Advertisements
Advertisements
Question
Calculate the mean deviation about the median of the observation:
22, 24, 30, 27, 29, 31, 25, 28, 41, 42
Solution
Formula used for mean deviation:
\[MD = \frac{1}{n} \sum^n_{i = 1} \left| d_i \right|\]
\[Here, \]
\[ d_i = x_i - M\]
M = Median
iv) Arranging the data in ascending order.
22, 24, 25, 27, 28, 29, 30, 31, 41, 42
\[\text{ Here } , n = 10 .\]
Also, median is the AM of the fifth and the sixth observation.
\[Median, M = \frac{28 + 29}{2} = 28 . 5\]
xi |
\[\left| d_i \right| = \left| x_i - M \right|\]
|
22 | 6.5 |
24 | 4.5 |
30 | 1.5 |
27 | 1.5 |
29 | 0.5 |
31 | 2.5 |
25 | 3.5 |
28 | 0.5 |
41 | 12.5 |
41 | 13.5 |
Total | 47 |
APPEARS IN
RELATED QUESTIONS
Find the mean deviation about the mean for the data.
4, 7, 8, 9, 10, 12, 13, 17
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.
xi | 5 | 7 | 9 | 10 | 12 | 15 |
fi | 8 | 6 | 2 | 2 | 2 | 6 |
Find the mean deviation about the mean for the data.
Income per day in ₹ | Number of persons |
0-100 | 4 |
100-200 | 8 |
200-300 | 9 |
300-400 | 10 |
400-500 | 7 |
500-600 | 5 |
600-700 | 4 |
700-800 | 3 |
Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age | Number |
16 - 20 | 5 |
21 - 25 | 6 |
26 - 30 | 12 |
31 - 35 | 14 |
36 - 40 | 26 |
41 - 45 | 12 |
46 - 50 | 16 |
51 - 55 | 9 |
Calculate the mean deviation about the median of the observation:
38, 70, 48, 34, 42, 55, 63, 46, 54, 44
Calculate the mean deviation about the median of the observation:
34, 66, 30, 38, 44, 50, 40, 60, 42, 51
Calculate the mean deviation from the mean for the data:
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
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 | 20 | 21 | 22 | 23 | 24 |
Frequency | 6 | 4 | 5 | 1 | 4 |
Find the mean deviation from the median for the data:
xi | 15 | 21 | 27 | 30 | 35 |
fi | 3 | 5 | 6 | 7 | 8 |
Compute the mean deviation from the median of the following distribution:
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 5 | 10 | 20 | 5 | 10 |
Compute mean deviation from mean of the following distribution:
Mark | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |
No. of students | 8 | 10 | 15 | 25 | 20 | 18 | 9 | 5 |
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 |
For a frequency distribution mean deviation from mean is computed by
A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean is
The mean deviation for n observations \[x_1 , x_2 , . . . , x_n\] from their mean \[\bar{X} \] is given by
Let \[x_1 , x_2 , . . . , x_n\] be n observations and \[X\] be their arithmetic mean. The standard deviation is given by
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 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 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 |
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 |
Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.
Mean deviation for n observations x1, x2, ..., xn from their mean `barx` is given by ______.
When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623
The mean deviations (in hours) from their mean is ______.
Find the mean deviation about the mean for the data.
xi | 5 | 10 | 15 | 20 | 25 |
fi | 7 | 4 | 6 | 3 | 5 |
If the mean deviation of number 1, 1 + d, 1 + 2d, ..., 1 + 100d from their mean is 255, then d is equal to ______.