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प्रश्न
Draw a ‘more than’ ogive for the data given below which gives the marks of 100 students.
| Marks | 0 – 10 | 10 – 20 | 20 – 30 | 30 - 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 |
| No of Students | 4 | 6 | 10 | 10 | 25 | 22 | 18 | 5 |
उत्तर
The frequency distribution table of more than type is as follows:
| Marks (upper class limits) | Cumulative frequency (cf) |
| More than 0 | 96 + 4 = 100 |
| More than 10 | 90 + 6 = 96 |
| More than 20 | 80 + 10 = 90 |
| More than 30 | 70 + 10 = 80 |
| More than 40 | 45 + 25 = 70 |
| More than 50 | 23 + 22 = 45 |
| More than 60 | 18 + 5 = 23 |
| More than 70 | 5 |
Taking lower class limits of on x-axis and their respective cumulative frequencies on y-axis,its ogive can be drawn as follows:
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संबंधित प्रश्न
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 following table gives the production yield per hectare of wheat of 100 farms of a village.
| Production Yield (kg/ha) | 50 –55 | 55 –60 | 60 –65 | 65- 70 | 70 – 75 | 75 80 |
| Number of farms | 2 | 8 | 12 | 24 | 238 | 16 |
Change the distribution to a ‘more than type’ distribution and draw its ogive. Using ogive, find the median of the given data.
The following frequency distribution gives the monthly consumption of electricity of 64 consumers of locality.
| Monthly consumption (in units) | 65 – 85 | 85 – 105 | 105 – 125 | 125 – 145 | 145 – 165 | 165 – 185 |
| Number of consumers | 4 | 5 | 13 | 20 | 14 | 8 |
Form a ‘ more than type’ cumulative frequency distribution.
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 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 |
Look at the following table below.
| Class interval | Classmark |
| 0 - 5 | A |
| 5 - 10 | B |
| 10 - 15 | 12.5 |
| 15 - 20 | 17.5 |
The value of A and B respectively are?
If the sum of all the frequencies is 24, then the value of z is:
| Variable (x) | 1 | 2 | 3 | 4 | 5 |
| Frequency | z | 5 | 6 | 1 | 2 |
Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class:
| Marks | Below 20 | Below 40 | Below 60 | Below 80 | Below 100 |
| Number of students | 17 | 22 | 29 | 37 | 50 |
Form the frequency distribution table for the data.
