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प्रश्न
The algebraic sum of the deviations of a frequency distribution from its mean is always ______.
पर्याय
Always positive
Always negative
Zero
A non-zero number
उत्तर
The algebraic sum of the deviations of a frequency distribution from its mean is always zero.
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संबंधित प्रश्न
The measurements (in mm) of the diameters of the head of the screws are given below:
| Diameter (in mm) | No. of Screws |
| 33 — 35 | 10 |
| 36 — 38 | 19 |
| 39 — 41 | 23 |
| 42 — 44 | 21 |
| 45 — 47 | 27 |
Calculate mean diameter of head of a screw by ‘Assumed Mean Method’.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Fine the mean heartbeats per minute for these women, choosing a suitable method.
| Number of heartbeats per minute | 65 - 68 | 68 - 71 | 71 - 74 | 74 - 77 | 77 - 80 | 80 - 83 | 83 - 86 |
| Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Find the value of p for the following distribution whose mean is 16.6
| x | 8 | 12 | 15 | P | 20 | 25 | 30 |
| f | 12 | 16 | 20 | 24 | 16 | 8 | 4 |
Find the mean of each of the following frequency distributions
| Class interval | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 |
| Frequency | 9 | 12 | 15 | 10 | 14 |
The measurements (in mm) of the diameters of the head of the screws are given below :
| Diameter (in mm) | no. of screws |
| 33 - 35 | 9 |
| 36 - 38 | 21 |
| 39 - 41 | 30 |
| 42 - 44 | 22 |
| 45 - 47 | 18 |
Calculate the mean diameter of the head of a screw by the ' Assumed Mean Method'.
The contents of 100 match box were checked to determine the number of match sticks they contained.
| Number of match sticks | Number of boxes |
| 35 | 6 |
| 36 | 10 |
| 37 | 18 |
| 38 | 25 |
| 39 | 21 |
| 40 | 12 |
| 41 | 8 |
(i) Calculate correct to one decimal place, the mean number of match sticks per box.
(ii) Determine how many matchsticks would have to be added. To the total contents of the 100 boxes to bring the mean up exactly 39 match sticks.
In a small scale industry, salaries of employees are given in the following distribution table:
|
Salary (in Rs.) |
4000 - 5000 |
5000 - 6000 |
6000 - 7000 |
7000 - 8000 |
8000 - 9000 |
9000 - 10000 |
|
Number of employees |
20 | 60 | 100 | 50 | 80 | 90 |
Then the mean salary of the employee is?
There is a grouped data distribution for which mean is to be found by step deviation method.
| Class interval | Number of Frequency (fi) | Class mark (xi) | di = xi - a | `u_i=d_i/h` |
| 0 - 100 | 40 | 50 | -200 | D |
| 100 - 200 | 39 | 150 | B | E |
| 200 - 300 | 34 | 250 | 0 | 0 |
| 300 - 400 | 30 | 350 | 100 | 1 |
| 400 - 500 | 45 | 450 | C | F |
| Total | `A=sumf_i=....` |
Find the value of A, B, C, D, E and F respectively.
The daily income of a sample of 50 employees are tabulated as follows:
| Income (in Rs) |
1 – 200 | 201 – 400 | 401 – 600 | 601 – 800 |
| Number of employees |
14 | 15 | 14 | 7 |
Find the mean daily income of employees.
Find the mean of the following frequency distribution:
| Class | 1 – 5 | 5 – 9 | 9 – 13 | 13 – 17 |
| Frequency | 4 | 8 | 7 | 6 |
