Advertisements
Advertisements
प्रश्न
Calculate the mean of the distribution, given below using the short cut method:
| Marks | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 | 51 – 60 | 61 – 70 | 71 – 80 |
| No. of students | 2 | 6 | 10 | 12 | 9 | 7 | 4 |
उत्तर १
| Marks | f | x | d = x – A = x – 45.5 | fd |
| 11 – 20 | 2 | 15.5 | –30 | –60 |
| 21 – 30 | 6 | 25.5 | –20 | –120 |
| 31 – 40 | 10 | 35..5 | –10 | –100 |
| 41 – 50 | 12 | A = 45.5 | 0 | 0 |
| 51 – 60 | 9 | 55.5 | 10 | 90 |
| 61 – 70 | 7 | 65.5 | 20 | 140 |
| 71 – 80 | 4 | 75.5 | 30 | 120 |
| ∑f = 50 | ∑fd = 70 |
∴ Mean = `A + (sumfd)/(sumf)`
= `45.5 + 70/50`
= `45.5 + 7/5`
=`(227.5 +7)/5`
= `234.5/5`
= 46.9
उत्तर २
| Class Interval (Inclusive form) |
Class Interval (Exclusive form) |
No. of Students `bb((f_i))` |
`bb(x_i)` | `bb(A = 45.5)` `bb(d_i = x - 45.5)` |
`bb(f_i d_i)` |
| 11 – 20 | 10.5 – 20.5 | 2 | 15.5 | –30 | `{:(-60),(-120),(-100):}} - 280` |
| 21 – 30 | 20.5 – 30.5 | 6 | 25.5 | –20 | |
| 31 – 40 | 30.5 – 40.5 | 10 | 35.5 | –10 | |
| 41 – 50 | 40.5 – 50.5 | 12 | 45.5 | 0 | 0 |
| 51 – 60 | 50.5 – 60.5 | 9 | 55.5 | 10 | `{:(90),(140),(120):}} 350` |
| 61 – 70 | 60.5 – 70.5 | 7 | 65.5 | 20 | |
| 71 – 80 | 70.5 – 80.5 | 4 | 75.5 | 30 | |
| `sumf_i = 50` | `sumf_i d_i = 70` |
Assumed mean (A) = 45.5
`sumf_i = 50, sumf_i d_i = 70`
Mean = `A + (sumf_i d_i)/(sumf_i)`
= `45.5 + (70)/(50)`
= 45.5 + 1.4
= 46.9
संबंधित प्रश्न
The median is always one of the numbers in a data.
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 |
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.
The following are the marks obtained by 70 boys in a class test:
| Marks | No. of boys |
| 30 – 40 | 10 |
| 40 – 50 | 12 |
| 50 – 60 | 14 |
| 60 – 70 | 12 |
| 70 – 80 | 9 |
| 80 – 90 | 7 |
| 90 – 100 | 6 |
Calculate the mean by:
Step-deviation method
Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks:
0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7 and 8.
Find the mean of first 10 prime numbers.
Find the mean of the following frequency distribution :
| Class | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
| Frequency | 6 | 10 | 8 | 12 | 4 |
The mean of the following frequency distribution is 25.8 and the sum of all the frequencies is 50. Find x and y.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 7 | x | 15 | y | 10 |
Estimate the median, the lower quartile and the upper quartile of the following frequency distribution by drawing an ogive:
| Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency | 4 | 12 | 21 | 18 | 15 | 7 | 3 |
Find the median of 26, 33, 41, 18, 30, 22, 36, 45 and 24
