Members of a wrestling team were weighed twice. Was the standard devia

Question

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.

CrackverbalGMAT · Accepted Answer

In this question, a very subtle trap is laid out in the data given in the first statement. The first statement says that half of the team lost 1 pound in the second weighing and the other half gained 1 pound. All well till now.
But, it does not tell us which half lost weight and which one gained. It does not tell us if weights of some persons are equal or not. This essentially means, each one of us can assume our own values and end up getting different answers.

For example, if we assume that there are 4 students in the team with weights of 10, 20, 30 and 40. Mean = 25. Deviations of the individual numbers from the mean are -15, -5, 5 and 15.
Square of these deviations = 225, 25, 25 and 225.

Now, in the second weighing, if the first two are reduced by 1 each and the next two are increased by 1 each, then,
Mean = 25
Deviations = -16, -6, 6 and 16. These deviations will correspond to an increase in the SD.

On the other hand, if the first two are increased by 1 each and the next two are reduced by 1 each, then,
Mean = 25
Deviations = -14, 4, 4, 14. These deviations will correspond to a decrease in the SD.

Clearly, we have a YES and NO situation for the question asked. We don’t even have to take any other case.
Statement I alone is insufficient. Answer options A and D can be eliminated. Possible answer options are B, C or E.

From statement II, we can infer that the weights of the team members are all same (or ZERO, which is illogical considering that we are talking about the weights of members of a Wrestling team

). Only when all the values are equal will the SD be Zero.

But, statement II doesn’t give us any information about the second weighing. Hence, statement II alone is insufficient. Answer option B can be eliminated.

When we combine both the statements, we know that all the values in the data set are equal. Also, the SD in the first weighing is ZERO. If one half of the values is now increased by 1 and the other half reduced by 1, the deviations wrt mean will increase. This will cause the SD to increase in the second weighing.
Values in which half are increased and in which half are decreased does not matter since all the values are equal.
The combination of statements is sufficient.

The correct answer option is C.

Hope this helps!

). Only when all the values are equal will the SD be Zero. But, statement II doesn’t give us any information about the second weighing. Hence, statement II alone is insufficient. Answer option B can be eliminated. When we combine both the statements, we know that all the values in the data set are equal. Also, the SD in the first weighing is ZERO. If one half of the values is now increased by 1 and the other half reduced by 1, the deviations wrt mean will increase. This will cause the SD to increase in the second weighing. Values in which half are increased and in which half are decreased does not matter since all the values are equal. The combination of statements is sufficient. The correct answer option is C. Hope this helps!

push12345 · Answer

I would disagree with above solution

We dont know previous mean so how can we conclude that sd for first statement will increase

Eg 48,48,52,52
Mean is 50
Taking square of difference between mean and terms

Variance=4+4+4+4/4=16/4=4
Taking root for sd so sd=2

Applying first statement

I)49,49,51,51
Here sd will decrease as gap between mean and numbers decrease

II)47,47,53,53
Here sd will increase as difference between number and mean increase

So first statement is not sufficient

We require both statements to answer question

So C is answer

Give kudos if it helps

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DavidTutorexamPAL · Answer

Questions involving standard deviations often rely on basic properties of the standard deviation and not calculations.
We'll look for such a solution, a Logical approach.

(1) Since the standard deviation reflects 'how far apart the datapoints are from each other', or, in equations, is calculated based on the differences of all values from the mean, then this statement is exactly what we need! If half the team lost 1 pound and the other gained 1 pound, the mean stayed the same but all the weights were further apart from it (so the std increased).
Sufficient.

(2) Since this gives us no information on the second weighing, it can't be enough. (Trivia: the standard deviation is 0 only when all of the numbers are equal!)
Insufficient.

(A) is our answer.

jackspire · Answer

In the first case, the mean remains same and individual values are going away from the mean. So standard deviation, which is a representation of differences in individual values from the mean, increases. So, it's sufficient to answer.

From second option we can't obtain the standard deviation of second weighing. It's not sufficient to answer.

So, the answer is A.

SiffyB · Answer

Bunuel , the source of this question is 800Score CAT test.

Bunuel · Answer

Thank you! Added the source.

krndatta · Answer

Bunuel

Can you provide an explanation for the above question? 
Not very clear with the explanations provided.

Thanks

bumpbot · Answer

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Members of a wrestling team were weighed twice. Was the standard devia

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