Select the outlier in the data set. 10, 13, 15, 17, 24, 37, 63, 81, 804

Arrange Data in Ascending Order: Arrange the data set in ascending order. The data set in ascending order is: 10, 13, 15, 17, 24, 37, 63, 81, 804.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 24. Next, find Q1, the median of the lower half of the data set (not including the median). The lower half is 10, 13, 15, 17, so Q1 is the average of 13 and 15, which is 14. Then, find Q3, the median of the upper half of the data set (not including the median). The upper half is 37, 63, 81, 804, so Q3 is the average of 63 and 81, which is 72. Now, calculate the IQR: Q3 - Q1 = 72 - 14 = 58.Determine Outlier Boundaries: Determine the outlier boundaries. The lower boundary for outliers is Q1 - 1.5 * IQR, which is 14 - 1.5 * 58 = 14 - 87 = -73. The upper boundary for outliers is Q3 + 1.5 * IQR, which is 72 + 1.5 * 58 = 72 + 87 = 159. Any data point below -73 or above 159 is considered an outlier.Identify Outliers: Identify any outliers based on the boundaries. Looking at the data set, the value 804 is above the upper boundary of 159, so it is an outlier.

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Select the outlier in the data set. $\newline$ $10, 13, 15, 17, 24, 37, 63, 81, 804$

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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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10, 13, 15, 17, 24, 37, 63, 81, 804

Arrange Data in Ascending Order: Arrange the data set in ascending order. $\newline$ The data set in ascending order is: $10, 13, 15, 17, 24, 37, 63, 81, 804$ .
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 $24$ . $\newline$ Next, find $Q_1$ , the median of the lower half of the data set (not including the median). The lower half is $10$ , $13$ , $15$ , $17$ , so $Q_1$ is the average of $13$ and $15$ , which is $24$ $0$ . $\newline$ Then, find $24$ $1$ , the median of the upper half of the data set (not including the median). The upper half is $24$ $2$ , $24$ $3$ , $24$ $4$ , $24$ $5$ , so $24$ $1$ is the average of $24$ $3$ and $24$ $4$ , which is $24$ $9$ . $\newline$ Now, calculate the IQR: $Q_1$ $0$ .
Determine Outlier Boundaries: Determine the outlier boundaries. $\newline$ The lower boundary for outliers is $Q1 - 1.5 \times IQR$ , which is $14 - 1.5 \times 58 = 14 - 87 = -73$ . $\newline$ The upper boundary for outliers is $Q3 + 1.5 \times IQR$ , which is $72 + 1.5 \times 58 = 72 + 87 = 159$ . $\newline$ Any data point below $-73$ or above $159$ is considered an outlier.
Identify Outliers: Identify any outliers based on the boundaries. $\newline$ Looking at the data set, the value $804$ is above the upper boundary of $159$ , so it is an outlier.