Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Let \[\bar{x}\] be the mean of the data set.
\[\bar{ x } = \frac{22 + 24 + 30 + 27 + 29 + 31 + 25 + 28 + 41 + 42}{10} = 29 . 9\]
|
\[x_i\]
|
\[\left| d_i \right| = \left| x_i - 29 . 9 \right|\]
|
| 22 | 7.9 |
| 24 | 5.9 |
| 30 | 0.1 |
| 27 | 2.9 |
| 29 | 0.9 |
| 31 | 1.1 |
| 25 | 4.9 |
| 28 | 1.9 |
| 41 | 11.9 |
| 42 | 12.1 |
| Total | 48.8 |
\[MD = \frac{1}{10} \times 48 . 8 = 4 . 88\]
\[\bar{ x } - M . D . = 29 . 9 - 4 . 88 = 25 . 02, \]
\[\text{ and } \bar { x } + M . D . = 29 . 9 + 4 . 88 = 34 . 78\]
There are 5 observations between 25.02 and 34.78.
APPEARS IN
संबंधित प्रश्न
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 |
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 about the median of the observation:
3011, 2780, 3020, 2354, 3541, 4150, 5000
Calculate the mean deviation about the median of the observation:
38, 70, 48, 34, 42, 55, 63, 46, 54, 44
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 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
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 |
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 median for the data:
| xi | 74 | 89 | 42 | 54 | 91 | 94 | 35 |
| fi | 20 | 12 | 2 | 4 | 5 | 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 |
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 |
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 |
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
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 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
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 for n observations \[x_1 , x_2 , . . . , x_n\] from their mean \[\bar{X} \] is given by
Find the mean deviation about the median of the following distribution:
| Marks obtained | 10 | 11 | 12 | 14 | 15 |
| No. of students | 2 | 3 | 8 | 3 | 4 |
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 |
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.
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`
Find the mean deviation about the mean for the data.
| xi | 5 | 10 | 15 | 20 | 25 |
| fi | 7 | 4 | 6 | 3 | 5 |
If the mean deviation of number 1, 1 + d, 1 + 2d, ..., 1 + 100d from their mean is 255, then d is equal to ______.
