Advertisements
Advertisements
Question
Find the correct answer from the alternatives given.
The formula to find mean from a grouped frequency table is \[X = A + \frac{\sum f_i u_i}{\sum f_i} \times hg\] .
Options
\[\frac{x_i + A}{g}\]
\[\left( x_i - A \right)\]
\[\frac{x_i - A}{g}\]
- \[\frac{A - x_i}{g}\]
Solution
To find mean of a grouped frequency table using
In this formula, \[u_i = \frac{x_i - A}{g}\]
RELATED QUESTIONS
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
| Number of days | 0 - 6 | 6 - 10 | 10 -14 | 14 -20 | 20 -28 | 28 -38 | 38 -40 |
| Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students
| Age (in years) | 15 | 16 | 17 | 18 | 19 | 20 |
| No. of students | 3 | 8 | 10 | 10 | 5 | 4 |
Find the mean of each of the following frequency distributions
| Class interval | 0 - 8 | 8 - 16 | 16 - 24 | 24 - 32 | 32 - 40 |
| Frequency | 6 | 7 | 10 | 8 | 9 |
Find the mean of each of the following frequency distributions
| Class interval | 0 - 8 | 8 - 16 | 16 - 24 | 24 - 32 | 32 - 40 |
| Frequency | 5 | 6 | 4 | 3 | 2 |
Find the mean of each of the following frequency distributions
| Classes | 25 - 29 | 30 - 34 | 35 - 39 | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 |
| Frequency | 14 | 22 | 16 | 6 | 5 | 3 | 4 |
Find the mean of the following data, using direct method:
| Class | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 |
| Frequency | 6 | 9 | 15 | 12 | 8 |
During a medical check-up, the number of heartbeats per minute of 30 patients were recorded and summarized as follows:
| Number of heartbeats per minute |
65 – 68 | 68 – 71 | 71 – 74 | 74 – 77 | 77 – 80 | 80 – 83 | 83 - 86 |
| Number of patients |
2 | 4 | 3 | 8 | 7 | 4 | 2 |
Find the mean heartbeats per minute for these patients, choosing a suitable method.
Find the mean age from the following frequency distribution:
| Age (in years) | 25 – 29 | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 | 55 – 59 |
| Number of persons | 4 | 14 | 22 | 16 | 6 | 5 | 3 |
If the mean of the following frequency distribution is 18, find the missing frequency.
| Class interval | 11 – 13 | 13 – 15 | 15 – 17 | 17 – 19 | 19 – 21 | 21 – 23 | 23 – 25 |
| Frequency | 3 | 6 | 9 | 13 | f | 5 | 4 |
If the mean of 6, 7, x, 8, y, 14 is 9, then ______.
If the mean of first n natural numbers is \[\frac{5n}{9}\], then n =
The mean of first n odd natural numbers is \[\frac{n^2}{81}\],then n =
Find the mean of the following distribution:
| x | 4 | 6 | 9 | 10 | 15 |
| f | 5 | 10 | 10 | 7 | 8 |
Find the mean of the following frequency distribution:
| Class Interval | Frequency |
| 0 - 50 | 4 |
| 50 - 100 | 8 |
| 100 - 150 | 16 |
| 150 - 200 | 13 |
| 200 - 250 | 6 |
| 250 - 300 | 3 |
There is a grouped data distribution for which mean is to be found by step deviation method.
| Class interval | Number of Frequency (fi) | Class mark (xi) | di = xi - a | `u_i=d_i/h` |
| 0 - 100 | 40 | 50 | -200 | D |
| 100 - 200 | 39 | 150 | B | E |
| 200 - 300 | 34 | 250 | 0 | 0 |
| 300 - 400 | 30 | 350 | 100 | 1 |
| 400 - 500 | 45 | 450 | C | F |
| Total | `A=sumf_i=....` |
Find the value of A, B, C, D, E and F respectively.
Find the values of x and y if the mean and total frequency of the distribution are 25 and 50 respectively.
| Class Interval | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| Frequency | 7 | x | 5 | y | 4 | 2 |
If the mean of 9, 8, 10, x, 14 is 11, find the value of x.
The mean of the following frequency distribution is 25. Find the value of f.
| Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
| Frequency | 5 | 18 | 15 | f | 6 |
Find the mean of the following data using assumed mean method:
| Class | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 |
| Frequency | 8 | 7 | 10 | 13 | 12 |
