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Select the outlier in the data set. $\newline$ $75, 84, 86, 89, 90, 92, 95, 97, 878$

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Calculate quartiles and interquartile range

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Q. Select the outlier in the data set.

\newline

75, 84, 86, 89, 90, 92, 95, 97, 878

Arrange Data in Ascending Order: Arrange the data set in ascending order. $\newline$ The data set in ascending order is: $75, 84, 86, 89, 90, 92, 95, 97, 878$ .
Calculate Interquartile Range: 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 $90$ . $\newline$ Next, find $Q_1$ , the median of the lower half of the data set (excluding $Q_2$ ). The lower half is $75$ , $84$ , $86$ , $89$ . The median of this half is the average of $84$ and $86$ , which is $90$ $0$ . $\newline$ Then, find $90$ $1$ , the median of the upper half of the data set (excluding $Q_2$ ). The upper half is $90$ $3$ , $90$ $4$ , $90$ $5$ , $90$ $6$ . The median of this half is the average of $90$ $4$ and $90$ $5$ , which is $90$ $9$ . $\newline$ Now, calculate the IQR: $Q_1$ $0$ .
Determine Outlier Boundaries: Determine the outlier boundaries. $\newline$ The lower boundary for outliers is $Q1 - 1.5 \times IQR = 85 - 1.5 \times 11 = 85 - 16.5 = 68.5$ . $\newline$ The upper boundary for outliers is $Q3 + 1.5 \times IQR = 96 + 1.5 \times 11 = 96 + 16.5 = 112.5$ .
Identify Outliers: Identify any values outside the outlier boundaries. $\newline$ The value $878$ is well above the upper boundary of $112.5$ , so it is considered an outlier.

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