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प्रश्न
Using step-deviation method, find mean for the following frequency distribution:
| Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
| Frequency | 3 | 4 | 7 | 6 | 8 | 2 |
उत्तर
| Class | x | u = d/i | f | fu |
| 0 – 15 | 7.5 | –3 | 3 | –9 |
| 15 – 30 | 22.5 | –2 | 4 | –8 |
| 30 – 45 | 37.5 | –1 | 7 | –7 |
| 45 – 60 | 52.5 | 0 | 6 | 0 |
| 60 – 75 | 67.5 | 1 | 8 | 8 |
| 75 – 90 | 82.5 | 2 | 2 | 4 |
| 30 | –12 |
Mean = `A + (sumfu)/(sumf) xx i`
= `52.5 + (-12)/30 xx 15`
= 52.5 – 6
= 46.50
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संबंधित प्रश्न
Consider the following distribution of daily wages of 50 worker of a factory.
|
Daily wages (in Rs) |
100 − 120 |
120 − 140 |
140 −1 60 |
160 − 180 |
180 − 200 |
|
Number of workers |
12 |
14 |
8 |
6 |
10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
If the mean of the following data is 15, find p.
| x | 5 | 10 | 15 | 20 | 25 |
| f | 6 | P | 6 | 10 | 5 |
The arithmetic mean of the following data is 25, find the value of k.
| x1 | 5 | 15 | 25 | 35 | 45 |
| f1 | 3 | k | 3 | 6 | 2 |
Find the mean of each of the following frequency distributions
| Class interval | 10 - 30 | 30 - 50 | 50 - 70 | 70 - 90 | 90 - 110 | 110 - 130 |
| Frequency | 5 | 8 | 12 | 20 | 3 | 2 |
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2.
| Class | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 |
| Frequency | 5 | f1 | 10 | f2 | 7 | 8 |
Find the mean using direct method:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency | 7 | 5 | 6 | 12 | 8 | 2 |
The mean of first n odd natural numbers is \[\frac{n^2}{81}\],then n =
The weekly wages of 120 workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.
|
Weekly wages
(Rupees)
|
0 - 2000 | 2000 - 4000 | 4000 - 6000 | 6000 - 8000 |
| No. of workers | 15 | 35 | 50 | 20 |
Find the mean of: 5, 2.4, 6.2, 8.9, 4.1 and 3.4
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
| Length (in mm) | Number of leaves |
| 118−126 | 3 |
| 127–135 | 5 |
| 136−144 | 9 |
| 145–153 | 12 |
| 154–162 | 5 |
| 163–171 | 4 |
| 172–180 | 2 |
Find the mean length of the leaves.
