Advertisements
Advertisements
प्रश्न
The mean deviation of the data 2, 9, 9, 3, 6, 9, 4 from the mean is ______.
विकल्प
2.23
2.57
3.23
3.57
उत्तर
The mean deviation of the data 2, 9, 9, 3, 6, 9, 4 from the mean is 2.57.
Explanation:
M.D. `(barx) = |x_i - barx|/n`
= `(4 + 3 + 3 + 3 + 0 + 3 + 2)/7`
= 2.57
APPEARS IN
संबंधित प्रश्न
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 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 from the mean for the data:
38, 70, 48, 40, 42, 55, 63, 46, 54, 44a
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 | 10 | 30 | 50 | 70 | 90 |
fi | 4 | 24 | 28 | 16 | 8 |
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 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 |
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 |
Find the mean deviation about the mean of the distribution:
Size | 20 | 21 | 22 | 23 | 24 |
Frequency | 6 | 4 | 5 | 1 | 4 |
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 |
Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd 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 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.
Mean deviation for n observations x1, x2, ..., xn from their mean `barx` is given by ______.
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.