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प्रश्न
Find the mean of the distribution:
| Class | 1 – 3 | 3 – 5 | 5 – 7 | 7 – 10 |
| Frequency | 9 | 22 | 27 | 17 |
उत्तर
We first, find the class mark xi of each class and then proceed as follows.
| Class | Class marks `(bb(x_i))` |
Frequency `(bb(f_i))` |
`bb(f_ix_i)` |
| 1 – 3 | 2 | 9 | 18 |
| 3 – 5 | 4 | 22 | 88 |
| 5 – 7 | 6 | 27 | 162 |
| 7 – 10 | 8.5 | 17 | 144.5 |
| `sumf_i = 75` | `sumf_ix_i = 412.5` |
Therefore, mean `(barx) = (sumf_ix_i)/(sumf_i)`
= `412.5/75`
= 5.5
Hence, mean of the given distribution is 5.5.
संबंधित प्रश्न
The following table gives the frequency distribution of trees planted by different Housing Societies in a particular locality:
| No. of Trees | No. of Housing Societies |
| 10-15 | 2 |
| 15-20 | 7 |
| 20-25 | 9 |
| 25-30 | 8 |
| 30-35 | 6 |
| 35-40 | 4 |
Find the mean number of trees planted by Housing Societies by using ‘Assumed Means Method’
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
| concentration of SO2 (in ppm) | Frequency |
| 0.00 − 0.04 | 4 |
| 0.04 − 0.08 | 9 |
| 0.08 − 0.12 | 9 |
| 0.12 − 0.16 | 2 |
| 0.16 − 0.20 | 4 |
| 0.20 − 0.24 | 2 |
Find the mean concentration of SO2 in the air.
If the mean of the following data is 15, find p.
| x | 5 | 10 | 15 | 20 | 25 |
| f | 6 | P | 6 | 10 | 5 |
Find the value of p, if the mean of the following distribution is 20.
| x | 15 | 17 | 19 | 20+P | 23 |
| f | 2 | 3 | 4 | 5P | 6 |
The following table gives the number of children of 150 families in a village. Find the average number of children per family.
| No. of children (x) | 0 | 1 | 2 | 3 | 4 | 5 |
| No. of families (f) | 10 | 21 | 55 | 42 | 15 | 7 |
The following distribution gives the number of accidents met by 160 workers in a factory during a month.
| No. of accidents(x) | 0 | 1 | 2 | 3 | 4 |
| No. of workers (f) | 70 | 52 | 34 | 3 | 1 |
Find the average number of accidents per worker.
Find the mean of each of the following frequency distributions
| Class interval | 0 - 8 | 8 - 16 | 16 - 24 | 24 - 32 | 32 - 40 |
| Frequency | 5 | 6 | 4 | 3 | 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 |
Find the mean of each of the following frequency distributions
| Classes | 25 - 29 | 30 - 34 | 35 - 39 | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 |
| Frequency | 14 | 22 | 16 | 6 | 5 | 3 | 4 |
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 of the following data, using direct method:
| Class | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
| Frequency | 6 | 10 | 8 | 12 | 4 |
The daily expenditure of 100 families are given below. Calculate `f_1` and `f_2` if the mean daily expenditure is ₹ 188.
|
Expenditure (in Rs) |
140-160 | 160-180 | 180-200 | 200-220 | 220-240 |
|
Number of families |
5 | 25 | `f_1` | `f_2` | 5 |
Find the mean of the following data, using assumed-mean method:
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 - 120 |
| Frequency | 20 | 35 | 52 | 44 | 38 | 31 |
Find the arithmetic mean of the following frequency distribution using step-deviation method:
| Age (in years) | 18 – 24 | 24 – 30 | 30 – 36 | 36 – 42 | 42 – 48 | 48 – 54 |
| Number of workers | 6 | 8 | 12 | 8 | 4 | 2 |
Find the correct answer from the alternatives given.
The formula to find mean from a grouped frequency table is \[X = A + \frac{\sum f_i u_i}{\sum f_i} \times hg\] .
Consider the following distribution of daily wages of 50 workers of a factory:
| Daily wages (in ₹) |
500-520 | 520-540 | 540-560 | 560-580 | 580-600 |
| 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 first n natural number is 15, then n =
The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distance.
|
Distance (km)
|
200 - 210 | 210 - 220 | 220 - 230 | 230 - 240 | 240 - 250 |
| No. of buses | 40 | 60 | 80 | 50 | 20 |
If the mean of the following distribution is 7.5, find the missing frequency ‘f’:
| Variable : | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Frequency: | 20 | 17 | f | 10 | 8 | 6 | 7 | 6 |
A frequency distribution of the life times of 400 T.V., picture tubes leased in tube company is given below. Find the average life of tube:
| Life time (in hrs) | Number of tubes |
| 300 - 399 | 14 |
| 400 - 499 | 46 |
| 500 - 599 | 58 |
| 600 - 699 | 76 |
| 700 - 799 | 68 |
| 800 - 899 | 62 |
| 900 - 999 | 48 |
| 1000 - 1099 | 22 |
| 1100 - 1199 | 6 |
Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is ______.
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?
The median from the table is?
| Value | Frequency |
| 7 | 2 |
| 8 | 1 |
| 9 | 4 |
| 10 | 5 |
| 11 | 6 |
| 12 | 1 |
| 13 | 3 |
The mean of the following distribution is
| xi | 10 | 13 | 16 | 19 |
| fi | 2 | 5 | 7 | 6 |
The average weight of a group of 25 men was calculated to be 78.4 kg. It was discovered later that one weight was wrongly entered as 69 kg instead of 96 kg. What is the correct average?
In the formula `barx = a + h((sumf_iu_i)/(sumf_i))`, for finding the mean of grouped frequency distribution, ui = ______.
The following table gives the number of pages written by Sarika for completing her own book for 30 days:
| Number of pages written per day |
16 – 18 | 19 – 21 | 22 – 24 | 25 – 27 | 28 – 30 |
| Number of days | 1 | 3 | 4 | 9 | 13 |
Find the mean number of pages written per day.
Find the mean of the following frequency distribution:
| Class: | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 |
| Frequency: | 4 | 10 | 5 | 6 | 5 |
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 |
