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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: More than type cumulative frequency distribution.
उत्तर
Also, we observe that all 300 patients which take medical treatment more than or equal to 10. Since, thete are 60 patients which take medical treatment in the interval 10 – 20, this means that there are 300 – 60 = 240 patients which take medical treatment more than or equal to 20. Continuing in the same manner.
| More than type | |
| Age (in year) | Number of patients |
| More than or equals 10 | 60 + 42 + 55 + 70 + 53 + 20 = 300 |
| More than or equals 20 | 42 + 55 + 70 + 53 + 20 = 240 |
| More than or equals 30 | 55 + 70 + 53 + 20 = 198 |
| More than or equals 40 | 70 + 53 + 20 = 143 |
| More than or equals 50 | 53 + 20 = 73 |
| More than or equals 60 | 60 |
| More than or equals 70 | 0 |
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संबंधित 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 | 6 | 42 | 55 | 70 | 53 | 20 |
Form a ‘less than type’ cumulative frequency distribution.
The median of the distribution given below is 14.4 . Find the values of x and y , if the total frequency is 20.
| Class interval : | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
| Frequency : | 4 | x | 5 | y | 1 |
For a frequency distribution, mean, median and mode are connected by the relation
The median of a given frequency distribution is found graphically with the help of
The mode of a frequency distribution can be determined graphically from ______.
The mean of a discrete frequency distribution xi / fi, i = 1, 2, ......, n is given by
If \[u_i = \frac{x_i - 25}{10}, \Sigma f_i u_i = 20, \Sigma f_i = 100, \text { then }\]`overlineX`
In the formula `barx=a+h((sumf_iu_i)/(sumf_i))`, for finding the mean of grouped frequency distribution ui = ______.
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 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?
