warlock wrote:can anyone explain me this problem

thank you

is it A?

yes...what is the logic involved?

warlock wrote:yes...what is the logic involved?

A says that 30 % was removed from every tank.. so standard deviation remains the same and it will be 70% of 10 i.e., 7

with B we can't say anything as we don't know individual tank's volumes..

In this case only average and median will change, SD will remain the same

Hi....gotta doubt here with the actual value of the new SD....I know its not required in this problem but just wanna extrapolate

Heres how i approached it
V = Data Value
A = Average

S.D. = Sqrt{(V1-A)^2+(V2-A)^2+(V3-A)^2+(V4-A)^2+(V5-A)^2+(V6-A)^2}/6

Now, as the new values r 70% of the original....the average will also b 70% of the original
Let V' = New data values
& A' = New average value
So, V1' = 70% of V1, V2' = 70% of V2, etc....

So taking 70% common from the equation inside the sqrt,
We will have

New S.D.' = Sqrt (70%) of Old S.D.

Hence new SD=Sqrt(70) of 10...

Is this right...pls correct me if i'm wrong...

Thanks