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

Arrange Data in Ascending Order: Arrange the data set in ascending order. 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. First, find the median (Q2), which is the middle value when the data is ordered. For our data set, the median is 90. Next, find Q1, the median of the lower half of the data set (excluding Q2). The lower half is 75, 84, 86, 89. The median of this half is the average of 84 and 86, which is (84 + 86) / 2 = 85. Then, find Q3, the median of the upper half of the data set (excluding Q2). The upper half is 92, 95, 97, 878. The median of this half is the average of 95 and 97, which is (95 + 97) / 2 = 96. Now, calculate the IQR: Q3 - Q1 = 96 - 85 = 11.Determine Outlier Boundaries: Determine the outlier boundaries. The lower boundary for outliers is Q1 - 1.5 * IQR = 85 - 1.5 * 11 = 85 - 16.5 = 68.5. The upper boundary for outliers is Q3 + 1.5 * IQR = 96 + 1.5 * 11 = 96 + 16.5 = 112.5.Identify Outliers: Identify any values outside the outlier boundaries. The value 878 is well above the upper boundary of 112.5, so it is considered an outlier.

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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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Grade 6

Calculate quartiles and interquartile range

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Q. Select the outlier in the data set.

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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.