Advertisements
Advertisements
प्रश्न
Calculate the mean deviation of the following income groups of five and seven members from their medians:
| I Income in Rs. |
II Income in Rs. |
| 4000 4200 4400 4600 4800 |
300 4000 4200 4400 4600 4800 5800 |
Advertisements
उत्तर
Calculate the mean deviation for the first data set.
The data is already arranged in ascending order.
For this data set, n is equal to 5.
Also, median,
M = 4400
\[MD = \frac{1}{n} \sum^n_{i = 1} \left| d_i \right|, \text{ where } \left| d_i \right| = \left| x_i - M \right|\]
|
\[x_i\]
|
\[\left| d_i \right| = \left| x_i - M \right|\]
|
| 4000 | 400 |
| 4200 | 200 |
| 4400 | 0 |
| 4600 | 200 |
| 4800 | 400 |
| Total | 1200 |
\[MD = \frac{1}{5} \times 1200 = 240\]
Therefore, for the income of families in the first group, the mean deviation from the median is Rs 240.
Now, consider the second data set. This is also arranged in ascending order.
Here,
n = 7.
Also, median,
| xi |
\[\left| d_i \right| = \left| x_i - M \right|\]
|
| 300 | 4100 |
| 4000 | 400 |
| 4200 | 200 |
| 4400 | 0 |
| 4600 | 200 |
| 4800 | 400 |
| 5800 | 1400 |
| Total | 6700 |
\[MD = \frac{1}{7} \times 6700 = 957 . 14\]
Therefore, for the income of families in the second group, the mean deviation from the median is Rs 957.14.
APPEARS IN
संबंधित प्रश्न
Find the mean deviation about the mean for the data.
4, 7, 8, 9, 10, 12, 13, 17
Find the mean deviation about the mean for the data.
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Find the mean deviation about the median for the data.
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Find the mean deviation about the median for the data.
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Find the mean deviation about the mean for the data.
| xi | 5 | 10 | 15 | 20 | 25 |
| fi | 7 | 4 | 6 | 3 | 5 |
Calculate the mean deviation about the median of the observation:
34, 66, 30, 38, 44, 50, 40, 60, 42, 51
Calculate the mean deviation about the median of the observation:
38, 70, 48, 34, 63, 42, 55, 44, 53, 47
Calculate the mean deviation from the mean for the data:
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Calculate the mean deviation from the mean for the data:
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
In 34, 66, 30, 38, 44, 50, 40, 60, 42, 51 find the number of observations lying between
\[\bar{ X } \] + M.D, where M.D. is the mean deviation from the mean.
In 38, 70, 48, 34, 63, 42, 55, 44, 53, 47 find the number of observations lying between
\[\bar { X } \] − M.D. and
\[\bar { X } \] + M.D, where M.D. is the mean deviation from the mean.
Find the mean deviation from the mean for the data:
| xi | 5 | 7 | 9 | 10 | 12 | 15 |
| fi | 8 | 6 | 2 | 2 | 2 | 6 |
Find the mean deviation from the mean for the data:
| xi | 5 | 10 | 15 | 20 | 25 |
| fi | 7 | 4 | 6 | 3 | 5 |
Find the mean deviation from the mean for the data:
| xi | 10 | 30 | 50 | 70 | 90 |
| fi | 4 | 24 | 28 | 16 | 8 |
Find the mean deviation from the mean for the data:
| Size | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 |
| Frequency | 3 | 3 | 4 | 14 | 7 | 4 | 3 | 4 |
Find the mean deviation from the median for the data:
| xi | 15 | 21 | 27 | 30 | 35 |
| fi | 3 | 5 | 6 | 7 | 8 |
Find the mean deviation from the median for the data:
| xi | 74 | 89 | 42 | 54 | 91 | 94 | 35 |
| fi | 20 | 12 | 2 | 4 | 5 | 3 | 4 |
Find the mean deviation from the mean for the data:
| Classes | 95-105 | 105-115 | 115-125 | 125-135 | 135-145 | 145-155 |
| Frequencies | 9 | 13 | 16 | 26 | 30 | 12 |
The age distribution of 100 life-insurance policy holders is as follows:
| Age (on nearest birth day) | 17-19.5 | 20-25.5 | 26-35.5 | 36-40.5 | 41-50.5 | 51-55.5 | 56-60.5 | 61-70.5 |
| No. of persons | 5 | 16 | 12 | 26 | 14 | 12 | 6 | 5 |
Calculate the mean deviation from the median age
Find the mean deviation from the mean and from median of the following distribution:
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| No. of students | 5 | 8 | 15 | 16 | 6 |
Calculate mean deviation about median age for the age distribution of 100 persons given below:
| Age: | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 |
| Number of persons | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
Calculate the mean deviation about the mean for the following frequency distribution:
| Class interval: | 0–4 | 4–8 | 8–12 | 12–16 | 16–20 |
| Frequency | 4 | 6 | 8 | 5 | 2 |
The mean deviation from the median is
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
Find the mean deviation about the mean of the following data:
| Size (x): | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 |
| Frequency (f): | 3 | 3 | 4 | 14 | 7 | 4 | 3 | 4 |
The mean deviation of the data 2, 9, 9, 3, 6, 9, 4 from the mean is ______.
Find the mean deviation about the mean of the distribution:
| Size | 20 | 21 | 22 | 23 | 24 |
| Frequency | 6 | 4 | 5 | 1 | 4 |
Find the mean deviation about the median of the following distribution:
| Marks obtained | 10 | 11 | 12 | 14 | 15 |
| No. of students | 2 | 3 | 8 | 3 | 4 |
Find the mean and variance of the frequency distribution given below:
| `x` | 1 ≤ x < 3 | 3 ≤ x < 5 | 5 ≤ x < 7 | 7 ≤ x < 10 |
| `f` | 6 | 4 | 5 | 1 |
Calculate the mean deviation from the median of the following data:
| Class interval | 0 – 6 | 6 – 12 | 12 – 18 | 18 – 24 | 24 – 30 |
| Frequency | 4 | 5 | 3 | 6 | 2 |
Mean deviation for n observations x1, x2, ..., xn from their mean `barx` is given by ______.
Let X = {x ∈ N: 1 ≤ x ≤ 17} and Y = {ax + b: x ∈ X and a, b ∈ R, a > 0}. If mean and variance of elements of Y are 17 and 216 respectively then a + b is equal to ______.
If the mean deviation of number 1, 1 + d, 1 + 2d, ..., 1 + 100d from their mean is 255, then d is equal to ______.
