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?
The mean number of plants per house is 8.1 plants per house. I used the Direct Method for finding the mean because the values of class marks and frequencies are small numbers, making the calculations simple and straightforward without needing simplification techniques like the Assumed Mean Method or Step Deviation Method.
Explanation
This question from Exercise 13.1 tests the student's ability to calculate the mean of grouped frequency distribution. The Direct Method is appropriate here as the class marks (1, 3, 5, 7, 9, 11, 13) and frequencies are manageable numbers. The student needs to find class marks (midpoints), multiply by frequencies, sum them, and divide by total frequency.
Solution Steps
Step 1: Find the class mark (midpoint) for each class interval using the formula: Class Mark = (Lower Limit + Upper Limit) / 2
For 0-2: x₁ = (0+2)/2 = 1; For 2-4: x₂ = (2+4)/2 = 3; For 4-6: x₃ = (4+6)/2 = 5; For 6-8: x₄ = (6+8)/2 = 7; For 8-10: x₅ = (8+10)/2 = 9; For 10-12: x₆ = (10+12)/2 = 11; For 12-14: x₇ = (12+14)/2 = 13
Step 2: Calculate the product of frequency (f) and class mark (x) for each class: f₁x₁ = = 1; f₂x₂ = = 6; f₃x₃ = = 5; f₄x₄ = = 35; f₅x₅ = = 54; f₆x₆ = = 22; f₇x₇ = = 39
Step 3: Find the sum of all frequencies: Σf = 1+2+1+5+6+2+3 = 20
Step 4: Find the sum of all f×x values: Σfx = 1+6+5+35+54+22+39 = 162
Step 5: Apply the Direct Method formula: Mean = Σfx / Σf = 162/20 = 8.1