Advertisements
Advertisements
प्रश्न
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.
उत्तर
Marks | Number of students | C.I. | No. of students |
Below 10 | 10 | 0 – 10 | 10 |
Below 20 | 50 | 10 – 20 | 50 – 10 = 40 |
Below 30 | 130 | 20 – 30 | 130 – 50 = 80 |
Below 40 | 270 | 30 – 40 | 270 – 130 = 140 |
Below 50 | 440 | 40 – 50 | 440 – 270 = 170 |
Below 60 | 570 | 50 – 60 | 570 – 440 = 130 |
Below 70 | 670 | 60 – 70 | 670 – 570 = 100 |
Below 80 | 740 | 70 – 80 | 740 – 670 = 70 |
Below 90 | 780 | 80 – 90 | 780 – 740 = 40 |
Below 100 | 800 | 90 – 100 |
800 – 780 = 20 |
APPEARS IN
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.
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.
What is the lower limit of the modal class of the following frequency distribution?
Age (in years) | 0 - 10 | 10- 20 | 20 -30 | 30 – 40 | 40 –50 | 50 – 60 |
Number of patients | 16 | 13 | 6 | 11 | 27 | 18 |
The monthly pocket money of 50 students of a class are given in the following distribution
Monthly pocket money (in Rs) | 0 - 50 | 50 – 100 | 100 – 150 | 150 -200 | 200 – 250 | 250 - 300 |
Number of Students | 2 | 7 | 8 | 30 | 12 | 1 |
Find the modal class and give class mark of the modal class.
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 |
The mode of a frequency distribution can be determined graphically from ______.
Consider the following frequency distributions
Class | 65 - 85 | 85 - 105 | 105 - 125 | 125 - 145 | 145 - 165 | 165 - 185 | 185-205 |
Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
The difference of the upper limit of the median class and the lower limit of the modal class is?
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.