The given data is
4, 7, 8, 9, 10, 12, 13, 17
Mean of the data, \(\bar{x}=\frac{4+7+8+10+12+13+17}{8}=\frac{80}{8}=10\)
The deviations of the respective observations from the mean \(\bar{x},i.e.x_i-\bar{x},\) are:
6, 3, 2, 1, 0, 2, 3, 7
The absolute values of the deviations, i.e. \(|x_i-\bar{x}|\), are:
6, 3, 2, 1, 0, 2, 3, 7
The required mean deviation about the mean is
M.D.\(\bar{x}=\frac{\sum_{i=1}^{8}|x_i-\bar{x}|}{8}=\frac{6+3+2+1+0+2+3+7}{8}\)
\(=\frac{24}{8}=3\)
Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
---|---|---|---|---|---|---|
Frequency | 11 | 8 | 15 | 7 | 10 | 9 |
Variance of the following discrete frequency distribution is:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 \\ \hline \text{Frequency (}f_i\text{)} & 2 & 3 & 5 & 3 & 2 \\ \hline \end{array} \]
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: