Answer :
The outliers in the given data set are 106 and 126. To determine the outliers in a data set, we typically use the concept of the interquartile range (IQR) and the 1.5 IQR rule.
The IQR is the range between the first quartile (Q1) and the third quartile (Q3) of the data set.
First, we need to find the quartiles of the data set. Arranging the data in ascending order, we have:
27, 31, 35, 43, 49, 53, 61, 65, 66, 74, 106, 126
The first quartile, Q1, is the median of the lower half of the data set, which is 43.
The third quartile, Q3, is the median of the upper half of the data set, which is 66.
Next, we calculate the IQR by subtracting Q1 from Q3: IQR = Q3 - Q1 = 66 - 43 = 23.
According to the 1.5 IQR rule, any value that is more than 1.5 times the IQR away from either Q1 or Q3 is considered an outlier. In this case, any value below Q1 - 1.5 * IQR or above Q3 + 1.5 * IQR is an outlier.
Calculating the outlier boundaries:
Lower bound = Q1 - 1.5 * IQR = 43 - 1.5 * 23 = 8.5
Upper bound = Q3 + 1.5 * IQR = 66 + 1.5 * 23 = 106.5
From the given data set, the values 106 and 126 are greater than the upper bound, indicating that they are outliers. Therefore, the outliers in the data set are 106 and 126. The correct answer is option d: 106 and 126.
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The value 126 is above the upper bound. Hence, the correct option is e) 126.
To identify outliers in a dataset, one common method is to use the interquartile range (IQR).
1. First, we need to arrange the data in ascending order:
27, 31, 35, 43, 49, 53, 61, 65, 66, 74, 106, 126
2. Calculate the first quartile (Q1) and the third quartile (Q3):
- Q1 is the median of the lower half of the data.
- Q3 is the median of the upper half of the data.
Since the dataset has 12 data points, the lower half is from 27 to 61, and the upper half is from 66 to 126.
[tex]\[ Q1 = \frac{35 + 43}{2} = 39 \]\[ Q3 = \frac{66 + 74}{2} = 70 \][/tex]
3. Calculate the interquartile range (IQR):
[tex]\[ \text{IQR} = Q3 - Q1 = 70 - 39 = 31 \][/tex]
4. Calculate the lower and upper bounds for outliers:
[tex]\[ \text{Lower Bound} = Q1 - 1.5 \times \text{IQR} = 39 - 1.5 \times 31 = -4.5 \] \[ \text{Upper Bound} = Q3 + 1.5 \times \text{IQR} = 70 + 1.5 \times 31 = 116.5 \][/tex]
5. Any data point outside the range (-4.5, 116.5) is considered an outlier.
From the dataset, we can see that the value 126 is above the upper bound, so it is an outlier. Therefore, the correct answer is option e) 126.
