Question -
| xi | 15 | 21 | 27 | 30 |
| fi | 3 | 5 | 6 | 7 |
(ii)
| xi | 74 | 89 | 42 | 54 | 91 | 94 | 35 |
| fi | 20 | 12 | 2 | 4 | 5 | 3 | 4 |
(iii)
| Marks obtained | 10 | 11 | 12 | 14 | 15 |
| No. of students | 2 | 3 | 8 | 3 | 4 |
Answer -
(i)
To find the mean deviation from the median, firstly let uscalculate the median.
We know, N = 21
Median = (21)/2 = 10.5
So, the median Corresponding to 10.5 is 27
| xi | fi | Cumulative Frequency | |di| = |xi – M| | fi |di| |
| 15 | 3 | 3 | 15 | 45 |
| 21 | 5 | 8 | 9 | 45 |
| 27 | 6 | 14 | 3 | 18 |
| 30 | 7 | 21 | 0 | 0 |
| | | | |
| Total = 21 | Total = 46 | | Total = 108 |
N = 21
MD=
= 1/21 × 108
= 5.14
∴ Themean deviation is 5.14
(ii)
To find the mean deviation from the median, firstly let uscalculate the median.
We know, N = 50
Median = (50)/2 = 25
So, the median Corresponding to 25 is 74
| xi | fi | Cumulative Frequency | |di| = |xi – M| | fi |di| |
| 74 | 20 | 4 | 39 | 156 |
| 89 | 12 | 6 | 32 | 64 |
| 42 | 2 | 10 | 20 | 80 |
| 54 | 4 | 30 | 0 | 0 |
| 91 | 5 | 42 | 15 | 180 |
| 94 | 3 | 47 | 17 | 85 |
| 35 | 4 | 50 | 20 | 60 |
| Total = 50 | Total = 189 | | Total = 625 |
N = 50
MD=
= 1/50 × 625
= 12.5
∴ Themean deviation is 12.5
(iii)
To find the mean deviation from the median, firstly let uscalculate the median.
We know, N = 20
Median = (20)/2 = 10
So, the median Corresponding to 10 is 12
| xi | fi | Cumulative Frequency | |di| = |xi – M| | fi |di| |
| 10 | 2 | 2 | 2 | 4 |
| 11 | 3 | 5 | 1 | 3 |
| 12 | 8 | 13 | 0 | 0 |
| 14 | 3 | 16 | 2 | 6 |
| 15 | 4 | 20 | 3 | 12 |
| Total = 20 | | | Total = 25 |
N = 20
MD=
= 1/20 × 25
= 1.25
∴ Themean deviation is 1.25