Advertisements
Advertisements
प्रश्न
Calculate the mean deviation from the mean for the data:
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Advertisements
उत्तर
Formula used for finding the mean deviation about the mean is given below:
\[MD = \frac{1}{n} \sum^n_{i = 1} \left| d_i \right| , \text{ where } \left| d_i \right| = \left| x_i - x \right|\]
iv)
Let x be the mean of the given data.
\[x = \frac{36 + 746 + 42 + 60 + 45 + 53 + 46 + 51 + 59}{10} = 50\]
|
|
|
| 36 | 14 |
| 72 | 22 |
| 46 | 4 |
| 42 | 8 |
| 60 | 10 |
| 45 | 5 |
| 53 | 3 |
| 46 | 4 |
| 51 | 1 |
| 49 | 1 |
| Total | 72 |
\[MD = \frac{1}{10} \times 72 = 7 . 2\]
APPEARS IN
संबंधित प्रश्न
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 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 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 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 from the mean for the data:
4, 7, 8, 9, 10, 12, 13, 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 |
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
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 the mean also.
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:
| 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 | 10 | 30 | 50 | 70 | 90 |
| fi | 4 | 24 | 28 | 16 | 8 |
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 |
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 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 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 series a, a + d, a + 2d, ..., a + 2n from its mean is
The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is
The mean deviation of the data 2, 9, 9, 3, 6, 9, 4 from the mean is ______.
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 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 |
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`
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 ______.
If the mean deviation of number 1, 1 + d, 1 + 2d, ..., 1 + 100d from their mean is 255, then d is equal to ______.
