Advertisements
Advertisements
Question
Calculate the mean deviation from the mean for the data:
38, 70, 48, 40, 42, 55, 63, 46, 54, 44a
Advertisements
Solution
Formula used for finding the mean deviation about the mean is given below:
\[MD = \frac{1}{n} \sum^n_{i = 1} \left| d_i \right| , \text{ where } \left| d_i \right| = \left| x_i - x \right|\]
Let x be the mean of the given data.
\[x = \frac{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}{10} = 50\]
|
|
\[\left| d_i \right| = \left| x_i - \bar{x} \right|\]
|
| 38 | 12 |
| 70 | 20 |
| 48 | 2 |
| 40 | 10 |
| 42 | 8 |
| 55 | 5 |
| 63 | 13 |
| 46 | 4 |
| 54 | 4 |
| 44 | 6 |
| Total | 84 |
\[MD = \frac{1}{10} \times 84 = 8 . 4\]
APPEARS IN
RELATED QUESTIONS
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.
| 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:
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 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:
| Size | 20 | 21 | 22 | 23 | 24 |
| Frequency | 6 | 4 | 5 | 1 | 4 |
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 | 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 |
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
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 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 ______.
Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.
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 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.
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 |
