Find the mean deviation about the median for the data
| xi | 15 | 21 | 27 | 30 | 35 |
| fi | 3 | 5 | 6 | 7 | 8 |
The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.
| \(x_i\) | \(f_i\) | \(c.f\) |
| 15 | 3 | 3 |
| 21 | 5 | 8 |
| 27 | 6 | 14 |
| 30 | 7 | 21 |
| 35 | 8 | 29 |
Here, N = 29, which is odd.
∴ Median= \((\frac{29+1}{2})^{th}\) observation = 15th observation.
This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.
∴ Median = 30
The absolute values of the deviations from median, i.e \(|x_i-M|,\) are
| \(|x_i,M|\) | 15 | 9 | 3 | 0 | 5 |
| \(f_i\) | 3 | 5 | 6 | 7 | 8 |
| \(f_i|x-M|\) | 45 | 45 | 18 | 0 | 40 |
\(\sum_{I=1}^{5}f_i=29\) , \(\sum_{I=1}^{5}f_i|x_i-M|=148\)
\(=M.D.(M)=\frac{1}{N}\sum_{i=1}^{5}f_i|x_i-M|=\frac{1}{29}×148=5.1\)
\(f(x) = \begin{cases} x^2, & \quad 0≤x≤3\\ 3x, & \quad 3≤x≤10 \end{cases}\)
The relation g is defined by
\(g(x) = \begin{cases} x^2, & \quad 0≤x≤2\\ 3x, & \quad 2≤x≤10 \end{cases}\)
Show that f is a function and g is not a function.
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: