A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
| Number of plants | 0 − 2 | 2 − 4 | 4 − 6 | 6 − 8 | 8 − 10 | 10 − 12 | 12 − 14 |
| Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Which method did you use for finding the mean, and why?
To find the class mark (xi) for each interval, the following relation is used.
Class mark \((x_i)\) = \(\frac {\text{Upper \,limit + Lower \,limit}}{2}\)
\(x_i\) and \(f_ix_i\) can be calculated as follows.
Number of plants | Number of houses (\(\bf{f_i}\)) | \(\bf{x_i}\) | \(\bf{f_ix_i}\) |
|---|---|---|---|
0-2 | 1 | 1 | 1 x 1 =1 |
2-4 | 2 | 3 | 2 x 3 = 6 |
4-6 | 1 | 5 | 2 x 3 = 5 |
6-8 | 5 | 7 | 5 x 7 = 35 |
8-10 | 6 | 9 | 6 x 9 = 54 |
10-12 | 2 | 11 | 2 x 11 = 22 |
12-14 | 3 | 13 | 3 x 13 = 39 |
Total | 20 |
| 162 |
From the table, it can be observed that
\(\sum f_i = 20\)
\(\sum f_ix_i = 162\)
Mean, \(\overset{-}{x} = \frac{\sum f_ix_i}{\sum f_i}\)
x = \(\frac{162 }{20}\)
x = 8.1
Therefore, mean number of plants per house is 8.1.
Here, direct method has been used as the values of class marks (\(x_i\)) and \(f_i\) are small.
Consider the following distribution of daily wages of 50 workers of a factory
| Daily wages (in Rs) | 500 - 520 | 520 -540 | 540 - 560 | 560 - 580 | 580 -600 |
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
| Daily pocket | 11 - 13 | 13 - 15 | 15 - 17 | 17 - 19 | 19 - 21 | 21 - 23 | 23 - 25 |
| Number of workers | 7 | 6 | 9 | 13 | f | 5 | 4 |
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
| Number of heartbeats per minute | 65 - 68 | 68 - 71 | 71 - 74 | 74 - 77 | 77 - 80 | 80 - 83 | 83 - 86 |
| Number of boxs | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
| Number of heartbeats per minute | 50-52 | 53-55 | 56-58 | 59-61 | 62-64 |
| Number of boxs | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
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.
| Case No. | Lens | Focal Length | Object Distance |
|---|---|---|---|
| 1 | \(A\) | 50 cm | 25 cm |
| 2 | B | 20 cm | 60 cm |
| 3 | C | 15 cm | 30 cm |