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प्रश्न
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: Less than type cumulative frequency distribution.
उत्तर
We observe that the number of patients which take medical treatment in a hospital on a particular day less than 10 is 0. Similarly, less than 20 include the number of patients which take medical treatment from 0 – 10 as well as the number of patients which take medical treatment from 10 – 20.
So, the total number of patients less than 20 is 0 + 60 = 60, we say that the cumulative frequency of the class 10 – 20 is 60. Similarly, for other class.
| Less than type | |
| Age (in year) | Number of patients |
| Less than 10 | 0 |
| Less than 20 | 60 + 0 = 60 |
| Less than 30 | 60 + 42 = 102 |
| Less than 40 | 102 + 55 = 157 |
| Less than 50 | 157 + 70 = 227 |
| Less than 60 | 227 + 53 = 280 |
| Less than 70 | 280 + 20 = 300 |
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संबंधित प्रश्न
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.
Write the median class of the following distribution:
| Class | 0 – 10 | 10 -20 | 20- 30 | 30- 40 | 40-50 | 50- 60 | 60- 70 |
| Frequency | 4 | 4 | 8 | 10 | 12 | 8 | 4 |
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 median of a given frequency distribution is found graphically with the help of
If \[u_i = \frac{x_i - 25}{10}, \Sigma f_i u_i = 20, \Sigma f_i = 100, \text { then }\]`overlineX`
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?
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?
Consider the following distribution:
| Marks obtained | Number of students |
| More than or equal to 0 | 63 |
| More than or equal to 10 | 58 |
| More than or equal to 20 | 55 |
| More than or equal to 30 | 51 |
| More than or equal to 40 | 48 |
| More than or equal to 50 | 42 |
The frequency of the class 30 – 40 is:
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 |
