Answers:
- Range = 12.5
- Interquartile Range (IQR) = 7
- Interpretation: Both determine how spread out the data set is. The larger the value, the more spread out. The range handles the entire set, while the IQR only focuses on the middle 50% of the data.
- There are no outliers.
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Further explanation:
When going from smallest to largest, the data set sorts to
{23.5, 24, 25, 27, 29.25, 30.75, 31.25, 32, 33.5, 36}
The range is the difference of the smallest and largest items.
range = max - min
range = 36 - 23.5
range = 12.5
This is the gap between the smallest and largest values. It helps tell us how spread out the data set is. The larger the range, the more spread out the data set will be.
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There are n = 10 items in the data set.
The median is between slot 5 and 6 because n/2 = 10/2 = 5
The values in slots 5 and 6 are 29.25 and 30.75 in that order.
The midpoint of them is (29.25+30.75)/2 = 30 which is the median
Next, we'll form two smaller subsets L and U
L = lower subset of items below the median
L = {23.5, 24, 25, 27, 29.25}
U = upper subset of items above the median
U = {30.75, 31.25, 32, 33.5, 36}
The medians of sets L and U are 25 and 32 respectively. These are the Q1 and Q3 values
Q1 = first quartile = 25
Q3 = third quartile = 32
IQR = interquartile range
IQR = Q3 - Q1
IQR = 32 - 25
IQR = 7
The interquartile range is 7. It also tells us how spread out the data set is, but it only focuses on the middle 50% of the data.
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Now let's compute the lower fence (LF) and upper fence (UF). This will be helpful to see if we have any outliers.
We'll start with LF
LF = Q1 - 1.5*IQR
LF = 25 - 1.5*7
LF = 14.5
Is any value smaller than LF? The answer is no because the 23.5 is the min. We don't have any small outliers.
We'll need the upper fence UF as well
UF = Q3 + 1.5*IQR
UF = 32 + 1.5*7
UF = 42.5
Is any value above this upper fence? The answer is no because 36 is the max. We don't have any large outliers.
Every value in the data set is between LF = 14.5 and UF = 42.5; therefore, there are no outliers.
