Advertisements
Advertisements
Question
The mean deviation from the median is
Options
equal to that measured from another value
maximum if all observations are positive
greater than that measured from any other value.
less than that measured from any other value.
Solution
less than that measured from any other value.
In a frequency distribution, the sum of absolute values of deviations from the mean and mode is always more than the sum of the deviations from the median.
APPEARS IN
RELATED QUESTIONS
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 mean for the data.
xi | 5 | 10 | 15 | 20 | 25 |
fi | 7 | 4 | 6 | 3 | 5 |
Find the mean deviation about the mean for the data.
xi | 10 | 30 | 50 | 70 | 90 |
fi | 4 | 24 | 28 | 16 | 8 |
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 |
Find the mean deviation about the mean for the data.
Height in cms | Number of boys |
95 - 105 | 9 |
105 - 115 | 13 |
115 - 125 | 26 |
125 - 135 | 30 |
135 - 145 | 12 |
145 - 155 | 10 |
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:
3011, 2780, 3020, 2354, 3541, 4150, 5000
Calculate the mean deviation from the mean for the data:
4, 7, 8, 9, 10, 12, 13, 17
Calculate the mean deviation from the mean for the data:
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Calculate the mean deviation from the mean for the data:
38, 70, 48, 40, 42, 55, 63, 46, 54, 44a
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 |
In 22, 24, 30, 27, 29, 31, 25, 28, 41, 42 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 | 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 | 74 | 89 | 42 | 54 | 91 | 94 | 35 |
fi | 20 | 12 | 2 | 4 | 5 | 3 | 4 |
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 |
Find the mean deviation from the mean for the data:
Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Frequencies | 6 | 8 | 14 | 16 | 4 | 2 |
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 |
For a frequency distribution mean deviation from mean is computed by
The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the 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
The mean deviation of the data 2, 9, 9, 3, 6, 9, 4 from the mean is ______.
Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the item and the sum of the squares of the items.
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is ______.
Mean deviation for n observations x1, x2, ..., xn from their mean `barx` is given by ______.
The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is ______.
The sum of squares of the deviations of the values of the variable is ______ when taken about their arithmetic mean.
The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is ______.