Answer :
Sure! Let's address each part of the problem in a detailed, step-by-step manner.
### 3. Finding the Range and Interquartile Range (IQR)
Range:
The range of a data set is the difference between the maximum and minimum values.
- Identify the maximum value in the data set: 17
- Identify the minimum value in the data set: -4
- Calculate the range:
[tex]\[ \text{Range} = \text{Maximum} - \text{Minimum} = 17 - (-4) = 17 + 4 = 21 \][/tex]
Thereby, the range of the data is 21.
Interquartile Range (IQR):
The interquartile range is the range between the first quartile (Q1) and the third quartile (Q3).
- First, sort the data:
[tex]\[ -4, -3, -3, -1, 0, 1, 4, 11, 12, 17 \][/tex]
- Find Q1 (the 25th percentile) and Q3 (the 75th percentile):
- Q1 (25th percentile): [tex]\[ Q1 = -1 \][/tex]
- Q3 (75th percentile): [tex]\[ Q3 = 10.75 \][/tex]
- Calculate the IQR:
[tex]\[ \text{IQR} = Q3 - Q1 = 10.75 - (-1) = 10.75 + 1 = 11.75 \][/tex]
So, the interquartile range (IQR) of the data is 11.75.
### 4. Finding the Mean Absolute Deviation (MAD) and Outliers
Mean Absolute Deviation (MAD):
The mean absolute deviation is the average of the absolute deviations from the median of the data set.
- Compute the median of the data set. For data sorted as:
[tex]\[ -4, -3, -3, -1, 0, 1, 4, 11, 12, 17 \][/tex]
The median (the middle value) is the average of the 5th and 6th values:
[tex]\[ \text{Median} = \frac{0 + 1}{2} = \frac{1}{2} = 0.5 \][/tex]
- Calculate the absolute deviations from the median (0.5):
[tex]\[ \left| -4 - 0.5 \right|, \left| -3 - 0.5 \right|, \left| -3 - 0.5 \right|, \left| -1 - 0.5 \right|, \left| 0 - 0.5 \right|, \left| 1 - 0.5 \right|, \left| 4 - 0.5 \right|, \left| 11 - 0.5 \right|, \left| 12 - 0.5 \right|, \left| 17 - 0.5 \right| \][/tex]
These deviations are:
[tex]\[ 4.5, 3.5, 3.5, 1.5, 0.5, 0.5, 3.5, 10.5, 11.5, 16.5 \][/tex]
- Compute the mean of these absolute deviations:
[tex]\[ \text{MAD} = \frac{4.5 + 3.5 + 3.5 + 1.5 + 0.5 + 0.5 + 3.5 + 10.5 + 11.5 + 16.5}{10} = 5.6 \][/tex]
Thus, the mean absolute deviation (MAD) for the data is 5.6.
Outliers:
To determine the outliers, we use the IQR method with a multiplier of 1.5.
- Calculate the lower bound:
[tex]\[ \text{Lower Bound} = Q1 - 1.5 \times \text{IQR} = -1 - 1.5 \times 11.75 = -1 - 17.625 = -18.625 \][/tex]
- Calculate the upper bound:
[tex]\[ \text{Upper Bound} = Q3 + 1.5 \times \text{IQR} = 10.75 + 1.5 \times 11.75 = 10.75 + 17.625 = 28.375 \][/tex]
Any data point outside the range of \([-18.625, 28.375]\) is considered an outlier.
Upon examination, none of the data points fall outside this range, so there are no outliers in the given data set.
### Summary:
- The range of the data is 21.
- The interquartile range (IQR) is 11.75.
- The mean absolute deviation (MAD) is 5.6.
- There are no outliers in the data set.
### 3. Finding the Range and Interquartile Range (IQR)
Range:
The range of a data set is the difference between the maximum and minimum values.
- Identify the maximum value in the data set: 17
- Identify the minimum value in the data set: -4
- Calculate the range:
[tex]\[ \text{Range} = \text{Maximum} - \text{Minimum} = 17 - (-4) = 17 + 4 = 21 \][/tex]
Thereby, the range of the data is 21.
Interquartile Range (IQR):
The interquartile range is the range between the first quartile (Q1) and the third quartile (Q3).
- First, sort the data:
[tex]\[ -4, -3, -3, -1, 0, 1, 4, 11, 12, 17 \][/tex]
- Find Q1 (the 25th percentile) and Q3 (the 75th percentile):
- Q1 (25th percentile): [tex]\[ Q1 = -1 \][/tex]
- Q3 (75th percentile): [tex]\[ Q3 = 10.75 \][/tex]
- Calculate the IQR:
[tex]\[ \text{IQR} = Q3 - Q1 = 10.75 - (-1) = 10.75 + 1 = 11.75 \][/tex]
So, the interquartile range (IQR) of the data is 11.75.
### 4. Finding the Mean Absolute Deviation (MAD) and Outliers
Mean Absolute Deviation (MAD):
The mean absolute deviation is the average of the absolute deviations from the median of the data set.
- Compute the median of the data set. For data sorted as:
[tex]\[ -4, -3, -3, -1, 0, 1, 4, 11, 12, 17 \][/tex]
The median (the middle value) is the average of the 5th and 6th values:
[tex]\[ \text{Median} = \frac{0 + 1}{2} = \frac{1}{2} = 0.5 \][/tex]
- Calculate the absolute deviations from the median (0.5):
[tex]\[ \left| -4 - 0.5 \right|, \left| -3 - 0.5 \right|, \left| -3 - 0.5 \right|, \left| -1 - 0.5 \right|, \left| 0 - 0.5 \right|, \left| 1 - 0.5 \right|, \left| 4 - 0.5 \right|, \left| 11 - 0.5 \right|, \left| 12 - 0.5 \right|, \left| 17 - 0.5 \right| \][/tex]
These deviations are:
[tex]\[ 4.5, 3.5, 3.5, 1.5, 0.5, 0.5, 3.5, 10.5, 11.5, 16.5 \][/tex]
- Compute the mean of these absolute deviations:
[tex]\[ \text{MAD} = \frac{4.5 + 3.5 + 3.5 + 1.5 + 0.5 + 0.5 + 3.5 + 10.5 + 11.5 + 16.5}{10} = 5.6 \][/tex]
Thus, the mean absolute deviation (MAD) for the data is 5.6.
Outliers:
To determine the outliers, we use the IQR method with a multiplier of 1.5.
- Calculate the lower bound:
[tex]\[ \text{Lower Bound} = Q1 - 1.5 \times \text{IQR} = -1 - 1.5 \times 11.75 = -1 - 17.625 = -18.625 \][/tex]
- Calculate the upper bound:
[tex]\[ \text{Upper Bound} = Q3 + 1.5 \times \text{IQR} = 10.75 + 1.5 \times 11.75 = 10.75 + 17.625 = 28.375 \][/tex]
Any data point outside the range of \([-18.625, 28.375]\) is considered an outlier.
Upon examination, none of the data points fall outside this range, so there are no outliers in the given data set.
### Summary:
- The range of the data is 21.
- The interquartile range (IQR) is 11.75.
- The mean absolute deviation (MAD) is 5.6.
- There are no outliers in the data set.
