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If the average (arithmetic mean) of a, b, c, and d is m, is their stan

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If the average (arithmetic mean) of a, b, c, and d is m, is their stan [#permalink]

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New post 01 Jan 2018, 02:20
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C
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E

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[GMAT math practice question]

If the average (arithmetic mean) of \(a, b, c\), and \(d\) is \(m\), is their standard deviation greater than \(1\)?

1) \(a=1\)
2) \(m=4\)

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If the average (arithmetic mean) of a, b, c, and d is m, is their stan [#permalink]

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New post 01 Jan 2018, 04:41
MathRevolution wrote:
[GMAT math practice question]

If the average (arithmetic mean) of \(a, b, c\), and \(d\) is \(m\), is their standard deviation greater than \(1\)?

1) \(a=1\)
2) \(m=4\)


Statement 1: Nothing mentioned about other variables. if \(a=b=c=d\), then \(SD=0\) but for other values \(SD>0\). Insufficient

Statement 2: Same as Statement 1, nothing mentioned about other variable. Insufficient

Combining 1 & 2: We know \(SD\) is the variation from mean. So even if we assume that \(b\), \(c\), & \(d\) are equal to mean (which will not be in this case) resulting in \(0\) variation, \(a\) will have variation.

So variation of \(a\) from mean \(= 1-4=-3\) and other variations \(= 0\)(assuming)

\(Variance =\) average of square of variations \(= \frac{(-3)^2+0^2+0^2+0^2}{4}=2.25\)

So minimum \(SD= \sqrt{2.25}>1\) (square root of any number greater than \(1\) will be greater than \(1\)). Hence Sufficient

Option C
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Re: If the average (arithmetic mean) of a, b, c, and d is m, is their stan [#permalink]

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New post 03 Jan 2018, 01:38
=>

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.

The average of \(a,b,c\) and \(d\) is
\(\frac{( a + b + c + d )}{4} = m\).
Since we have 5 variables and 1 equation, E is most likely to be the answer. So, we should consider conditions 1) and 2) together first.

Conditions 1) and 2):
Standard deviation
= \(\frac{{√(a–m)^2 + (b-m)^2 + (c-m)^2 + (d-m)^2}}{4}\)
≥ \(√ (\frac{3^2}{4} )\)
= \(√ (\frac{9}{4} )\)
= \(\frac{3}{2}\)
> \(1\)
Both conditions 1) and 2) together are sufficient.

Since this is a statistics question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1):
If \(a = b = c = d = 1\), the standard deviation is \(0<1\), and the answer is ‘no’.
If \(a = 1, b = 4, c = 4, d = 7\), the standard deviation is \(4.5 > 1\), and the answer is ‘yes’.
Since we do not have a unique answer, condition 1) is not sufficient.

Condition 2):
If \(a = b = c = d = 4, m=4\), the standard deviation is \(0<1\), and the answer is
‘no’.
If \(a = 1, b = 4, c = 4, d = 7, m=4\), the standard deviation is \(4.5 > 1\), and the answer is ‘yes’.
Since we do not have a unique answer, condition 2) is not sufficient.

Therefore, C is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

Answer: C
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Re: If the average (arithmetic mean) of a, b, c, and d is m, is their stan   [#permalink] 03 Jan 2018, 01:38
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