Mean Absolute Deviation
- Introduction
- Formula for Mean Absolute Deviation
- Steps to Calculate Mean Absolute Deviation
- Applications of Mean Absolute Deviation
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Introduction
The mean absolute deviation (MAD) is a measure of variability that indicates the average distance between observations and their mean. Larger values signify that the data points spread out further from the average. Conversely, lower values correspond to data points bunching closer to it. The mean absolute deviation is also known as the mean deviation and average absolute deviation. MAD uses the original units of the data, which simplifies interpretation.
To calculate the Mean Absolute Deviation:
`1`. Find the mean (average) of the dataset.
`2`. For each data point, find the absolute difference between that data point and the mean.
`3`. Take the average of all these absolute differences.
Formula for Mean Absolute Deviation
MAD is calculated using the following formula:
\[ \text{MAD} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \text{mean}| \]
Where:
- \( n \) is the number of data points in the dataset.
- \( x_i \) represents each individual data point.
- "mean" is the mean (average) of the dataset.
The formula for the mean absolute deviation (MAD) for grouped data involves considering the frequency of each group.
Let \( x_i \) represent the midpoint of each class interval,
\( f_i \) represent the frequency of each class interval, and
\( \bar{x} \) represent the mean of the grouped data.
Below is the formula for calculating MAD for grouped data.
\[\text{MAD} = \frac{\sum f_i |x_i - \bar{x}|}{N}\]
In this formula:
- \( |x_i - \bar{x}| \) represents the absolute deviation of each midpoint from the mean.
- \( f_i \) represents the frequency of each class interval.
- \( N \) represents the total number of data points or the sum of frequencies.
This formula calculates the weighted average of the absolute deviations from the mean, where each deviation is weighted by the frequency of its corresponding class interval.
MAD provides a measure of how spread out the data points are around the mean, with a larger MAD indicating greater variability in the data. It is particularly useful when dealing with datasets with outliers since it gives equal weight to all deviations from the mean, regardless of their direction.
Steps to Calculate Mean Absolute Deviation
Step `1`. Find the Mean (Average)
Let's say we have a dataset with \( n \) data points \( x_1, x_2, ..., x_n \). The mean of this dataset is given by:
\[ \text{Mean} = \frac{1}{n} \sum_{i=1}^{n} x_i \]
Step `2`. Calculate deviations
Now, we find the deviation of each data point from the mean. The deviation of the \( i^{th} \) data point \( x_i \) from the mean is:
\[ \text{Deviation}_i = x_i - \text{Mean} \]
Step `3`. Find absolute deviations
We take the absolute value of each deviation to ensure we capture the magnitude of the difference, regardless of direction:
\[ \text{Absolute Deviation}_i = |x_i - \text{Mean}| \]
Step `4`. Calculate sum of absolute deviations
Next, we sum up all these absolute deviations:
\[ \text{Sum of Absolute Deviations} = \sum_{i=1}^{n} |x_i - \text{Mean}| \]
Step `5`. Calculate the Mean Absolute Deviation:
Finally, we calculate the Mean Absolute Deviation by taking the average of these absolute deviations:
\[ \text{MAD} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \text{Mean}| \]
This is the derived formula for Mean Absolute Deviation. It measures the average absolute deviation of each data point from the mean of the dataset.
Applications of Mean Absolute Deviation
Mean Absolute Deviation (MAD) has several applications across various fields:
`1`. Finance and Economics- In finance, MAD is used to measure the dispersion or variability of returns on investments. It helps investors understand the risk associated with an investment by quantifying how much the returns deviate from the mean return.
- Economists use MAD to analyse economic data such as inflation rates, GDP growth rates, and unemployment rates to assess the stability and variability of these indicators over time.
`2`. Quality Control- MAD is used in quality control to measure the variability of a process or product. For example, in manufacturing, MAD can be used to assess the consistency of product dimensions or the quality of output over time.
`3`. Retail and Supply Chain Management- Retailers use MAD to analyse sales data and forecast demand variability for products. It helps them optimise inventory management and ensure sufficient stock levels to meet customer demand while minimising excess inventory costs.
- Supply chain managers use MAD to evaluate the variability in lead times, transportation costs, and order fulfilment times. This information is crucial for optimising supply chain operations and reducing the risk of disruptions.
