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प्रश्न
The contents of 100 match boxes were checked to determine the number of matches they contained.
| No. of matches | 35 | 36 | 37 | 38 | 39 | 40 | 41 |
| No. of boxes | 6 | 10 | 18 | 25 | 21 | 12 | 8 |
- Calculate, correct to one decimal place, the mean number of matches per box.
- Determine, how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean up to exactly 39 matches.
उत्तर
| No. of matches (x) |
No. of boxes (f) |
fx |
| 35 | 6 | 210 |
| 36 | 10 | 360 |
| 37 | 18 | 666 |
| 38 | 25 | 950 |
| 39 | 21 | 819 |
| 40 | 12 | 480 |
| 41 | 8 | 328 |
| Total | 100 | 3813 |
i. `barx = (sumfx)/(sumf)`
= `3813/100`
= 38.13
ii. In the second case,
New mean = 39 matches
Total contents = 39 × 100 = 3900
But total number of matches already given = 3813
Number of new matches to be added = 3900 – 3813 = 87
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संबंधित प्रश्न
Find the mean of the following distribution by step deviation method:
| Class Interval | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 10 | 6 | 8 | 12 | 5 | 9 |
By drawing an ogive, estimate the median for the following frequency distribution:
| Weight (kg) | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 |
| No. of boys | 11 | 25 | 12 | 5 | 2 |
The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:
| Expenditure (Rs) | No. of students |
| 20 – 25 | 4 |
| 25 – 30 | 7 |
| 30 – 35 | 23 |
| 35 – 40 | 18 |
| 40 – 45 | 6 |
| 45 – 50 | 2 |
The weights of 11 students in a class are 36 kg, 45 kg, 44 kg, 37 kg, 36 kg, 41 kg, 45 kg, 43 kg, 39 kg, 42 kg and 40 kg. Find the median of their weights.
Find the median of all prime numbers between 20 and 50
Estimate the median, the lower quartile and the upper quartile of the following frequency distribution by drawing an ogive:
| Marks(more than) | 90 | 80 | 70 | 60 | 50 | 40 | 30 | 20 | 10 | 0 |
| No. of students | 6 | 13 | 22 | 34 | 48 | 60 | 70 | 78 | 80 | 80 |
Out of 10 students, who appeared in a test, three secured less than 30 marks and 3 secured more than 75 marks. The marks secured by the remaining 4 students are 35, 48, 66 and 40. Find the median score of the whole group.
Find the mean of 35, 44, 31, 57, 38, 29, 26,36, 41 and 43.
Find the median of first nine even natural numbers.
An incomplete frequency distribution is given below
| Variate | Frequency |
| 10 – 20 | 12 |
| 20 – 30 | 30 |
| 30 – 40 | ? |
| 40 – 50 | 65 |
| 50 – 60 | 45 |
| 60 – 70 | 25 |
| 70 – 80 | 18 |
| Total | 229 |
Median value is 46, the missing frequency is:
