Answer :
In this data set, the only value that is an outlier is 100, since it is above the upper bound of 98.25.
To identify the outliers in the data set, we can use the concept of the interquartile range (IQR) and the 1.5×IQR criterion.
First, we need to find the first and third quartiles (Q1 and Q3) of the data set. To do this, we can order the data set from smallest to largest:
54, 72, 75, 81, 81, 82, 83, 83, 83, 85, 100
The median of the data set is the middle value, which is 82.
The lower half of the data set will consists of:
54, 72, 75, 81, 81
The median of the lower half is (72 + 75)/2 = 73.5, which is the value halfway between the two middle values.
The upper half of the data set will consists of:
83, 83, 83, 85, 100
The median of the upper half is (83 + 85)/2 = 84, which is the value halfway between the two middle values.
Therefore, the first quartile (Q1) is 73.5 and the third quartile (Q3) is 84.
The interquartile range (IQR) is the difference between Q3 and Q1:
IQR = Q3 - Q1 = 84 - 73.5 = 10.5
To identify the outliers in the data set using the 1.5×IQR criterion, we need to calculate the lower and upper bounds:
Lower bound = Q1 - 1.5×IQR = 73.5 - 1.5×10.5 = 57.75
Upper bound = Q3 + 1.5×IQR = 84 + 1.5×10.5 = 98.25
Any data point that is below the lower bound or above the upper bound is considered an outlier.
To know more about interquartile range here
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