Find the mean deviation about the mean for the given data.
| \(x_i\) | 5 | 10 | 15 | 20 | 25 |
| \(f_i\) | 7 | 4 | 6 | 3 | 5 |
| \(x_i\) | \(f_i\) | \(f_ix_i\) | \(|x_i-\bar{x}|\) | \(f_i|x_i-\bar{x}|\) |
| 5 | 7 | 35 | 9 | 63 |
| 10 | 4 | 40 | 4 | 16 |
| 15 | 6 | 90 | 1 | 6 |
| 20 | 3 | 60 | 6 | 18 |
| 25 | 5 | 125 | 11 | 55 |
| 25 | 350 | 158 |
\(N=\sum_{I=1}^{5}f_i=25\)
\(N=\sum_{I=1}^{5}f_ix_i=350\)
∴ \(\bar{x}=\frac{1}{N}\sum_{I=1}^{5}f_ix_i=\frac{1}{25}×350=14\)
∴ \(=MD\bar{(x)}=\frac{1}{N}\sum_{i=1}^{5}f_i|x_i-\bar{x}|=\frac{1}{25}×158=6.32\)
Let the Mean and Variance of five observations $ x_i $, $ i = 1, 2, 3, 4, 5 $ be 5 and 10 respectively. If three observations are $ x_1 = 1, x_2 = 3, x_3 = a $ and $ x_4 = 7, x_5 = b $ with $ a>b $, then the Variance of the observations $ n + x_n $ for $ n = 1, 2, 3, 4, 5 $ is
Find the variance of the following frequency distribution:
| Class Interval | ||||
| 0--4 | 4--8 | 8--12 | 12--16 | |
| Frequency | 1 | 2 | 2 | 1 |
Observe the given sequence of nitrogenous bases on a DNA fragment and answer the following questions: 
(a) Name the restriction enzyme which can recognise the DNA sequence.
(b) Write the sequence after restriction enzyme cut the palindrome.
(c) Why are the ends generated after digestion called as ‘Sticky Ends’?
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: