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प्रश्न
The sum of squares of the deviations of the values of the variable is ______ when taken about their arithmetic mean.
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उत्तर
The sum of squares of the deviations of the values of the variable is minimum when taken about their arithmetic mean.
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संबंधित प्रश्न
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 median for the data.
| xi | 15 | 21 | 27 | 30 | 35 |
| fi | 3 | 5 | 6 | 7 | 8 |
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 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 about the median of the observation:
22, 24, 30, 27, 29, 31, 25, 28, 41, 42
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:
38, 70, 48, 40, 42, 55, 63, 46, 54, 44a
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 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.
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 |
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 | 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
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 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 |
For a frequency distribution mean deviation from mean is computed by
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 |
Find the mean deviation about the mean of the distribution:
| Size | 20 | 21 | 22 | 23 | 24 |
| Frequency | 6 | 4 | 5 | 1 | 4 |
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 |
While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
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 ______.
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 ______.
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 ______.
