Advertisements
Advertisements
प्रश्न
The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is
पर्याय
25
5
1.2
0
उत्तर
1.2
\[\text{ Mean } \left( X \right) = \frac{3 + 4 + 5 + 6 + 7}{5}\]
\[ = \frac{25}{5}\]
\[ = 5\]
Taking the absolute value of deviation of each term from the mean, we get:
\[MD = \frac{\left| (3 - 5) \right| + \left| (4 - 5) \right| + \left| (5 - 5) \right| + \left| (6 - 5) \right| + \left| (7 - 5) \right|}{5}\]
\[ = \frac{2 + 1 + 0 + 1 + 2}{5}\]
\[ = \frac{6}{5}\]
\[ = 1 . 2\]
APPEARS IN
संबंधित प्रश्न
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 mean for the data.
| xi | 5 | 10 | 15 | 20 | 25 |
| fi | 7 | 4 | 6 | 3 | 5 |
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 |
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:
34, 66, 30, 38, 44, 50, 40, 60, 42, 51
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:
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
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 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.
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 | 20 | 21 | 22 | 23 | 24 |
| Frequency | 6 | 4 | 5 | 1 | 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 | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |
| Frequencies | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |
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 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 |
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 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
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 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 even number.
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 |
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.
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`
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 ______.
Find the mean deviation about the mean for the data.
| xi | 5 | 10 | 15 | 20 | 25 |
| fi | 7 | 4 | 6 | 3 | 5 |
