Advertisements
Advertisements
Question
The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is
Options
25
5
1.2
0
Solution
1.2
\[\text{ Mean } \left( X \right) = \frac{3 + 4 + 5 + 6 + 7}{5}\]
\[ = \frac{25}{5}\]
\[ = 5\]
Taking the absolute value of deviation of each term from the mean, we get:
\[MD = \frac{\left| (3 - 5) \right| + \left| (4 - 5) \right| + \left| (5 - 5) \right| + \left| (6 - 5) \right| + \left| (7 - 5) \right|}{5}\]
\[ = \frac{2 + 1 + 0 + 1 + 2}{5}\]
\[ = \frac{6}{5}\]
\[ = 1 . 2\]
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 mean for the data.
xi | 5 | 10 | 15 | 20 | 25 |
fi | 7 | 4 | 6 | 3 | 5 |
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 |
Find the mean deviation about the median for the data.
xi | 15 | 21 | 27 | 30 | 35 |
fi | 3 | 5 | 6 | 7 | 8 |
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 the mean for the data.
Height in cms | Number of boys |
95 - 105 | 9 |
105 - 115 | 13 |
115 - 125 | 26 |
125 - 135 | 30 |
135 - 145 | 12 |
145 - 155 | 10 |
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:
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
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 |
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 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 | 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 | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Frequencies | 6 | 8 | 14 | 16 | 4 | 2 |
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 |
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 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.
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
Let \[x_1 , x_2 , . . . , x_n\] be n observations and \[X\] be their arithmetic mean. The standard deviation 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 |
Find the mean deviation about the mean of the distribution:
Size | 20 | 21 | 22 | 23 | 24 |
Frequency | 6 | 4 | 5 | 1 | 4 |
Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the item and the sum of the squares of the items.
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 |
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`
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 ______.