Consider the following grouped frequency distribution :Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 1 2 4 6 4 3
What is mean deviation about the median ?
10.5
The question asks us to find the mean deviation about the median for the given grouped frequency distribution.
Mean deviation about the median is a measure of dispersion that calculates the average of the absolute deviations of the observations from the median. For a grouped frequency distribution, the formula is:
\(\text{Mean Deviation (M.D.)} = \frac{\sum_{i=1}^{n} f_i |x_i - \text{Median}|}{\sum_{i=1}^{n} f_i}\)
Where:
First, we need to find the median of the grouped data. We construct a table with cumulative frequencies.
| Class | Frequency (f) | Cumulative Frequency (CF) |
|---|---|---|
| 0-10 | 1 | 1 |
| 10-20 | 2 | 1 + 2 = 3 |
| 20-30 | 4 | 3 + 4 = 7 |
| 30-40 | 6 | 7 + 6 = 13 |
| 40-50 | 4 | 13 + 4 = 17 |
| 50-60 | 3 | 17 + 3 = 20 |
The total frequency is \(N = \sum f_i = 20\).
We need to find the class containing the \(\left(\frac{N}{2}\right)\)-th observation. \(\frac{N}{2} = \frac{20}{2} = 10\).
The cumulative frequency just greater than or equal to 10 is 13, which corresponds to the class 30-40. So, the median class is 30-40.
Now, we use the formula for the median of a grouped frequency distribution:
\(\text{Median} = L + \frac{\frac{N}{2} - CF}{f} \times h\)
Where:
Substitute these values into the formula:
\(\text{Median} = 30 + \frac{10 - 7}{6} \times 10\)
\(\text{Median} = 30 + \frac{3}{6} \times 10\)
\(\text{Median} = 30 + 0.5 \times 10\)
\(\text{Median} = 30 + 5\)
\(\text{Median} = 35\)
Now that we have the median (35), we need to calculate the mean deviation about the median. We'll add columns to our table for midpoints (\(x_i\)), absolute deviations from the median (\(|x_i - 35|\)), and the product of frequency and absolute deviation (\(f_i |x_i - 35|\)).
| Class | Frequency (f) | Midpoint (\(x_i\)) | \(|x_i - 35|\) | \(f_i |x_i - 35|\) |
|---|---|---|---|---|
| 0-10 | 1 | \(\frac{0+10}{2} = 5\) | $|5 - 35| = 30$ | \(1 \times 30 = 30\) |
| 10-20 | 2 | \(\frac{10+20}{2} = 15\) | $|15 - 35| = 20$ | \(2 \times 20 = 40\) |
| 20-30 | 4 | \(\frac{20+30}{2} = 25\) | $|25 - 35| = 10$ | \(4 \times 10 = 40\) |
| 30-40 | 6 | \(\frac{30+40}{2} = 35\) | $|35 - 35| = 0$ | \(6 \times 0 = 0\) |
| 40-50 | 4 | \(\frac{40+50}{2} = 45\) | $|45 - 35| = 10$ | \(4 \times 10 = 40\) |
| 50-60 | 3 | \(\frac{50+60}{2} = 55\) | $|55 - 35| = 20$ | \(3 \times 20 = 60\) |
Sum of \(f_i |x_i - 35|\) is \(\sum f_i |x_i - 35| = 30 + 40 + 40 + 0 + 40 + 60 = 210\).
Total frequency \(N = \sum f_i = 20\).
Now, calculate the Mean Deviation about the Median:
\(\text{M.D.} = \frac{\sum f_i |x_i - \text{Median}|}{N} = \frac{210}{20}\)
\(\text{M.D.} = 10.5\)
The mean deviation about the median for the given grouped frequency distribution is 10.5.
| Concept | Description | Formula (Grouped Data) |
|---|---|---|
| Mean Deviation | Average of absolute deviations from a central value (Mean or Median). | \(\frac{\sum f_i |x_i - A|}{N}\) where A is Mean or Median |
| Median (Grouped) | The middle value; divides the data into two equal halves. | \(L + \frac{\frac{N}{2} - CF}{f} \times h\) |
| Midpoint (\(x_i\)) | Average of the lower and upper limits of a class. | \(\frac{\text{Lower Limit} + \text{Upper Limit}}{2}\) |
Mean deviation is a measure of dispersion, which tells us how spread out the data is. Other common measures of dispersion include range, quartile deviation, variance, and standard deviation.
Mean deviation is less commonly used compared to standard deviation because it uses absolute values, which can be difficult to handle in further mathematical calculations.
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