Standard Deviation

would go for A

the question asks how many numbers must be there?

for a collection of 4 odd numbers, if the range is 4, then the 2nd and the 3rd numbers must be the same. so in effect there are 3 distinct numbers (1st, 2nd/3rd and 4th)

[Post Edited]
Would go for C

All possible sets of odd numbers with range of 4:

x is odd:
1) x, x, x+2, x+4
2) x, x+2, x+2, x+4
3) x, x+2, x+4, x+4
4) x, x, x, x+4
5) x, x+4, x+4, x+4

All the above sets will have different SD

I would go for A as well.


I just picked numbers for this one. To maximize the the standard dev. I picked 1, 1, 1, 5

5-1 = 4 : range

avg: 8/4 = 2

greatest stand dev would be 5-2 = 3

So 3 is a Definite possibility

What is the OA?

It is 3

Range is the maximum distance between two observations inn a set but Sd is the summation of the differences squared (w.r.t mean) divided by the no of elements ( general ) ..........or its the spread w.r.t the mean

so its < or = Range

agoyal2 wrote:[Post Edited]
Would go for C

All possible sets of odd numbers with range of 4:

x is odd:
1) x, x, x+2, x+4
2) x, x+2, x+2, x+4
3) x, x+2, x+4, x+4
4) x, x, x, x+4
5) x, x+4, x+4, x+4

All the above sets will have different SD

OA given is B, i am not sure how.. but my approach is same as yours and the question is asking how many SDs are there.. so i'm 99 percent sure about C as the answer. please lemme know if i am missing something or or OA has some proof..

OK, I was getting impatient and googled the solution. I am not sure if that is against the rules of the forum but here is goes:

Possible Set (mean) [Differences from mean]

1 1 1 5 (2) [1, 1, 1, 3]
1 1 3 5 (2.5) [1.5, 1.5, 0.5, 2.5]
1 1 5 5 (3) [2, 2, 2, 2]
1 3 3 5 (3) [2, 0, 0, 2]
1 3 5 5 (3.5) [2.5, 0.5, 1.5, 1.5]
1 5 5 5 (4) [3, 1 , 1, 1]

There are 4 unique set of differences and hence 4 SDs.