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प्रश्न
Find the median of the following data by making a ‘less than ogive’.
| Marks | 0 - 10 | 10-20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 | 80-90 | 90-100 |
| Number of Students | 5 | 3 | 4 | 3 | 3 | 4 | 7 | 9 | 7 | 8 |
उत्तर
The frequency distribution table of less than type is given as follows:
| Marks (upper class limits) | Cumulative frequency (cf) |
| Less than 10 | 5 |
| Less than 20 | 5 + 3 = 8 |
| Less than 30 | 8 + 4 = 12 |
| Less than 40 | 12 + 3 = 15 |
| Less than 50 | 15 + 3 = 18 |
| Less than 60 | 18 + 4 = 22 |
| Less than 70 | 22 + 7 = 29 |
| Less than 80 | 29 + 9 = 38 |
| Less than 90 | 38 + 7 = 45 |
| Less than 100 | 45 + 8 = 53 |
Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows:
Here, N = 53 ⇒ `N/2` = 26.5.
Mark the point A whose ordinate is 26.5 and
its x-coordinate is 66.4.
Thus, median of the data is 66.4.
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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 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 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.
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 |
In the formula `barx=a+h((sumf_iu_i)/(sumf_i))`, for finding the mean of grouped frequency distribution ui = ______.
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 arithmetic mean of the following frequency distribution is 53. Find the value of k.
| Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
| Frequency | 12 | 15 | 32 | k | 13 |
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 |
For the following distribution:
| C.I. | 0 - 5 | 6 - 11 | 12 - 17 | 18 - 23 | 24 - 29 |
| f | 13 | 10 | 15 | 8 | 11 |
the upper limit of the median class is?
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?
