To identify the outlier in the dataset \( {5, 10, 12, 14, 18, 24, 60} \), we can use the 1.5 IQR (interquartile range) rule.
Step 1: Calculate the quartiles and IQR
- Sort the data: The data is already sorted: \( 5, 10, 12, 14, 18, 24, 60 \)
- Find Q1 (first quartile): The median of the first half of the data (5, 10, 12) is 10.
- Find Q3 (third quartile): The median of the second half of the data (18, 24, 60) is 24.
- Calculate IQR: \[ \text{IQR} = Q3 - Q1 = 24 - 10 = 14 \]
Step 2: Determine the lower and upper bounds for outliers
- Lower bound: \[ Q1 - 1.5 \times \text{IQR} = 10 - 1.5 \times 14 = 10 - 21 = -11 \]
- Upper bound: \[ Q3 + 1.5 \times \text{IQR} = 24 + 1.5 \times 14 = 24 + 21 = 45 \]
Step 3: Identify the outlier
Any value below -11 or above 45 is considered an outlier. Thus, in the dataset \( {5, 10, 12, 14, 18, 24, 60} \), the outlier is 60.
Step 4: Impact on statistics if the outlier is excluded
Now we will assess the effect on the median, mean, mode, and range of the dataset if the outlier \( 60 \) is excluded.
Original Data: \( 5, 10, 12, 14, 18, 24, 60 \)
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Median:
- With outlier: \( 14 \) (Middle value)
- Without outlier \( {5, 10, 12, 14, 18, 24} \): Median is \( \frac{12 + 14}{2} = 13 \) (Median decreases)
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Mean:
- With outlier: \( \frac{5 + 10 + 12 + 14 + 18 + 24 + 60}{7} = \frac{133}{7} \approx 19 \)
- Without outlier: \( \frac{5 + 10 + 12 + 14 + 18 + 24}{6} = \frac{89}{6} \approx 14.83 \) (Mean decreases)
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Mode:
- There is no repeated value; thus, the mode is none in both cases. (No effect)
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Range:
- With outlier: \( 60 - 5 = 55 \)
- Without outlier: \( 24 - 5 = 19 \) (Range decreases)
Summary Chart
| Statistic | Outlier Excluded | Impact | |----------------|--------------------------|---------------| | Median | 13 | Decrease | | Mean | 14.83 | Decrease | | Mode | None | No effect | | Range | 19 | Decrease |
Final Answers
- A (Median): Decrease
- B (Mean): Decrease
- C (Mode): No effect
- D (Range): Decrease
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