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# A certain list of 100 data has an average (arithmetic mean)

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Intern
Joined: 31 May 2006
Posts: 44
Followers: 0

Kudos [?]: 5 [0], given: 0

A certain list of 100 data has an average (arithmetic mean) [#permalink]  04 Nov 2006, 18:48
A certain list of 100 data has an average (arithmetic mean) of 6 and a standard deviation of d, where d is positive. Which of the following pairs of data, when added to the list, must result in a list of 102 data with standard deviation less than d?

a) -6 and 0
b) 0 and 0
c) 0 and 6
d) 0 and 12
e) 6 and 6

Manager
Joined: 22 Jul 2006
Posts: 50
Followers: 0

Kudos [?]: 1 [0], given: 0

it was probably bcs i knew the answer, but here is how i found it..!

assuming the no.s are a1, a2...a100

(a1-6)^2 + (a2-6)^2 + (a3-6)^2....+(a100-6)^2 = 100*d^2

When u add the two new no.s the AM is still going to be around 6

So let us find the SD with option 1: -6,0
(a1-6)^2 + (a2-6)^2 + (a3-6)^2....+(a100-6)^2 + (-6-6)^2 + (0-6)^2

So ur object should be two get the last two terms to a minimum when it is option 5: 6,6 it becomes 0,0..which will result in an SD < d..!
Senior Manager
Joined: 05 Oct 2006
Posts: 268
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Kudos [?]: 4 [0], given: 0

hey sam,

u got any material on std dev.
i wasn't aware of (

(a1-mean^2)+........= 100*std dev^2

choice e looks like it because nothing is added on left hand side so for 102.D^2 = 100*d^2...D<d
Manager
Joined: 22 Jul 2006
Posts: 50
Followers: 0

Kudos [?]: 1 [0], given: 0

(a1-mean^2)+........= 100*std dev^2

I got it by simplifying

{[(a1-mean^2)+.......+(a100-mean^2)]/100}^1/2= std dev

Yes ur right regarding the second part, the numerator is constant but denoninator has increased to 102, hence the new SD decreases..
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