Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
Find the mean deviation about the mean for the data.
4, 7, 8, 9, 10, 12, 13, 17
Find the mean deviation about the median for the data.
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
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 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:
38, 70, 48, 34, 63, 42, 55, 44, 53, 47
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:
38, 70, 48, 40, 42, 55, 63, 46, 54, 44a
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.
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 | 20 | 21 | 22 | 23 | 24 |
| Frequency | 6 | 4 | 5 | 1 | 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 |
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
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 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 |
Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even number.
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 ______.
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 ______.
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 ______.
