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16 Sep 2014, 01:31
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Is the standard deviation of list \(A=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x\}\) greater than the standard deviation of list \(B=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x, \ y \}\)?
(1) The standard deviation of list A is positive.
(2) \(y=3\).
(1) The standard deviation of list A is positive.
(2) \(y=3\).
16 Sep 2014, 01:31
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Official Solution:
Is the standard deviation of list \(A=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x\}\) greater than the standard deviation of list \(B=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x, \ y \}\)?
First, note that the mean of list A is \(\frac{(3-2x) +(3-x)+3+(3+x)+(3+2x)}{5}=3\). Additionally, it's important to note that list B is simply list A with the addition of the element \(y\).
(1) The standard deviation of list A is positive.
We know that the standard deviation of any list is greater than or equal to zero. The standard deviation of a list is zero only when the list consists of identical elements. So, this statement implies that list A does NOT consist of identical elements or that \(x\) does not equal zero. However, this statement is not sufficient to answer the question.
(2) \(y=3\).
We know that the mean of list A is 3. If \(x \neq 0\) (e.g., if \(x=1\)), then \(A=\{1, \ 2, \ 3, \ 4, \ 5\}\) and \(B = \{1, \ 2, \ 3, \ 3, \ 4, \ 5\}\). The standard deviation of B would be smaller than the standard deviation of A, since the elements of B are less widespread than the elements of list A. However, if \(x=0\), then \(A=\{3, \ 3, \ 3, \ 3, \ 3\}\) and \(B=\{3, \ 3, \ 3, \ 3, \ 3, \ 3\}\), so both will have a standard deviation of zero. This statement alone is not sufficient.
(1)+(2) We know that the mean of list A is 3 and it does not consist of identical elements. Since \(y=3\), then list B is list A with the addition of the element 3. The addition of a new element equal to the mean in list B makes list B less widespread than list A, resulting in a decrease in the standard deviation. Consequently, \(SD(A) \gt SD(B)\). Combining both statements provides sufficient information to answer the question.
Answer: C
Is the standard deviation of list \(A=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x\}\) greater than the standard deviation of list \(B=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x, \ y \}\)?
First, note that the mean of list A is \(\frac{(3-2x) +(3-x)+3+(3+x)+(3+2x)}{5}=3\). Additionally, it's important to note that list B is simply list A with the addition of the element \(y\).
(1) The standard deviation of list A is positive.
We know that the standard deviation of any list is greater than or equal to zero. The standard deviation of a list is zero only when the list consists of identical elements. So, this statement implies that list A does NOT consist of identical elements or that \(x\) does not equal zero. However, this statement is not sufficient to answer the question.
(2) \(y=3\).
We know that the mean of list A is 3. If \(x \neq 0\) (e.g., if \(x=1\)), then \(A=\{1, \ 2, \ 3, \ 4, \ 5\}\) and \(B = \{1, \ 2, \ 3, \ 3, \ 4, \ 5\}\). The standard deviation of B would be smaller than the standard deviation of A, since the elements of B are less widespread than the elements of list A. However, if \(x=0\), then \(A=\{3, \ 3, \ 3, \ 3, \ 3\}\) and \(B=\{3, \ 3, \ 3, \ 3, \ 3, \ 3\}\), so both will have a standard deviation of zero. This statement alone is not sufficient.
(1)+(2) We know that the mean of list A is 3 and it does not consist of identical elements. Since \(y=3\), then list B is list A with the addition of the element 3. The addition of a new element equal to the mean in list B makes list B less widespread than list A, resulting in a decrease in the standard deviation. Consequently, \(SD(A) \gt SD(B)\). Combining both statements provides sufficient information to answer the question.
Answer: C
General Discussion
fitzpratik
24 Jul 2017, 13:23
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can someone please explain why mean of set A is 3?
thanks a ton
thanks a ton
