Members of a wrestling team were weighed twice

[gmat_winter]

Members of a wrestling team were weighed twice. Was the standard deviation of their weights at the second weighing greater than the standard deviation of their weights at the first weighing?

(1) The second weighing showed that, since the first weighing, half of the team had lost 1 pound, while the other half had gained 1 pound.

(2) The standard deviation of all the weights at the first weighing was 0 pounds.

oa C

[DavidG@VeritasPrep]

S1: Take a very simple case. Let's say we have two members on this team.

Case1
First weighing: 100, 100, and standard deviation is 0.
Second weighing: 99, 101, standard deviation is positive. So YES, standard deviation is greater for second weighing.

Case2
First weighing 101, 99, standard deviation is positive
Second weighing 100,100, standard deviation is 0, so NO, standard deviation is NOT greater for second weighing.

S1: Not Sufficient


S2: if the standard deviation of the first weighing was 0, it means every member of the team had the same weight initially.

Case1
First weighing: 100, 100, and standard deviation is 0.
Second weighing: 99, 101, standard deviation is positive. So YES, standard deviation is greater for second weighing.

Case 2
First weighing: 100, 100, and standard deviation is 0.
First weighing: 100, 100, and standard deviation is 0. So NO, standard deviation is not greater for second weighing.

S2: Not Sufficient.

Together:
Initially, every member of the team had the same weight, and the standard deviation was 0. If half the team lost a pound and half the team gain a pound, we'll then have a positive standard deviation. So we know the answer will be YES, the second weighing will have a higher standard deviation.

The answer is C

[GMATGuruNY]

Members of a wrestling team were weighed twice. Was the standard deviation of their weights at the second weighing greater than the standard deviation of their weights at the first weighing?

(1) The second weighing showed that, since the first weighing, half of the team had lost 1 pound, while the other half had gained 1 pound.

(2) The standard deviation of all the weights at the first weighing was 0 pounds.

SD (standard deviation) describes how much data points DEVIATE from the mean.

Was the standard deviation of their weights at the second weighing greater than the standard deviation of their weights at the first weighing?
If the weights at the second weighing are MORE SPREAD OUT, then they will deviate more from the mean, increasing the SD.
Question stem, rephrased:
Were the weights at the second weighing more spread out?

Let the team consists of 4 members A, B, C and D.

Statement 1: The second weighing showed that, since the first weighing, half of the team had lost 1 pound, while the other half had gained 1 pound.
Case 1:
1st weighing: A=10, B=10, C=10, D=10
2nd weighing: A=9, B=9, C=11, D=11
In this case, the weights at the 2nd weighing are more spread out.

Case 2:
1st weighing: A=9, B=9, C=10, D=10
2nd weighing: A=10, B=10, C=9, D=9
In this case, the weights at the 2nd weighing are NOT more spread out.
INSUFFICIENT.

Statement 2: The standard deviation of all the weights at the first weighing was 0 pounds.
Since there is no deviation from the mean, the 4 weights at the first weighing are EQUAL.
No information about the 2nd weighing.
INSUFFICIENT.

Statements combined:
The 4 weights at the 1st weighing are equal, whereas the 4 weights at the 2nd weighing are NOT equal.
Implication:
The weights at the 2nd weighing are MORE SPREAD OUT.
SUFFICIENT.

The correct answer is C.