Calculate the mean deviation about median age for the age distribution of 100 persons given below
Age (in years) | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 |
Number | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
The table is formed as follows.
Age | Number \(f_i\) | Cumulative frequency (c.f.) | Mid point \(x_i\) | \(|x_i-Med.|\) | \(f_i|x_i-Med.|\)| |
15.5-20.5 | 5 | 5 | 18 | 20 | 100 |
20.5-25.5 | 6 | 11 | 23 | 15 | 90 |
25.5-30.5 | 12 | 23 | 28 | 10 | 120 |
30.5-35.5 | 14 | 37 | 33 | 5 | 70 |
35.5-40.5 | 26 | 63 | 38 | 0 | 0 |
40.5-45.5 | 12 | 75 | 43 | 5 | 60 |
45.5-50.5 | 16 | 91 | 48 | 10 | 160 |
50.5-55.5 | 9 | 100 | 53 | 15 | 135 |
100 | 735 |
The class interval containing the \((\frac{N}{2})^{th}\) or 50th item is 35.5 – 40.5.
Therefore, 35.5 – 40.5.is the median class.
It is known that,
Median= \(I+\frac{\frac{N}{2}-c}{f}h\)
Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100
\(Median=35.5+\frac{50-37}{26}×5=35.5+\frac{13×5}{26}=35.5+2.5=38\)
Thus, mean deviation about the median is given by,
\(M.D.(\bar{x})=\frac{1}{N}\sum_{i=1}^{8}f_i|x_i-M|=\frac{1}{100}×735.1=7.35\)
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: