PS--SD again

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nitya34: Master | Next Rank: 500 Posts; Posts: 179; Joined: Tue Jan 08, 2008 7:23 pm; Thanked: 11 times; GMAT Score:590

PS--SD again

by nitya34 » Tue Jul 21, 2009 6:12 am

Set X consists of 100 numbers. The average (arithmetic mean) of set X is 10, and the standard deviation is 4.6. Which of the following two numbers, when added to set X, will decrease the set’s standard deviation by the greatest amount?

-100 and -100;
-10 and -10;
0 and 0;
0 and 20;
10 and 10

OA-[spoiler]E.....>10 and 10[/spoiler]

source:www.magoosh.com

try the DS section of magoosh

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Source: Beat The GMAT — Problem Solving | Tue Jul 21, 2009 6:12 am

shibal: Master | Next Rank: 500 Posts; Posts: 345; Joined: Wed Mar 18, 2009 6:53 pm; Location: Sao Paulo-Brazil; Thanked: 12 times; GMAT Score:660

by shibal » Tue Jul 21, 2009 6:15 am

10 and 10 or -10 and -10?

IMO 10and10

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vzzai: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Tue Sep 02, 2008 12:13 am; Thanked: 1 times; GMAT Score:650

by vzzai » Fri Jun 03, 2011 9:16 pm

Could someone please provide detailed explanation?

I was under the opinion that if the mean is reduced by maximum then the SD will reduce by maximum. So, I was thinking that "-100 and -100" reduce the mean by greatest value. I'm wrong. Please explain.

Thank you,
Vj

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Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 am; Location: India; Thanked: 375 times; Followed by:53 members

by Frankenstein » Fri Jun 03, 2011 9:59 pm

Hi,
Let the elements be x1,x2,...x100, mean be 'm' and standard deviation be 'd'.
d^2=[(x1-m)^2+(x2-m)^2+...(x100-m)^2]/n = [(x1^2+x2^2+...x100^2)-nm^2]/n.
Lets denote (x1^2+x2^2+...x100^2) by p.
So, nd^2 = p-nm^2 => d^2=(p/n)-m^2 => (4.6)^2 = p/100 - 10^2 => p=12116
Options D and E will not change the mean but D adds 20^2 to p where as E adds (10^2)+(10^2) to p. So, 'd' will less in E compared to D.
For Op E, d^2= (p+200)/102 - 10^2 =~ (p/102 - 98)
For Op A, d^2= (p+2.10^4)/102 - (800/102)^2
For op B, d^2= (p+200)/102 - (980/102)^2
For op C, d^2= (p/102) - (1000/102)^2 =~ (p/102 - 96)

Clearly d^2 is smallest in op E.
Hence, E

There is theoretical approach to solve this. S.D is nothing but the variation from the mean. The lower the SD the closer are the points to the mean. Adding numbers far from the mean increases SD. As we come closer to the mean, the distribution curve shrinks making SD smaller as we move close to the mean. So, adding mean values to the set actually decreases SD to the maximum extent.

Cheers!

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cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 am; Location: India; Thanked: 310 times; Followed by:123 members; GMAT Score:750

by cans » Sat Jun 04, 2011 8:56 am

Set X consists of 100 numbers. The average (arithmetic mean) of set X is 10, and the standard deviation is 4.6. Which of the following two numbers, when added to set X, will decrease the set's standard deviation by the greatest amount?

-100 and -100;
-10 and -10;
0 and 0;
0 and 20;
10 and 10

avg of 100=10. Thus sum is 1000. sum of 102 numbers = (1000+a+b)/102
a)-100 and -100, mean is nearly equal to 8. But as two numbers -100,-100 are too far away from the mean, it will increase the s.d.
b)-10,-10 will effect mean a little, but because numbers are far from mean, thus s.d. will increase
similar for c) and d)
for e)10,10. Mean remains same and thus maximum decrease in s.d.
IMO E

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