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.
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 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 | 15 | 21 | 27 | 30 | 35 |
| fi | 3 | 5 | 6 | 7 | 8 |
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 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 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
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 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 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 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 |
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 from the median is
The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is
The mean deviation for n observations \[x_1 , x_2 , . . . , x_n\] from their mean \[\bar{X} \] is given by
The mean deviation of the data 2, 9, 9, 3, 6, 9, 4 from the mean is ______.
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 |
Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.
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.
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`
The sum of squares of the deviations of the values of the variable is ______ when taken about their arithmetic mean.
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 ______.
