For the data 2, 5, 6, 9, 12, we have the following five-number summary:
minimum = 2
first quartile = 3.5
median = 6
third quartile = 10.5
maximum = 12
IQR = 10.5 – 3.5 = 7, so 1.5·IQR = 10.5.
To determine if there are outliers we must consider the numbers that are 1.5·IQR or 10.5 beyond the quartiles.
first quartile – 1.5·IQR = 3.5 – 10.5 = –7
third quartile + 1.5·IQR = 10.5 + 10.5 = 21
Since none of the data are outside the interval from –7 to 21, there are no outliers. |