17}- which has the mean and median = 10

I think the answer should thus be D. Whats the source of this problem ?

Hi,
Mean of the given set is 10.
From(1):
The two numbers picked can be any of the following pairs: (1,19),(3,17),...(9,11).
So, depending on the pair picked, we get different values for standard deviation.
Not sufficient
From(2):
Same

Hence, E

But i did not understand 2 numbers are selected means? is it removing from the list ? Can someone please explain?

I guess the wording is a bit dubious. It is either removing two from the given 10 and calculating for the remaining 8 or selecting 2 from the 10 and calculating for those 2. Either way, the answwer will be E.

Thanks Frank,

I get how the mean and median will remain the same after removing the various pairs.

Quick question - Do we still call the set-"an evenly spaced set" after removing the other pairs (other than 1,19) ? I hope NO ?

bblast wrote: Quick question - Do we still call the set-"an evenly spaced set" after removing the other pairs (other than 1,19) ? I hope NO ?

You are right..A big 'NO'

Thanks guys for clarifying.

I guess the question meant selecting any 2 no's form list and calculating those 2 no's SD.

ktn11 wrote:If two numbers are selected from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19 randomly, what is the standard
deviation of the new list?
(1) The mean remains unchanged. (2) The median of the new list is 10.

OA:E

But i did not understand 2 numbers are selected means? is it removing from the list ? Can someone please explain?

The sum of 10 numbers is 100.

Condition 1-- mean remains unchanged after the two number are removed from the list..

This means that the new sum will be 80 of 8 numbers and the mean will still be 10.

we can remove from (11,9) (7,13) (3,17) (19,1) (5,15)

So many cases-- and SD will be different every time.. NOT SUFFICIENT

Condition 2. -- median of the list is 10. Since we are removing 2 numbers the median will be the mean of 4th and 5th number when the series is arranged in ascending order.
1,3,5,7,13,15,17,19 [removed 9 and 11]
1,3,5,9,11,15,17,19 [removed 7 and 13]
1,5,7,9,11,13,15,19 [removed 3 and 17]

So many cases-- and SD will be different every time.. NOT SUFFICIENT

Combine both, still NOT SUFFICIENT -- HENCE E

If two numbers are selected from 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19 randomly, what is the standard
deviation of the new list?
(1) The mean remains unchanged. (2) The median of the new list is 10.

It has to be E

From 1,
Mean=10, we can choose any from (1,19),(2,17) etc where SD is 18,15 - varies!! INSUFF

From 2,
Median=10, we can choose from (1,19),(2,17) etc where SD is 18,15 - varies!! INSUFF

From 1 and 2,
Mean=Median=10, agian we can choose from (1,19),(2,17) etc where SD is 18,15 - varies!! INSUFF

IMO:E