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

1) a, b and c are consecutive integers with a < b < c. 2) m = 2

OA:A

Given : \(\frac{a+b+c}{3}= m\) is standard deviation of a,b and c less than 1?

Statement 1 : \(a, b\) and \(c\) are consecutive integers with \(a < b < c\). It means \(b\) is mean, Standard Deviation, \(S.D =\sqrt{\frac{(a-b)^2+(b-b)^2+(c-b)^2}{3}}\)

\(a-b=-1,c-b=1\) as \(a,b,c\) are consecutive number.

Standard Deviation, \(S.D =\sqrt{\frac{(-1)^2+(0)^2+(1)^2}{3}} =\sqrt{\frac{2}{3}}<1\)

Statement 1 alone is sufficient.

2) \(m = 2\)

Now Plugging in number,taking \(a=1,b=2\) and \(c=3\) \(S.D<1\) as already seen in statement 1 that 3 consecutive number have S.D less than 1. or We could have taken \(a=2,b=2,c=2\), then also m would have been \(2\). In that case \(S.D=0\) Is S.D<1 for a,b,c : Yes

Now taking \(a=-4, b =2, c = 8\), mean would be 2 as required by Statement 2. Standard Deviation, \(S.D =\sqrt{\frac{(-4-2)^2+(2-2)^2+(8-2)^2}{3}}= \sqrt{\frac{72}{3}}=2\sqrt{6}>1\) Is S.D<1 for a,b,c : No

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

1) a, b and c are consecutive integers with a < b < c. 2) m = 2

OA:A

Given : \(\frac{a+b+c}{3}= m\) is standard deviation of a,b and c less than 1?

Statement 1 : \(a, b\) and \(c\) are consecutive integers with \(a < b < c\). It means \(b\) is mean, Standard Deviation, \(S.D =\sqrt{\frac{(a-b)^2+(b-b)^2+(c-b)^2}{3}}\)

\(a-b=-1,c-b=1\) as \(a,b,c\) are consecutive number.

Standard Deviation, \(S.D =\sqrt{\frac{(-1)^2+(0)^2+(1)^2}{3}} =\sqrt{\frac{2}{3}}<1\)

Statement 1 alone is sufficient.

2) \(m = 2\)

Now Plugging in number,taking \(a=1,b=2\) and \(c=3\) \(S.D<1\) as already seen in statement 1 that 3 consecutive number have S.D less than 1. or We could have taken \(a=2,b=2,c=2\), then also m would have been \(2\). In that case \(S.D=0\) Is S.D<1 for a,b,c : Yes

Now taking \(a=-4, b =2, c = 8\), mean would be 2 as required by Statement 2. Standard Deviation, \(S.D =\sqrt{\frac{(-4-2)^2+(2-2)^2+(8-2)^2}{3}}= \sqrt{\frac{72}{3}}=2\sqrt{6}>1\) Is S.D<1 for a,b,c : No

Statement 2 alone is not sufficient.

'

Bismarck what if you took 20, 21, 22 for statement one

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

1) a, b and c are consecutive integers with a < b < c. 2) m = 2

OA:A

Given : \(\frac{a+b+c}{3}= m\) is standard deviation of a,b and c less than 1?

Statement 1 : \(a, b\) and \(c\) are consecutive integers with \(a < b < c\). It means \(b\) is mean, Standard Deviation, \(S.D =\sqrt{\frac{(a-b)^2+(b-b)^2+(c-b)^2}{3}}\)

\(a-b=-1,c-b=1\) as \(a,b,c\) are consecutive number.

Standard Deviation, \(S.D =\sqrt{\frac{(-1)^2+(0)^2+(1)^2}{3}} =\sqrt{\frac{2}{3}}<1\)

Statement 1 alone is sufficient.

2) \(m = 2\)

Now Plugging in number,taking \(a=1,b=2\) and \(c=3\) \(S.D<1\) as already seen in statement 1 that 3 consecutive number have S.D less than 1. or We could have taken \(a=2,b=2,c=2\), then also m would have been \(2\). In that case \(S.D=0\) Is S.D<1 for a,b,c : Yes

Now taking \(a=-4, b =2, c = 8\), mean would be 2 as required by Statement 2. Standard Deviation, \(S.D =\sqrt{\frac{(-4-2)^2+(2-2)^2+(8-2)^2}{3}}= \sqrt{\frac{72}{3}}=2\sqrt{6}>1\) Is S.D<1 for a,b,c : No

Statement 2 alone is not sufficient.

'

Bismarck what if you took 20, 21, 22 for statement one

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

1) a, b and c are consecutive integers with a < b < c. 2) m = 2

Target question:Is the standard deviation of a, b and c less than 1?

Statement 1: a, b and c are consecutive integers with a < b < c. It's important to know that standard deviation is a measure of dispersion (how spread apart the values are). So, ANY 3 consecutive integers will have the same standard deviation. For example, the standard deviation of {1,2,3} = the standard deviation of {6,7,8} = the standard deviation of {23,24,25} etc So, IF we calculate the standard deviation of {1,2,3} then THAT value will provide sufficient info to the answer to the target question Since we COULD answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: m = 2 There are several values a, b and c that satisfy statement 2. Here are two: Case a: a = 2, b = 2 and c = 2 (mean = 2 and standard deviation = 0). In this case, the answer to the target question is YES, the standard deviation IS less than 1 Case b: a = -100, b = 0 and c = 106 (mean = 2 and standard deviation = some number much greater than 1. In this case, the answer to the target question is NO, the standard deviation is NOT less than 1 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Re: If the average (arithmetic mean) of a, b and c is m, is their standard
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10 Aug 2018, 02:31

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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.

We have \(m = \frac{( a + b + c )}{3}\) from the original condition. Since we have 4 variables and 1 equation, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2) Since \(a, b\) and \(c\) are consecutive integers, we can assume \(a = b – 1\) and \(c = b + 1\) and \(b = m\). Since \(m = 2\) by condition 2, we have \(a = 1, b = 2, c = 3.\) The standard deviation is

\(\sqrt{\frac{a-b^2+(b-b)^2+(c-b)^2}{3}}\)

=\(\sqrt{\frac{(-1)^2+0^2+1^2}{3}}\) = \(\sqrt{\frac{2}{3}}\) which is less than 1. Thus, both conditions are sufficient, when considered together.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1) Since \(a, b\) and \(c\) are consecutive integers, we can assume that \(a = b – 1\) and \(c = b + 1\) and \(b = m\). Then \(a – b = -1\) and \(c – b = 1\), and the standard deviation is \(\sqrt{\frac{a-b^2+(b-b)^2+(c-b)^2}{3}}\)

=\(\sqrt{\frac{(-1)^2+0^2+1^2}{3}}\) = \(\sqrt{\frac{2}{3}}\) which is less than 1. Thus, condition 1) is sufficient on its own.

Condition 2) Condition 2) is not sufficient since it gives us no information about \(a, b\) and \(c\).

Therefore, A is the answer.

Answer: A

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.
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