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What is the standard deviation of x1, x2, …, xn?

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What is the standard deviation of x1, x2, …, xn?

by Max@Math Revolution » Thu Apr 16, 2020 6:51 am

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[GMAT math practice question]

What is the standard deviation of x1, x2, …, xn?

1) The average of x1, x2, …, xn is 1.
2) The average of x1^2, x2^2, …, xn^2 is 5.
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Source: — Data Sufficiency |
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Re: What is the standard deviation of x1, x2, …, xn?

by Max@Math Revolution » Sun Apr 19, 2020 7:30 pm

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. A standard deviation is the square root of an average of squares minus the square of the average. Thus, C is the answer.

The following is more detailed reasoning.

Assume m is the average of x1, x2, …, xn and p is the average of x1^2, x2^2, …, xn^2.
Then we have (x1 + x2 + … + xn) / n = m or (x1 + x2 + … + xn) = mn, and we have
(x1^2+x2^2+⋯+xn^2)/n= p or (x1^2+x2^2+⋯+xn^2)= pn.

Then we have
(x1-m)^2+(x2-m)^2+⋯+(xn-m)^2/n
= x1^2 + x2^2 +⋯+ xn^2- 2m(x1 + x2 + ⋯ + xn) + nm^2/n
= np - 2m∙nm + nm^2/n = p - m^2
The standard deviation is \(\sqrt{p-m^2}\) .
Since we have m = 1 and p = 5 from both conditions 1) and 2), we have the standard deviation \(\sqrt{5-1^2}=\sqrt{4}\) =2.
Since both conditions together yield a unique solution, they are sufficient.

Therefore, C is the answer.
Answer: C
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