Considering the standard deviation, the data-set has no outliers.
For the given data-set, the mean is:
(35+31+29+34+6+31+32+30)/8 = 28.5.
Hence the standard deviation is:
[tex]S = \sqrt{\frac{(35-28.5)^2+(31-28.5)^2+(29-28.5)^2+(34-28.5)^2+(6-28.5)^2+(31-28.5)^2+(32-28.5)^2+(30-28.5)^2}{8}} = 8.7[/tex]
A measure is said to be an outlier if it is more than 3 standard deviations for the mean. Hence the thresholds are:
Since all values are between 2.4 and 54.6, the data-set has no outliers.
More can be learned about the standard deviation of a data-set at https://brainly.com/question/24754716
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