Question -
Answer -
(iii) 34,66, 30, 38, 44, 50, 40, 60, 42, 51
We know that,
MD=
Where, |di| = |xi – x|
So, let ‘x’ be the mean of the given observation.
x = [34 + 66 + 30 + 38 + 44 + 50 + 40 + 60 + 42 + 51]/10
= 455/10
= 45.5
Number of observations, ‘n’ = 10
| xi | |di| = |xi – 45.5| |
| 34 | 11.5 |
| 66 | 20.5 |
| 30 | 15.5 |
| 38 | 7.5 |
| 44 | 1.5 |
| 50 | 4.5 |
| 40 | 5.5 |
| 60 | 14.5 |
| 42 | 3.5 |
| 51 | 5.5 |
| Total | 90 |
MD=
= 1/10 × 90
= 9
Now
–M.D = 45.5 – 9 = 36.5
+M.D = 45.5 + 9 = 54.5
So,There are total 6 observation between –M.D and +M.D
(iv) 22, 24,30, 27, 29, 31, 25, 28, 41, 42
We know that,
MD=
Where, |di| = |xi – x|
So, let ‘x’ be the mean of the given observation.
x = [22 + 24 + 30 + 27 + 29 + 31 + 25 + 28 + 41 + 42]/10
= 299/10
= 29.9
Number of observations, ‘n’ = 10
| xi | |di| = |xi – 29.9| |
| 22 | 7.9 |
| 24 | 5.9 |
| 30 | 0.1 |
| 27 | 2.9 |
| 29 | 0.9 |
| 31 | 1.1 |
| 25 | 4.9 |
| 28 | 1.9 |
| 41 | 11.1 |
| 42 | 12.1 |
| Total | 48.8 |
MD=
= 1/10 × 48.8
= 4.88
Now
So, There are total 5 observation between 
and
(v) 38, 70,48, 34, 63, 42, 55, 44, 53, 47
We know that,
MD=
Where, |di| = |xi – x|
So, let ‘x’ be the mean of the given observation.
x = [38 + 70 + 48 + 34 + 63 + 42 + 55 + 44 + 53 + 47]/10
= 494/10
= 49.4
Number of observations, ‘n’ = 10
| xi | |di| = |xi – 49.4| |
| 38 | 11.4 |
| 70 | 20.6 |
| 48 | 1.4 |
| 34 | 15.4 |
| 63 | 13.6 |
| 42 | 7.4 |
| 55 | 5.6 |
| 44 | 5.4 |
| 53 | 3.6 |
| 47 | 2.4 |
| Total | 86.8 |
MD=
= 1/10 × 86.8
= 8.68
Now
