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04 Sep 2009, 18:29
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B
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61% (00:52) correct
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If each term in set \(T\) is a multiple of 5, is the standard deviation of set \(T\) positive?
(1) Each term in set \(T\) is positive.
(2) Set \(T\) contains only one term.
M08-09
(1) Each term in set \(T\) is positive.
(2) Set \(T\) contains only one term.
M08-09
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22 Oct 2012, 09:18
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Official Solution:
If each term in set \(T\) is a multiple of 5, is the standard deviation of set \(T\) positive?
The standard deviation is a measure of the variation of the data points from the mean, a measure of how widespread a given set is. When the standard deviation is low, the data points tend to be close to the mean, while a high standard deviation implies that the data is spread out over a broader range of values. Essentially, it can be thought of as a measure of distance, and because distance can't be negative, the standard deviation of any dataset is always zero or greater than zero: \(SD \ge 0\).
Moreover, a set's standard deviation will be zero if and only if all elements in the set are identical, or equivalently, if the set consists of a single element.
(1) Each term in set \(T\) is positive.
If \(T=\{5\}\), the standard deviation is 0 since all elements are identical. However, if set \(T=\{5, 10\}\), the standard deviation is greater than zero, because the elements differ. Not sufficient.
(2) Set \(T\) contains only one term.
Any set with a single term always has a standard deviation of zero, as there's no variance between the single term and itself. Therefore, this statement alone is sufficient to answer the original question. The standard deviation of set \(T\) is not positive; it is zero.
Answer: B
If each term in set \(T\) is a multiple of 5, is the standard deviation of set \(T\) positive?
The standard deviation is a measure of the variation of the data points from the mean, a measure of how widespread a given set is. When the standard deviation is low, the data points tend to be close to the mean, while a high standard deviation implies that the data is spread out over a broader range of values. Essentially, it can be thought of as a measure of distance, and because distance can't be negative, the standard deviation of any dataset is always zero or greater than zero: \(SD \ge 0\).
Moreover, a set's standard deviation will be zero if and only if all elements in the set are identical, or equivalently, if the set consists of a single element.
(1) Each term in set \(T\) is positive.
If \(T=\{5\}\), the standard deviation is 0 since all elements are identical. However, if set \(T=\{5, 10\}\), the standard deviation is greater than zero, because the elements differ. Not sufficient.
(2) Set \(T\) contains only one term.
Any set with a single term always has a standard deviation of zero, as there's no variance between the single term and itself. Therefore, this statement alone is sufficient to answer the original question. The standard deviation of set \(T\) is not positive; it is zero.
Answer: B
04 Sep 2009, 19:57
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my answer is B..
1.) Let T = { 5,15}
Mean = 10
\(sd =\sqrt{\frac{((5 - 10)^2 + (15-10)^2)}{2}}\)
=>\(sd = 5\) > 0
but, if T = {5,5}
Then Sd = 0 ,, uncertainity,,,hence, insufficient..
2.) T has only one member..
sd = 0 hence, sufficient.. sd is not positive..
1.) Let T = { 5,15}
Mean = 10
\(sd =\sqrt{\frac{((5 - 10)^2 + (15-10)^2)}{2}}\)
=>\(sd = 5\) > 0
but, if T = {5,5}
Then Sd = 0 ,, uncertainity,,,hence, insufficient..
2.) T has only one member..
sd = 0 hence, sufficient.. sd is not positive..
