Problem Solving

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


My approach:

Though Not sure if a shorter method available but please advice if this one works!!

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

now we need to add 2 numbers and be as close as posible to the original SD which is sqrt(8)

assume A as the sum of the new variances ....

A= (X6 - mean)^2 + (X7 - mean)^2

we know that

(40 + A)/7 = 8


(X6 - mean)^2 + (X7 - mean)^2= 16
(X6 - 4)^2 + (X7 - 4)^2 = 16

by looking into the choices..... i get from choice D ==>8

and from choice E ==> 32

both 32 and 8 are equidistant from 16

SO MY QUESTION IS WHICH ONE SHOULD I GO FOR :O

Please refer

https://www.beatthegmat.com/shortcut-in- ... 10473.html

apoorva.srivastva wrote: assume A as the sum of the new variances ....

A= (X6 - mean)^2 + (X7 - mean)^2

we know that

(40 + A)/7 = 8


(X6 - mean)^2 + (X7 - mean)^2= 16
(X6 - 4)^2 + (X7 - 4)^2 = 16

I'm not understanding this part here. Can explain this part in more detail? I would like to learn this.

and as for picking between D and E, isn't 8 closer to 16 than 32?
|16-8|=8,
|32-16|=16

well i have tried adding the 6th and 7th elements... thats it... (x6 and x7)

A= (X6 - mean)^2 + (X7 - mean)^2

This question is about standard deviation, and you already calculated it to be sqrt(8). Forget variance.

Standard deviation is the measure of the spread from the mean to the points in the data set. This spread is about sqrt(8)= 2sqrt(2) ~ 2 * 1.4 = 2.8.

You noted that the mean is 4. You want to find the 2 points that are closest to a distance of 2.8 (standard deviation) from 4 (the mean).

Your ideal choice would be 4 - 2.8 = 1.2 and 4 + 2.8 = 6.8. Sadly, 1.2, 6.8 isn't a choice. 2, 6 are as close as we get so the choice is D.