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14 Aug 2019, 03:28
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C
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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.
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.
14 Aug 2019, 08:24
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The average of set R is 230, because it's an equally spaced set, so its average and median are equal. When we create set S, we're removing two things from R. If, from Statement 1, we do not change the average by removing two things, we must be removing two things that average out to 230. But without knowing which values we're removing, we can't tell what the standard deviation will be - if we take out the two extreme values to get this set:
225, 227, 229, 231, 233, 235
the values will be clustered more closely together (and we'll thus have a lower standard deviation) than if we take out the two middle values to get this set:
223, 225, 227, 233, 235, 237
Using only Statement 2, we only know one of the two values we're removing from the set. Again, if we take out the two extreme values, we can get this set:
225, 227, 229, 231, 233, 235
which is clustered more closely around its average than this other possible set:
225, 227, 229, 233, 235, 237
so we can get two sets with different standard deviations. Using both Statements, it must be true that we're removing the smallest and largest values if the average remains unchanged, so we know exactly what set S is, and can thus compute its standard deviation. So the answer is C.
225, 227, 229, 231, 233, 235
the values will be clustered more closely together (and we'll thus have a lower standard deviation) than if we take out the two middle values to get this set:
223, 225, 227, 233, 235, 237
Using only Statement 2, we only know one of the two values we're removing from the set. Again, if we take out the two extreme values, we can get this set:
225, 227, 229, 231, 233, 235
which is clustered more closely around its average than this other possible set:
225, 227, 229, 233, 235, 237
so we can get two sets with different standard deviations. Using both Statements, it must be true that we're removing the smallest and largest values if the average remains unchanged, so we know exactly what set S is, and can thus compute its standard deviation. So the answer is C.
19 Apr 2026, 03:42
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A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.
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