Advertisements
Advertisements
प्रश्न
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
पर्याय
2
2.57
3
3.57
उत्तर
The given observations are 3, 10, 10, 4, 7, 10, 5.
∴ Mean, \[x = \frac{3 + 10 + 10 + 4 + 7 + 10 + 5}{7} = \frac{49}{7} = 7\]
Now,
Mean deviation from mean, MD
\[= \frac{\left| 3 - 7 \right| + \left| 10 - 7 \right| + \left| 10 - 7 \right| + \left| 4 - 7 \right| + \left| 7 - 7 \right| + \left| 10 - 7 \right| + \left| 5 - 7 \right|}{7}\]
\[ = \frac{4 + 3 + 3 + 3 + 0 + 3 + 2}{7}\]
\[ = \frac{18}{7}\]
\[ = 2 . 57\]
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.
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 |
Find the mean deviation about median for the following data:
Marks | Number of girls |
0-10 | 6 |
10-20 | 8 |
20-30 | 14 |
30-40 | 16 |
40-50 | 4 |
50-60 | 2 |
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:
38, 70, 48, 34, 63, 42, 55, 44, 53, 47
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 | 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:
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 |
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 |
The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.
For a frequency distribution mean deviation from mean is computed by
The mean deviation of the series a, a + d, a + 2d, ..., a + 2n from its mean is
The mean deviation of the data 2, 9, 9, 3, 6, 9, 4 from the mean is ______.
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 |
Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.
Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even number.
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.
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 |
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is ______.
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 ______.
If `barx` is the mean of n values of x, then `sum_(i = 1)^n (x_i - barx)` is always equal to ______. If a has any value other than `barx`, then `sum_(i = 1)^n (x_i - barx)^2` is ______ than `sum(x_i - a)^2`
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 ______.