`4`. Meteorology and Climate Science- MAD is used in meteorology and climate science to analyse weather and climate data. It helps researchers assess the variability of temperature, precipitation, and other meteorological variables over time and across different geographic regions.
`5`. Education and Psychology- In educational research, MAD can be used to measure the variability in student performance on standardised tests or academic assessments. It helps educators identify areas where students may need additional support or intervention.
- Psychologists use MAD to analyse psychological test scores and assess the variability of responses within a sample population. It helps them understand the distribution of traits or behaviours within a group.
Overall, Mean Absolute Deviation is a versatile statistical measure that provides valuable insights into the variability and dispersion of data across a wide range of applications.
Solved Examples
Example `1`: Calculate the mean absolute deviation of the following data \( 3, 7, 9, 12, 15 \).
Solution:
Step `1`: Calculate the mean (average) of the data.
`\text{Mean} = \frac{3 + 7 + 9 + 12 + 15}{5} = \frac{46}{5} = 9.2`
Step `2`: Calculate the absolute deviations of each data point from the mean.
\( |3 - 9.2| = 6.2, \ |7 - 9.2| = 2.2, \ |9 - 9.2| = 0.2, \ |12 - 9.2| = 2.8, \ |15 - 9.2| = 5.8 \)
Step `3`: Find the mean of the absolute deviations.
`\text{MAD} = \frac{6.2 + 2.2 + 0.2 + 2.8 + 5.8}{5} = \frac{17.2}{5} = 3.44`
So, the mean absolute deviation is \( 3.44 \).
Example `2`: Calculate the mean absolute deviation of the following data \( 10, 12, 14, 18, 20 \).
Solution:
Step `1`: Calculate the mean (average) of the data.
`\text{Mean} = \frac{10 + 12 + 14 + 18 + 20}{5} = \frac{74}{5} = 14.8`
Step `2`: Calculate the absolute deviations of each data point from the mean.
\( |10 - 14.8| = 4.8, \ |12 - 14.8| = 2.8, \ |14 - 14.8| = 0.8, \ |18 - 14.8| = 3.2, \ |20 - 14.8| = 5.2 \)
Step `3`: Find the mean of the absolute deviations.
`\text{MAD} = \frac{4.8 + 2.8 + 0.8 + 3.2 + 5.2}{5} = \frac{16.8}{5} = 3.36`
So, the mean absolute deviation is \( 3.36 \).
Example `3`: Calculate the mean absolute deviation of the following data \( 2, 4, 6, 8, 10 \).
Solution:
Step `1`: Calculate the mean (average) of the data.
`\text{Mean} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6`
Step `2`: Calculate the absolute deviations of each data point from the mean.
\( |2 - 6| = 4, \ |4 - 6| = 2, \ |6 - 6| = 0, \ |8 - 6| = 2, \ |10 - 6| = 4 \)
Step `3`: Find the mean of the absolute deviations.
`\text{MAD} = \frac{4 + 2 + 0 + 2 + 4}{5} = \frac{12}{5} = 2.4`
So, the mean absolute deviation is \( 2.4 \).
Example `4`: Calculate the mean absolute deviation of the following data \( 5, 10, 15, 20, 25 \)
Solution:
Step `1`: Calculate the mean (average) of the data.
`\text{Mean} = \frac{5 + 10 + 15 + 20 + 25}{5} = \frac{75}{5} = 15`
Step `2`: Calculate the absolute deviations of each data point from the mean.
\( |5 - 15| = 10, \ |10 - 15| = 5, \ |15 - 15| = 0, \ |20 - 15| = 5, \ |25 - 15| = 10 \)
Step `3`: Find the mean of the absolute deviations.
`\text{MAD} = \frac{10 + 5 + 0 + 5 + 10}{5} = \frac{30}{5} = 6`
So, the mean absolute deviation is \( 6 \).
Example `5`: Consider the following grouped data representing the ages (in years) of students in a class:
Find the MAD for this grouped data.
Solution:
To find the MAD for grouped data, we'll follow these steps:
`1`. Calculate the midpoint \( x_i \) for each age range.
`2`. Find the mean \( \bar{x} \) of the grouped data.
`3`. Calculate the absolute deviation \( |x_i - \bar{x}| \) for each midpoint.
`4`. Multiply each absolute deviation by its corresponding frequency \( f_i \).
`5`. Sum up the products of absolute deviations and frequencies.
`6`. Divide the total by the total number of data points (or sum of frequencies).
