Find the mean deviation about median for the following data
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Number of Girls | 6 | 8 | 14 | 16 | 4 | 2 |
The following table is formed.
Marks | Number of boys \(f_i\) | Cumulative frequency \(c.f\) | Mid point \(x_i\) | \(|x_i-Med.|\) | \(f_i|x_i-Med.|\) |
0-10 | 6 | 6 | 5 | 22.85 | 137.1 |
10-20 | 8 | 14 | 15 | 12.85 | 102.8 |
20-30 | 14 | 28 | 25 | 2.85 | 39.9 |
30-40 | 16 | 44 | 35 | 7.15 | 114.4 |
40-50 | 4 | 48 | 45 | 17.15 | 68.6 |
50-60 | 2 | 50 | 55 | 27.15 | 54.3 |
- | 50 | - | - | - | 517.1 |
The class interval containing the \((\frac{N}{2})^{th}\) or 25th item is 20 – 30.
Therefore, 20 – 30 is the median class.
It is known that,
Median= \(I+\frac{\frac{N}{2}-c}{f}h\)
Here, l = 20, C = 14, f = 14, h = 10, and N = 50
∴ Median \(=20+\frac{25-14}{14}×10=20+\frac{110}{14}=20+7.85=27.85\)
Thus, mean deviation about the median is given by,
\(M.D.(M)=\frac{1}{N}\sum_{i=1}^{6}f_i|x_i-M|=\frac{1}{50}×517.1=10.34\)
What inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions?
A statistical measure that is used to calculate the average deviation from the mean value of the given data set is called the mean deviation.
The mean deviation for the given data set is calculated as:
Mean Deviation = [Σ |X – µ|]/N
Where,
Grouping of data is very much possible in two ways: