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22 Jul 2011, 15:02
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A data set consists of three integers. What is the standard deviation of this data set?
(1) The average (arithmetic mean) and the median are both 3.
(2) At least two of the numbers are 3.
Can we use negative numbers to calculate mean/median in such problems ?
(1) The average (arithmetic mean) and the median are both 3.
(2) At least two of the numbers are 3.
Can we use negative numbers to calculate mean/median in such problems ?
22 Jul 2011, 17:13
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Why would you use negative numbers?
In Stmnt 1. average of 3 numbers is 3 -> so at least one number has to be +ve
In Stmnt 2. two of the numbers are +ve.
In Stmnt 1. average of 3 numbers is 3 -> so at least one number has to be +ve
In Stmnt 2. two of the numbers are +ve.
23 Jul 2011, 22:19
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We can't rule out negative numbers in advance, but we don't need them.
(1) Since the median equals the mean, and we have only three numbers, we know that the two other numbers must be equidistant from the mean. So, to list a few examples, we could have:
3,3,3
2,3,4
0,3,6
and yes, -5,3,11
These all have different standard deviations. Insufficient.
(2) At least 2 of the numbers are 3, but the third number could be anything:
3,3,3
0,3,3
3,3,55
-1,000,047, 3, 3
. . . etc. These all have different standard deviations. Insufficient.
(1&2) Two of the numbers are 3, and the end numbers must be equidistant from the mean/median.
If the median and one other number are 3, then the third number would also have to be 3 to be equidistant from the median.
If the ends are both 3, the median would have to be 3. After all, there's nothing in between 3 and 3.
The set is 3,3,3. The SD is 0. Sufficient.
(1) Since the median equals the mean, and we have only three numbers, we know that the two other numbers must be equidistant from the mean. So, to list a few examples, we could have:
3,3,3
2,3,4
0,3,6
and yes, -5,3,11
These all have different standard deviations. Insufficient.
(2) At least 2 of the numbers are 3, but the third number could be anything:
3,3,3
0,3,3
3,3,55
-1,000,047, 3, 3
. . . etc. These all have different standard deviations. Insufficient.
(1&2) Two of the numbers are 3, and the end numbers must be equidistant from the mean/median.
If the median and one other number are 3, then the third number would also have to be 3 to be equidistant from the median.
If the ends are both 3, the median would have to be 3. After all, there's nothing in between 3 and 3.
The set is 3,3,3. The SD is 0. Sufficient.
