Answer :
An outlier is a value that lies far away from most of the values in a data set. The outlier(s) in the following data set is/are: 101.2.First, we can arrange the given data in ascending order:0.1, 0.25, 0.45, 0.68, 1, 1.38, 1.77, 2.5, 3.32, 4.23, 9.04, 101.2Outliers are values that fall outside the acceptable range of a set of data.
An outlier will be a value that is more than 1.5 times the difference between the median and the quartiles of the data. The formula for finding outliers is: IQR (Inter Quartile Range) = Q3 - Q1Q1 = Lower QuartileQ3 = Upper Quartile The lower quartile is the median of the lower half of the data, while the upper quartile is the median of the upper half of the data. To find the lower quartile: First, find the median of the data (the value that is halfway between the two middle values in the data set).
The median of the data is: 1.58To find the lower quartile, take the median of the lower half of the data.0.1, 0.25, 0.45, 0.68, 1, and 1.38Arrange these values in ascending order:0.1, 0.25, 0.45, 0.68, 1, and 1.38The median of these values is 0.565.To find the upper quartile, take the median of the upper half of the data.2.5, 3.32, 4.23, 9.04, and 101.2Arrange these values in ascending order:2.5, 3.32, 4.23, 9.04, and 101.2The median of these values is 6.635.IQR = Q3 - Q1IQR = 6.635 - 0.565IQR = 6.07Now, we can determine the upper and lower bounds for outliers.The upper bound for outliers is:Q3 + (1.5 × IQR)Upper bound = 6.635 + (1.5 × 6.07)Upper bound = 15.605The lower bound for outliers is:Q1 - (1.5 × IQR)Lower bound = 0.565 - (1.5 × 6.07)Lower bound = -8.94Any values that are above the upper bound or below the lower bound are outliers.101.2 is greater than the upper bound (15.605), so it is an outlier.
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