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

Arrange Data Set: Arrange the data set in ascending order. 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. First, find the median (Q2), which is the middle value when the data is ordered. For our data set, the median is 84. Next, find the first quartile (Q1), 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. Then, find the third quartile (Q3), 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 89, which is (87 + 89) / 2 = 88. The IQR is Q3 - Q1, which is 88 - 69 = 19.Determine Outlier Boundaries: Determine the outlier boundaries. The lower boundary for outliers is Q1 - 1.5 * IQR, which is 69 - 1.5 * 19 = 69 - 28.5 = 40.5. The upper boundary for outliers is Q3 + 1.5 * IQR, which is 88 + 1.5 * 19 = 88 + 28.5 = 116.5.Identify Outliers: Identify any values outside the outlier boundaries. The value 6 is below the lower boundary of 40.5, so it is an outlier. No values are above the upper boundary of 116.5.

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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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Grade 6

Calculate quartiles and interquartile range

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Q. Select the outlier in the data set.

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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 $116.5$ .