I'm not sure that this is the answer. However, can't you assume that since SD measures the variation between the numbers in the set and the mean ( which is 4), that the SD is 2. Therefore, adding 2 and 6 will keep the sd the same? Right?

raju232007 wrote:A set of data consists of the following 5 numbers: 0,2,4,6 and 8.Which two numbers,if added to create a set of 7 numbers will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?
A.-1 and 9
B. 4 and 4
C. 3 and 5
D. 2 and 6
E. 0 and 8


OA is D

A lot of calculations is necessary to solve this problem...Is there any shortcut method to solve this problem?

I go with D as well...

Though Not sure if a shorter method available but this one sure works..

0 2 4 6 8 mean 4
variance= ( X-m)^2 /N which is 2*(2^2+ 4^2)/5 = 8

now need 2 number to be added to maintain the variance

assume them as X and Y are variances ....

we know that

40 + X+Y/7 = 8

X+Y = 16

(X-4)^2 + (Y-4)^2 = 16

only D gives u the closest choice...

if u know the formula well.. all it takes is 1 minutes...

Hope that helps..

if i am right the formula for std deviation is sqrt(sum of (X-m)^2/N)...and so the standard deviation for the given 5 nos is sqrt(40^2/5)which is sqrt(320)...


Now we are supposed to find out which of the two numbers if included along with the 5 nos will work out to the closest std deviation i.e sqrt(320)

I calculated the ans using the longest method possible...and that is as follows

if we consider the first choice -1 and 9

Given set : 0 2 4 6 8
New set(X) : -1 0 2 4 6 8 9
Mean(m) : 4
X-m : -5 -4 -2 0 2 4 5
(X-m)^2 : 25 16 4 0 4 16 25
sum of (X-m)^2 :90
std deviation=sqrt(90^2)/5 = sqrt(1620)

In the same manner i calculated the std deviation for the other choices which took a lot of time and i found that choice D has a std deviation i.e sqrt(336) which is close to the desired value...i.e sqrt(320)

raju232007 wrote:if i am right the formula for std deviation is sqrt(sum of (X-m)^2/N)...and so the standard deviation for the given 5 nos is sqrt(40^2/5)which is sqrt(320)...


Now we are supposed to find out which of the two numbers if included along with the 5 nos will work out to the closest std deviation i.e sqrt(320)

I calculated the ans using the longest method possible...and that is as follows

if we consider the first choice -1 and 9

Given set : 0 2 4 6 8
New set(X) : -1 0 2 4 6 8 9
Mean(m) : 4
X-m : -5 -4 -2 0 2 4 5
(X-m)^2 : 25 16 4 0 4 16 25
sum of (X-m)^2 :90
std deviation=sqrt(90^2)/5 = sqrt(1620)

In the same manner i calculated the std deviation for the other choices which took a lot of time and i found that choice D has a std deviation i.e sqrt(336) which is close to the desired value...i.e sqrt(320)

Ur calculation of SD is wrong ..
ur right on the formula ..but execution is wrong ...

u know that the mean is 4
thus SD is

sq rt [(X-m)^2/N]
X is each number for instance for 0 2 4 6 8 X will take each of those numbers

thus its

sqrt[ (0-4)^2 + (2-4)^2 + ( 4 -4)^2 + (6-4)^2 + (8-4)^2 ] /5

16 +4+ 4 + 16 = 40/5 =8

thus its sqrt 8 = 2.8

hope that helps ..

Thanks for the clarification.... 8)

I still have a doubt on the OA

as i went on to solve my answers for the options provided are
sqrt(13)
sqrt(8)....this is eight
sqrt(6)
sqrt(7)(appro)
sqrt(10)(appro)

thus having 4 and 4 added to number stream does not make any change to the sd hence i would go for B ofcourse the next best choice is D

Regards

I am sorry i made the basic claculation error for b divided the resultant with 5...infact i fell into trap of calculating i should have right away eliminated B

my earlier next best choice ofcourse is the answer

oh i was just wondering whether the OA is ever wrong!!!! this one is real

Not sure how 2 and 6 satisfies the equation:

(X-4)^2 + (Y-4)^2 = 16 .

I thought 0 and 8 should satisfy? Can you please clarify?