Standard Deviation of a Subset

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R = {223, 225, 227, 229, 231, 233, 235, 237}

Given Set R, Set S is formed such that it contains exactly 6 distinct numbers that are members of Set R. What is the standard deviation of the numbers in Set S?

(1) The average (arithmetic mean) of the numbers that are members of Set S is equal to the average of the numbers that are members of Set R.

(2) Set S does not contain 223.

Review: Standard Deviation of a Subset




Explanation

We're told about a set "formed" from R. R has 8 elements...

{223, 225, 227, 229, 231, 233, 235, 237}

...and S will have 6 distinct elements, so choosing 6 elements to include in S is equivalent to choosing 2 elements of R to knock out. If we know exactly what two elements are knocked out, then we know exactly what's in S, and we'll know the standard deviation. There might be other ways to infer the standard deviation that involve knowing what the dispersion is. Let's turn to the data statements, evaluating them separately first.

Statement (1) tells us that the average of the two sets are equal. We can analyze this by cases. Set R is evenly distributed around 230; that's its average and hence must be the average of Set S, according to this data statement. Case I: we form Set S by knocking out the 1st and the 8th number. That set will have an average of 230. Case II: we form Set S by knocking out the 2nd and the 7th numbers. That set will also have an average of 230. The dispersions of the two versions of Set S are different: the Set S of Case I is more tightly bunched. Hence, we don't have sufficient information to determine the standard deviation definitively. Statement (1) is insufficient.

Statement (2) tells us one of the numbers knocked out of Set R in the formation of Set S, the first number. We are not told the other number that is knocked out. In one case, the other number that is knocked out is, say, the 2nd number.

{ ___, ___, 227, 229, 231, 233, 235, 237}

In another case, it could be the 7th number.

{ ___, 225, 227, 229, 231, 233, ___, 237}

The set in the former case is more tightly packed than the set in the latter case, so two allowed cases will yield different possible standard deviations. We have insufficient information to answer the question, so Statement (2) alone is insufficient.

Combining the two statements, we need to form Set S such that 223 is one of the numbers we knock out and the average of Set S is 230. There is only one possible way to do that, and that is for 237 to be the other number that we knock out; otherwise the average will be disrupted, because the "pull" of the numbers on either side of 230 will not be balanced. Therefore, the elements of Set S are uniquely determined, and that means the standard deviation could be computed directly. We have sufficient information to answer the question.

The correct answer is (C).


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