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

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Math Expert
Joined: 02 Sep 2009
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Members of a wrestling team were weighed twice. Was the standard devia  [#permalink]

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02 Jun 2018, 12:46
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Difficulty:

75% (hard)

Question Stats:

21% (01:33) correct 79% (01:29) wrong based on 34 sessions

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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.

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Re: Members of a wrestling team were weighed twice. Was the standard devia  [#permalink]

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02 Jun 2018, 13:01
Bunuel wrote:
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.

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.

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Re: Members of a wrestling team were weighed twice. Was the standard devia  [#permalink]

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02 Jun 2018, 15:56
Bunuel wrote:
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.

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.

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Re: Members of a wrestling team were weighed twice. Was the standard devia  [#permalink]

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04 Jul 2018, 18:51
2
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

Give kudos if it helps

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Re: Members of a wrestling team were weighed twice. Was the standard devia &nbs [#permalink] 04 Jul 2018, 18:51
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