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Select the outlier in the data set. $\newline$ $6, 76, 78, 81, 83, 89, 91, 93, 99$

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Calculate quartiles and interquartile range

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Q. Select the outlier in the data set.

\newline

6, 76, 78, 81, 83, 89, 91, 93, 99

Arrange Data Set: Arrange the data set in ascending order. $\newline$ The data set is already in ascending order: $6$ , $76$ , $78$ , $81$ , $83$ , $89$ , $91$ , $93$ , $99$ .
Calculate IQR: Calculate the interquartile range (IQR) of the data set. $\newline$ First, find the median ( $Q_2$ ), which is the middle value of the ordered data set. For our data set, the median is $83$ . $\newline$ Next, find the first quartile ( $Q_1$ ), which is the median of the lower half of the data set. The lower half is $6, 76, 78, 81$ . The median of this half is the average of $76$ and $78$ , which is $(76 + 78) / 2 = 77$ . $\newline$ Then, find the third quartile ( $Q_3$ ), which is the median of the upper half of the data set. The upper half is $89, 91, 93, 99$ . The median of this half is the average of $91$ and $83$ $0$ , which is $83$ $1$ . $\newline$ The IQR is $83$ $2$ , which is $83$ $3$ .
Determine Outlier Boundaries: Determine the outlier boundaries. $\newline$ The lower boundary for outliers is $Q1 - 1.5 \times IQR$ , which is $77 - 1.5 \times 15 = 77 - 22.5 = 54.5$ . $\newline$ The upper boundary for outliers is $Q3 + 1.5 \times IQR$ , which is $92 + 1.5 \times 15 = 92 + 22.5 = 114.5$ .
Identify Outliers: Identify any values outside the outlier boundaries. $\newline$ The value $6$ is below the lower boundary of $54.5$ , so it is an outlier. $\newline$ No values are above the upper boundary of $114.5$ .

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