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Select the outlier in the data set. $\newline$ $6, 62, 76, 83, 84, 86, 87, 89, 97$

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Calculate quartiles and interquartile range

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Q. Select the outlier in the data set.

\newline

6, 62, 76, 83, 84, 86, 87, 89, 97

Arrange Data Set: Arrange the data set in ascending order. $\newline$ The data set is already in ascending order: $6, 62, 76, 83, 84, 86, 87, 89, 97$ .
Calculate IQR: Calculate the interquartile range (IQR) of the data set. $\newline$ First, find the median ( $Q_2$ ), which is the middle value when the data is ordered. For our data set, the median is $84$ . $\newline$ Next, find the first quartile ( $Q_1$ ), which is the median of the lower half of the data set. The lower half is $6, 62, 76, 83$ . The median of this half is the average of $62$ and $76$ , which is $(62 + 76) / 2 = 69$ . $\newline$ Then, find the third quartile ( $Q_3$ ), which is the median of the upper half of the data set. The upper half is $86, 87, 89, 97$ . The median of this half is the average of $87$ and $84$ $0$ , which is $84$ $1$ . $\newline$ The IQR is $84$ $2$ , which is $84$ $3$ .
Determine Outlier Boundaries: Determine the outlier boundaries. $\newline$ The lower boundary for outliers is $Q1 - 1.5 \times IQR$ , which is $69 - 1.5 \times 19 = 69 - 28.5 = 40.5$ . $\newline$ The upper boundary for outliers is $Q3 + 1.5 \times IQR$ , which is $88 + 1.5 \times 19 = 88 + 28.5 = 116.5$ .
Identify Outliers: Identify any values outside the outlier boundaries. $\newline$ The value $6$ is below the lower boundary of $40.5$ , so it is an outlier. $\newline$ No values are above the upper boundary of $116.5$ .

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