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प्रश्न
In the formula
पर्याय
lower limits of classes
upper limits ofclasses
mid-points of classes
frequency of the class marks
उत्तर
The given formula represents the formula to find the mean by assumed mean method.
Here, di = \[x_i - a\]
where xi is the ith observation and a is assumed mean.
So, di's are the deviation from a of mid-points of the classes.
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संबंधित प्रश्न
Find the mean of the following data:-
| x | 19 | 21 | 23 | 25 | 27 | 29 | 31 |
| f | 13 | 15 | 16 | 18 | 16 | 15 | 13 |
If the mean of the following data is 20.6. Find the value of p.
| x | 10 | 15 | P | 25 | 35 |
| f | 3 | 10 | 25 | 7 | 5 |
If the mean of the following data is 15, find p.
| x | 5 | 10 | 15 | 20 | 25 |
| f | 6 | P | 6 | 10 | 5 |
The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:
| Marks(x) | 15 | 20 | 22 | 24 | 25 | 30 | 33 | 38 | 45 |
| Frequency (f) | 5 | 8 | 11 | 20 | 23 | 18 | 13 | 3 | 1 |
Find the average number of marks.
Find the mean of the following frequency distribution table using a suitable method:
| Class | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 - 70 |
| Frequency | 25 | 40 | 42 | 33 | 10 |
The following table shows the age distribution of patients of malaria in a village during a particular month:
| Age (in years) | 5 – 14 | 15 – 24 | 25 – 34 | 35 – 44 | 45 – 54 | 55 - 64 |
| No. of cases | 6 | 11 | 21 | 23 | 14 | 5 |
Find the average age of the patients.
The mean of first n odd natural number is
If the mean of the following distribution is 7.5, find the missing frequency ‘f’:
| Variable : | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Frequency: | 20 | 17 | f | 10 | 8 | 6 | 7 | 6 |
If the mean of first n natural numbers is `(5n)/9,` then n =?
The median from the table is?
| Value | Frequency |
| 7 | 2 |
| 8 | 1 |
| 9 | 4 |
| 10 | 5 |
| 11 | 6 |
| 12 | 1 |
| 13 | 3 |
