4.36,72,46,42, 60, 45, 53, 46,51,49
The given data:36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Arranging in ascending order, we get
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Since the number of observations is 10 which is even, so median is given as
M = ( n 2 ) th + ( n 2 + 1 ) th 2 (1)
M = ( 10 2 ) th + ( 10 2 + 1 ) th 2 = 5 t h + 6 t h 2
From the given data,
M = 46 + 49 2 = 95 2 = 47.5
Therefore, the median of the given data is 47.5.
The formula to calculate the deviations of each observation from the median M is, x i − M
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
x i M | x i − M | 36 47.5 11.5 42 47.5 5.5 45 47.5 2.5 46 47.5 1.5 46 47.5 1.5 49 47.5 1.5 51 47.5 3.5 53 47.5 5.5 60 47.5 12.5 72 47.5 24.5
The formula to calculate the mean deviation about the median is as follows:
M .D .= ∑ i = 1 n | x i − M | n (4)
Note: Here we consider the absolute value of the deviation of each observation from the medium.
Therefore,
M .D .= 11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5 10 = 70 10 = 7
Therefore, the mean deviation of the given data about the median is 7.
The given data:36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Arranging in ascending order, we get
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Since the number of observations is 10 which is even, so median is given as
From the given data,
Therefore, the median of the given data is 47.5.
The formula to calculate the deviations of each observation from the median
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
| | M | |
| 36 | 47.5 | 11.5 |
| 42 | 47.5 | 5.5 |
| 45 | 47.5 | 2.5 |
| 46 | 47.5 | 1.5 |
| 46 | 47.5 | 1.5 |
| 49 | 47.5 | 1.5 |
| 51 | 47.5 | 3.5 |
| 53 | 47.5 | 5.5 |
| 60 | 47.5 | 12.5 |
| 72 | 47.5 | 24.5 |
The formula to calculate the mean deviation about the median is as follows:
Note: Here we consider the absolute value of the deviation of each observation from the medium.
Therefore,
Therefore, the mean deviation of the given data about the median is 7.
