Advertisements
Advertisements
Question
Calculate the mean of the following frequency distribution :
| Class: | 10-30 | 30-50 | 50-70 | 70-90 | 90-110 | 110-130 |
| Frequency: | 5 | 8 | 12 | 20 | 3 | 2 |
Solution
| Class | frequency (fi) | Class mark(xi) | fixi |
| 10-30 | 5 | `(10+30)/2 = 20` | 100 |
| 30-50 | 8 | `(30+50)/2 = 40` | 320 |
| 50-70 | 12 | `(50+70)/2 = 60` | 720 |
| 70-90 | 20 | `(70+90)/2 = 80` | 1600 |
| 90-110 | 3 | `(90+110)/2 = 100` | 300 |
| 110-130 | 2 | `(110+130)/2 = 120` | 240 |
| `sumf_i = 50` | `sumf_ix_i = 3280` |
Using: Mean = `(sumf_ix_i)/(sumf_i)`
substituting the values in the formula
mean = `3280/50 = 65.6`
APPEARS IN
RELATED QUESTIONS
During the medical check-up of 35 students of a class, their weights were recorded as follows:
| Weight (in kg | Number of students |
| Less than 38 | 0 |
| Less than 40 | 3 |
| Less than 42 | 5 |
| Less than 44 | 9 |
| Less than 46 | 14 |
| Less than 48 | 28 |
| Less than 50 | 32 |
| Less than 52 | 35 |
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.
The monthly profits (in Rs.) of 100 shops are distributed as follows:
| Profits per shop: | 0 - 50 | 50 - 100 | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 |
| No. of shops: | 12 | 18 | 27 | 20 | 17 | 6 |
Draw the frequency polygon for it.
The heights of 50 girls of Class X of a school are recorded as follows:
| Height (in cm) | 135 - 140 | 140 – 145 | 145 – 150 | 150 – 155 | 155 – 160 | 160 – 165 |
| No of Students | 5 | 8 | 9 | 12 | 14 | 2 |
Draw a ‘more than type’ ogive for the above data.
The following table, construct the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.
| Marks obtained (in percent) | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 | 51 – 60 | 61 – 70 | 71 – 80 |
| Number of Students | 141 | 221 | 439 | 529 | 495 | 322 | 153 |
(a) Convert the given frequency distribution into the continuous form.
(b) Find the median class and write its class mark.
(c) Find the modal class and write its cumulative frequency.
Calculate the missing frequency form the following distribution, it being given that the median of the distribution is 24
| Age (in years) | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Number of persons |
5 | 25 | ? | 18 | 7 |
For a frequency distribution, mean, median and mode are connected by the relation
If the median of the following frequency distribution is 32.5. Find the values of f1 and f2.
For one term, absentee record of students is given below. If mean is 15.5, then the missing frequencies x and y are.
| Number of days | 0 - 5 | 5 - 10 | 10 - 15 | 15 - 20 | 20 - 25 | 25 - 30 | 30 - 35 | 35 - 40 | TOTAL |
| Total Number of students | 15 | 16 | x | 8 | y | 8 | 6 | 4 | 70 |
The following is the distribution of weights (in kg) of 40 persons:
| Weight (in kg) | 40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 | 60 – 65 | 65 – 70 | 70 – 75 | 75 – 80 |
| Number of persons | 4 | 4 | 13 | 5 | 6 | 5 | 2 | 1 |
Construct a cumulative frequency distribution (of the less than type) table for the data above.
The following table shows the cumulative frequency distribution of marks of 800 students in an examination:
| Marks | Number of students |
| Below 10 | 10 |
| Below 20 | 50 |
| Below 30 | 130 |
| Below 40 | 270 |
| Below 50 | 440 |
| Below 60 | 570 |
| Below 70 | 670 |
| Below 80 | 740 |
| Below 90 | 780 |
| Below 100 | 800 |
Construct a frequency distribution table for the data above.
