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Sagot :
Answer: Mean
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Explanation:
Let's sort the list to get
2, 3, 4, 6, 7, 8, 8, 9, 10, 12
There are n = 10 items in this list. The median is between slot n/2 = 10/2 = 5 and slot 6. The values in these slots are 7 and 8 in that order.
The midpoint of those items is (7+8)/2 = 15/2 = 7.5
The median of this data set is 7.5
Now form the set of values smaller or lower than the median
L = {2,3,4,5,6,7}
The median of set L is 4, so this is the Q1 or first quartile value
Q1 = 4
Now form the set of numbers larger than the median, which I'll call set U for "upper set".
U = {8, 8, 9, 10, 12}
The median of set U is 9, telling us that Q3 = 9
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We have this five number summary
- min = 2
- Q1 = 4
- median = 7.5
- Q3 = 9
- max = 12
The min and max should be fairly self-explanatory. They are the smallest and largest items respectively.
Based on that, let's now compute the IQR (interquartile range)
IQR = Q3 - Q1
IQR = 9 - 4
IQR = 5
It allows us to compute the lower fence (LF) and the upper fence (UF)
For more information, search out "tukey fence".
LF = Q1 - 1.5*IQR = 4 - 1.5*5 = -3.5
UF = Q3 + 1.5*IQR = 9 + 1.5*5 = 16.5
These fences help us determine if we have any outliers or not.
If all of the data points are between LF and UF, then we have no outliers. Otherwise, we'll have outliers.
In this case, there are no values smaller than -3.5; also, there are no values larger than 16.5 either. Therefore, this data set has no outliers
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If there are no outliers, then the mean is the best measure of center of a set of numbers. If there were outliers, then the median is the next best thing. This is because the mean is affected by outliers. The more distant, the worst the mean is at being the center.
For a real world example, check out concepts like the median home price.
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