Advertisements
Advertisements
प्रश्न
The ages of 40 students are given in the following table :
| Age (in years) | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| Frequency | 2 | 4 | 6 | 9 | 8 | 7 | 4 |
Find the arithmetic mean.
उत्तर
| Age in yrs `bb(x_i)` |
Frequency `bb((f_i))` |
`bb(f_i x_i)` |
| 12 | 2 | 24 |
| 13 | 4 | 52 |
| 14 | 6 | 84 |
| 15 | 9 | 135 |
| 16 | 8 | 128 |
| 17 | 7 | 119 |
| 18 | 4 | 72 |
| Total | 40 | 614 |
`barx = (f_ix_i)/(Σ f)`
= `614/40`
= 15.35
APPEARS IN
संबंधित प्रश्न
1) Using step–deviation method, calculate the mean marks of the following distribution.
2) State the modal class.
| Class Interval | 50 - 55 | 55 - 60 | 60 - 65 | 65 - 70 | 70 - 75 | 75 - 80 | 80 - 85 | 85 – 90 |
| Frequency | 5 | 20 | 10 | 10 | 9 | 6 | 12 | 8 |
The mean of the following distribution is 52 and the frequency of class interval 30-40 is ‘f’. Find ‘f’.
| Class Interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 5 | 3 | f | 7 | 2 | 6 | 13 |
If the mean of 11 , 14 , p , 26 , 10 , 12 , 18 , 6 is 15, find p.
Find the mean of the following frequency distribution by the short cut method :
| Class | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 |
| Frequency | 7 | 10 | 14 | 17 | 15 | 11 | 6 |
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 |
The mean of marks scored by 100 students was found to be 40, later on, it was discovered that a score of 53 was misread as 83. Find the correct mean.
Find the mean of: first five odd natural numbers
The median of the data 24, 29, 34, 38, 35 and 30, is ___________
The median first 6 odd natural numbers is ____________
In a given data, arranged in ascending or descending order, the middle most observation is called ______.
