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Question
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?
Solution
To find the class mark (xi) for each interval, the following relation is used.
Class mark (xi) = `("Upper class limit + Lower class limit")/2`
xi and fixi can be calculated as follows.
| Number of plant | Number of houses (fi) |
xi | fixi |
| 0 − 2 | 1 | 1 | 1 × 1 = 1 |
| 2 − 4 | 2 | 3 | 2 × 3 = 6 |
| 4 − 6 | 1 | 5 | 1 × 5 = 5 |
| 6 − 8 | 5 | 7 | 5 × 7 = 35 |
| 8 − 10 | 6 | 9 | 6 × 9 = 54 |
| 10 − 12 | 2 | 11 | 2 ×11 = 22 |
| 12 − 14 | 3 | 13 | 3 × 13 = 39 |
| Total | 20 | 162 |
From the table, it can be observed that
`sum f_i = 20`
`sumf_ix_i = 162`
Mean `barx = (sumf_ix_i)/(sumf_i)`
= `162/20 = 8.1`
Therefore, mean number of plants per house is 8.1.
Here, the direct method has been used as the values of class marks (xi) and fi are small.
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