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 |
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 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 | 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 |
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:
(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 |
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:
xi | 5 | 10 | 15 | 20 | 25 |
fi | 7 | 4 | 6 | 3 | 5 |
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 mean for the data:
Classes | 95-105 | 105-115 | 115-125 | 125-135 | 135-145 | 145-155 |
Frequencies | 9 | 13 | 16 | 26 | 30 | 12 |
Find the mean deviation from the mean for the data:
Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Frequencies | 6 | 8 | 14 | 16 | 4 | 2 |
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 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 |
For a frequency distribution mean deviation from mean is computed by
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 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
Let \[x_1 , x_2 , . . . , x_n\] be n observations and \[X\] be their arithmetic mean. The standard deviation is given by
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.
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 |
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 |
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is ______.
The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations 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.