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प्रश्न
The following table gives the age of 50 student of a class. Find the arithmetic mean of their ages.
| Age-years | 16 – 18 | 18 – 20 | 20 – 22 | 22 – 24 | 24 – 26 |
| No. of students | 2 | 7 | 21 | 17 | 3 |
उत्तर
| Age in years C.I. |
`x_i` | Number of student (`f_i`) |
`x_i f_i` |
| 16 – 18 | 17 | 2 | 34 |
| 18 – 20 | 19 | 7 | 133 |
| 20 – 22 | 21 | 21 | 441 |
| 22 – 24 | 23 | 17 | 391 |
| 24 – 26 | 25 | 3 | 75 |
| Total | 50 | 1074 |
`barx = (f_ix_i)/(sumf)`
= `1074/50`
= 21.48
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संबंधित प्रश्न
The marks of 10 students of a class in an examination arranged in ascending order is as follows:
13, 35, 43, 46, x, x + 4, 55, 61, 71, 80
If the median marks is 48, find the value of x. Hence, find the mode of the given data.
Find the mode and the median of the following frequency distributions.
| x | 10 | 11 | 12 | 13 | 14 | 15 |
| f | 1 | 4 | 7 | 5 | 9 | 3 |
Find the mean of the following frequency distribution :
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 4 | 7 | 6 | 3 | 5 |
Find the mean of the following frequency distribution by the step deviation method :
| Class | 100-110 | 110-120 | 120-130 | 130-140 | 140-150 |
| Frequency | 15 | 18 | 32 | 25 | 10 |
Estimate the median, the lower quartile and the upper quartile of the following frequency distribution by drawing an ogive:
| Marks (less than) | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
| No. of students | 5 | 15 | 30 | 54 | 72 | 86 | 94 | 100 |
If the mean of 8, 10, 7, x + 2 and 6 is 9, find the value of x.
The mean of 5 numbers is 27. If one new number is included, the new mean is 25. Find the included number.
Height (in cm) of 7 boys of a locality are 144 cm, 155 cm, 168 cm, 163 cm, 167 cm, 151 cm and 158 cm. Find their mean height.
Find the median of 17, 23, 36, 12, 18, 23, 40 and 20
Find the median of 80, 48, 66, 61, 75, 52, 45 and 70
