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 |
To find the class mark (\(x_i\)) for each interval, the following relation is used.
Class mark \((x_i)\) = \(\frac {\text{Upper \,limit + Lower \,limit}}{2}\)
Taking 17 as assumed mean (a), \(d_i\), and \(f_id_i\) can be calculated as follows.
| Number of days | Number of students (fi) | \(\bf{x_i}\) | \(\bf{d_i = x_i -17}\) | \(\bf{f_id_i}\) |
0 - 6 | 11 | 3 | -14 | -154 |
6 - 10 | 10 | 8 | -9 | -90 |
10 - 14 | 7 | 12 | -5 | -35 |
14 - 20 | 4 | 17 | 0 | 0 |
20 - 28 | 4 | 24 | 7 | 28 |
28 - 38 | 3 | 33 | 16 | 48 |
38 - 40 | 1 | 39 | 22 | 22 |
Total | 40 | -181 |
From the table, We obtain
\(\sum f_i = 40\)
\(\sum f_id_i = -181\)
Mean, \(\overset{-}{x} = a + (\frac{\sum f_id_i}{\sum f_i})\)
x = \(17 + (\frac{-181 }{40})\)
x = 17 - 4.525
x = 12.475 = 12.48
Therefore, the mean number of days is 12.48 days for which a student was absent.
Find mean of the following frequency table:

The following table shows the literacy rate (in percent) of 35 cities. Find the mean literacy rate.
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Literacy rate (in \%)} & 45-55 & 55-65 & 65-75 & 75-85 & 85-95 \\ \hline \text{Number of cities} & 3 & 10 & 11 & 8 & 3 \\ \hline \end{array}\]
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
| Literacy rate (in %) | 45 - 55 | 55 - 65 | 65 - 75 | 75 - 85 | 85 - 95 |
| Number of cities | 11 | 10 | 7 | 4 | 4 |
To find out the concentration of SO\(_2\) in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
| Concentration of SO\(\bf{_2}\) (in ppm) | Frequency |
0.00 - 0.04 0.04 - 0.08 0.08 - 0.12 0.12 - 0.16 0.16 - 0.20 0.20 - 0.24 | 4 9 9 2 4 2 |
The table below shows the daily expenditure on food of 25 households in a locality
| Daily expenditure (in Rs) | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 | 300 - 350 |
| Number of households | 4 | 5 | 12 | 2 | 2 |
Find the mean daily expenditure on food by a suitable method.
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD. 