PS--SD again

10 and 10 or -10 and -10?

IMO 10and10

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.

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.

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