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 : | 4 – 6 | 7 – 9 | 10 – 12 | 13 – 15 |
Frequency : | 5 | 4 | 9 | 10 |
Marks : | Below 10 | Below 20 | Below 30 | Below 40 | Below 50 |
Number of Students : | 3 | 12 | 27 | 57 | 75 |
\(\text{Length (in mm)}\) | 70-80 | 80-90 | 90-100 | 100-110 | 110-120 | 120-130 | 130-140 |
---|---|---|---|---|---|---|---|
\(\text{Number of leaves}\) | 3 | 5 | 9 | 12 | 5 | 4 | 2 |
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
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: