Advertisements
Advertisements
प्रश्न
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 |
Advertisements
उत्तर
Computation of mean deviation from the mean:
| Marks | Number of Students \[f_i\]
|
Midpoints \[x_i\]
|
\[f_i x_i\]
|
\[\left| x_i - X \right|\]
\[\left| x_i - 49 \right|\]
|
\[f_i \left| x_i - X \right|\]
|
| 10−20 | 8 | 15 | 120 | 34 | 272 |
| 20−30 | 10 | 25 | 250 | 24 | 240 |
| 30−40 | 15 | 35 | 525 | 14 | 210 |
| 40−50 | 25 | 45 | 1125 | 4 | 100 |
| 50−60 | 20 | 55 | 1100 | 6 | 120 |
| 60−70 | 18 | 65 | 1170 | 16 | 288 |
| 70−80 | 9 | 75 | 675 | 26 | 234 |
| 80−90 | 5 | 85 | 425 | 36 | 180 |
|
\[N = \sum^8_{i = 1} f_i = 110\]
|
\[\sum^8_{i = 1} f_i x_i = 5390\]
|
\[\sum^8_{i = 1} f_i \left| x_i - X \right| = 1644\]
|
and
\[\sum^8_{i = 1} f_i x_i = 5390\]
\[X = \frac{\sum^8_{i = 1} f_i x_i}{N}\]
\[ = \frac{5390}{110}\]
\[ = 49\]
\[\text{ Mean deviation } = \frac{\sum^8_{i = 1} f_i \left| x_i - X \right|}{N}\]
\[ = \frac{1644}{110}\]
\[ = 14 . 945\]
\[ \approx 14 . 95\]
APPEARS IN
संबंधित प्रश्न
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 |
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:
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
Find mean deviation from median
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:
| 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 | 15 | 21 | 27 | 30 | 35 |
| fi | 3 | 5 | 6 | 7 | 8 |
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 | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |
| Frequencies | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |
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 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.
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 of the data 2, 9, 9, 3, 6, 9, 4 from the mean is ______.
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 deviation about the median of the following distribution:
| Marks obtained | 10 | 11 | 12 | 14 | 15 |
| No. of students | 2 | 3 | 8 | 3 | 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 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.
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 ______.
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`
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 ______.
If the mean deviation of number 1, 1 + d, 1 + 2d, ..., 1 + 100d from their mean is 255, then d is equal to ______.