Let's proceed with the calculations:
`1`. Calculate the midpoint \( x_i \) for each age range:
For the first class interval `(5 - 10)`, the midpoint \( x_1 \) is: `x_1 = \frac{5 + 10}{2} = 7.5`.
Similarly, calculate midpoints for other class intervals.
`2`. Find the mean \( \bar{x} \) of the grouped data:
`\bar{x} = \frac{\sum x_i \cdot f_i}{\sum f_i} = \frac{7.5\times8+13\times12+18\times15+23\times10+28\times5}{50} = \frac{60+156+270+230+140}{50} = 17.12`
`3`. Calculate the absolute deviation \( |x_i - \bar{x}| \) for each midpoint:
`4`. Sum up the products of absolute deviations and frequencies:
Sum of \( \text{Absolute Deviation} \times \text{Frequency} \)
\( = 8\times 9.62 + 12\times 4.12 + 15\times 0.88 + 10\times 5.88 + 5\times 10.88 = 252.8\)
`5`. Divide the total by the total number of data points (or sum of frequencies):
`\text{MAD} = \frac{\sum \text{Absolute Deviation} \times \text{Frequency}}{\sum \text{Frequency}}`
`\text{MAD} = \frac{252.8}{50} = 5.056`
Hence, the MAD of the grouped data is \(5.056\).
Practice Problems
Q`1`: The mean absolute deviation (MAD) measures the `"____________________"`.
- Average deviation from the median
- Average deviation from the mode
- Average deviation from the mean
- Average deviation from the range
Answer: c
Q`2`: If a data set has a MAD of `5`, what does this imply?
- On average, each data point is `5` units away from the mean
- On average, each data point is `5%` away from the mean
- On average, each data point is `5` units away from the median
- On average, each data point is `5%` away from the median
Solution: a
Q`3`: Which of the following situations would likely result in a larger MAD?
- A data set with values `{1, 2, 3, 4, 5}`
- A data set with values `{100, 101, 102, 103, 104}`
- A data set with values `{1, 10, 15, 20, 25}`
- A data set with values `{10, 11, 12, 13, 100}`
Solution: d
Q`4`: If the mean of a data set is `50` and the MAD is `10`, what percentage of data points fall within the range of `40` to `60`?
- `25%`
- `50%`
- `75%`
- `90%`
Solution: c
Q`5`: Consider the following grouped data representing the scores of students in a math test:
What is the mean absolute deviation (MAD) for the given data?
- `9.6`
- `12.4`
- `15.2`
- `18.8`
Solution: b
Frequently Asked Questions
Q`1`: What is mean absolute deviation (MAD) and why is it important?
Answer: Mean absolute deviation (MAD) is a measure of the average deviation of a set of values from their mean. It is calculated by finding the average of the absolute differences between each value and the mean of the data set. MAD provides a measure of dispersion or variability within the data set. It's important because it gives insight into how spread out the values are from the average, helping to understand the consistency or variability of the data.
Q`2`: How do you calculate mean absolute deviation (MAD)?
Answer: To calculate the mean absolute deviation (MAD), follow these steps:
`1`. Find the mean (average) of the data set.
`2`. Subtract the mean from each data point, taking the absolute value of each difference.
`3`. Find the average of these absolute differences.
The formula for MAD is: `\text{MAD} = \frac{\sum |\text{data point} - \text{mean}|}{\text{number of data points}}`
Q`3`: What are the key differences between MAD and standard deviation?
Answer: While both MAD and standard deviation measure dispersion or variability within a data set, they differ in their calculation methods. MAD is the average of the absolute differences between each data point and the mean, making it less sensitive to outliers. Standard deviation, on the other hand, is the square root of the average of the squared differences between each data point and the mean, making it more sensitive to outliers. Standard deviation is also computationally more complex compared to MAD.
Q`4`: What does a high MAD value indicate about a data set?
Answer: A high MAD value indicates that the data points in the set are, on average, further away from the mean. This suggests greater variability or dispersion within the data set, meaning the data points are more spread out. In practical terms, it may imply less consistency or predictability in the values compared to a data set with a lower MAD.
Q`5`: Can MAD be negative?
Answer: No, MAD cannot be negative. By definition, MAD represents the average absolute deviation from the mean, which means it measures distance and cannot be negative. If MAD were negative, it would imply that the absolute deviations cancel each other out, which contradicts its purpose as a measure of variability.