To determine the outliers from the skewed dataset \(5, 6, 6, 7, 7, 8, 8, 8, 9, 9, 14, 16\), we can use the IQR (Interquartile Range) method.
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Order the dataset (not needed here, but it's given): \[ 5, 6, 6, 7, 7, 8, 8, 8, 9, 9, 14, 16 \]
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Find Q1 (the first quartile) and Q3 (the third quartile):
- The median (Q2) is the average of the 6th and 7th values: \[ Q2 = \frac{8 + 8}{2} = 8 \]
- For Q1, the lower half of the data is \(5, 6, 6, 7, 7\): \[ Q1 = 6 \]
- For Q3, the upper half of the data is \(8, 9, 9, 14, 16\): \[ Q3 = 9 \]
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Calculate the IQR: \[ \text{IQR} = Q3 - Q1 = 9 - 6 = 3 \]
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Determine the lower and upper bounds for outliers:
- Lower bound = \(Q1 - 1.5 \times \text{IQR} = 6 - 1.5 \times 3 = 6 - 4.5 = 1.5\)
- Upper bound = \(Q3 + 1.5 \times \text{IQR} = 9 + 1.5 \times 3 = 9 + 4.5 = 13.5\)
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Identify outliers:
- Any data point less than 1.5 or greater than 13.5 is considered an outlier.
- In the dataset, the values \(14\) and \(16\) are greater than \(13.5\), making them outliers.
- The value \(5\) is not considered an outlier since it is greater than \(1.5\).
Based on this analysis, the correct statement is:
- The outliers of the dataset are 14 and 16.