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प्रश्न
The table given below shows the weekly expenditures on food of some households in a locality
Weekly expenditure (in Rs) | Number of house holds |
100 – 200 | 5 |
200- 300 | 6 |
300 – 400 | 11 |
400 – 500 | 13 |
500 – 600 | 5 |
600 – 700 | 4 |
700 – 800 | 3 |
800 – 900 | 2 |
Draw a ‘less than type ogive’ and a ‘more than type ogive’ for this distribution.
उत्तर
The frequency distribution table of less than type is as follows:
Weekly expenditure (in Rs) (upper class limits) | Cumulative frequency (cf) |
Less than 200 | 5 |
Less than 300 | 5 + 6 = 11 |
Less than 400 | 11 + 11 = 22 |
Less than 500 | 22 + 13 = 35 |
Less than 600 | 35 + 5 = 40 |
Less than 700 | 40 + 4 = 44 |
Less than 800 | 44 + 3 = 47 |
Less than 900 | 47 + 2 = 49 |
Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows
Now,
The frequency distribution table of more than type is as follows:
Weekly expenditure (in Rs) (lower class limits) | Cumulative frequency (cf) |
More than 100 | 44 + 5 = 49 |
More than 200 | 38 + 6 = 44 |
More than 300 | 27 + 11 = 38 |
More than 400 | 14 + 13 = 27 |
More than 500 | 9 + 5 = 14 |
More than 600 | 5 + 4 = 9 |
More than 700 | 2 + 3 = 5 |
More than 800 | 2 |
Taking the lower class limits
on x-axis and their respective
cumulative frequencies on y-axis,
its ogive can be obtained 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 following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield (in kg/ha) | 50 − 55 | 55 − 60 | 60 − 65 | 65 − 70 | 70 − 75 | 75 − 80 |
Number of farms | 2 | 8 | 12 | 24 | 38 | 16 |
Change the distribution to a more than type distribution and draw ogive.
From the following frequency, prepare the ‘more than’ ogive.
Score | Number of candidates |
400 – 450 | 20 |
450 – 500 | 35 |
500 – 550 | 40 |
550 – 600 | 32 |
600 – 650 | 24 |
650 – 700 | 27 |
700 – 750 | 18 |
750 – 800 | 34 |
Total | 230 |
Also, find the median.
The marks obtained by 100 students of a class in an examination are given below:
Marks | Number of students |
0 – 5 | 2 |
5 – 10 | 5 |
10 – 15 | 6 |
15 – 20 | 8 |
20 – 25 | 10 |
25 – 30 | 25 |
30 – 35 | 20 |
35 – 40 | 18 |
40 – 45 | 4 |
45 – 50 | 2 |
Draw cumulative frequency curves by using (i) ‘less than’ series and (ii) ‘more than’ series.Hence, find the median.
The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:
Age (in years) | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 -70 |
Number of patients | 6 | 42 | 55 | 70 | 53 | 20 |
Form a ‘less than type’ cumulative frequency distribution.
The following table gives the life-time (in days) of 100 electric bulbs of a certain brand.
Life-tine (in days) | Less than 50 |
Less than 100 |
Less than 150 |
Less than 200 |
Less than 250 |
Less than 300 |
Number of Bulbs | 7 | 21 | 52 | 9 | 91 | 100 |
In the formula `barx=a+h((sumf_iu_i)/(sumf_i))`, for finding the mean of grouped frequency distribution ui = ______.
Consider the following frequency distribution :
Class: | 0-5 | 6-11 | 12-17 | 18-23 | 24-29 |
Frequency: | 13 | 10 | 15 | 8 | 11 |
The upper limit of the median class is
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 |
The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:
Age (in years) | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
Number of patients | 60 | 42 | 55 | 70 | 53 | 20 |
Form: More than type cumulative frequency distribution.