
A survey conducted by a group of students as part of their environment awareness program, 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
| 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 |
Answer
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Hint: Here we are given the number of pants and the number of houses falling in that range. To find mean we will directly use a formula to find mean. But we also have to find the middle value of the number of plant ranges. This middle value acts as a representative of the other frequencies falling in that class.
Complete step-by-step answer:
Now we will use the direct method for calculation. Let’s tabulate the data.
The middle value or class mark of a frequency range is given by,
Mid-value $ = \dfrac{{upper\,limit + lower\,limit}}{2}$
The frequency table is,
We know that the general formula to find the mean value is,
Mean $ = \dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{x_i}} }}$
Now, we will substitute the value for the sum of the product of frequency and midpoint and the value for the sum of total frequency.
$ \Rightarrow $ Mean $ = \dfrac{{162}}{{20}}$
Divide numerator by the denominator,
$\therefore $ Mean $ = 8.1$
Hence the mean number of plants per house is 8.1.
Note: Here in this problem data given of classes are in grouped form. No direct numbers are given so do find the middle values or class mark of the range. And then proceed for calculations. Always tabulate these types of problems.
In the mean formula, while computing $\sum {fx} $, don’t take the sum of $f$ and $x$ separately and then multiply them. It will be difficult. Students should carefully make the frequency distribution table; there are high chances of making mistakes while copying and computing data.
Complete step-by-step answer:
Now we will use the direct method for calculation. Let’s tabulate the data.
The middle value or class mark of a frequency range is given by,
Mid-value $ = \dfrac{{upper\,limit + lower\,limit}}{2}$
The frequency table is,
| No. of plants | No. of houses (${f_i}$) | Mid value (${x_i}$) | ${f_i}{x_i}$ |
| 0 – 2 | 1 | 1 | 1 |
| 2 – 4 | 2 | 3 | 6 |
| 4 – 6 | 1 | 5 | 5 |
| 6 – 8 | 5 | 7 | 35 |
| 8 – 10 | 6 | 9 | 54 |
| 10 – 12 | 2 | 11 | 22 |
| 12 – 14 | 3 | 13 | 39 |
| Total | $\sum {{f_i}} = 20$ | $\sum {{f_i}{x_i}} = 162$ |
We know that the general formula to find the mean value is,
Mean $ = \dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{x_i}} }}$
Now, we will substitute the value for the sum of the product of frequency and midpoint and the value for the sum of total frequency.
$ \Rightarrow $ Mean $ = \dfrac{{162}}{{20}}$
Divide numerator by the denominator,
$\therefore $ Mean $ = 8.1$
Hence the mean number of plants per house is 8.1.
Note: Here in this problem data given of classes are in grouped form. No direct numbers are given so do find the middle values or class mark of the range. And then proceed for calculations. Always tabulate these types of problems.
In the mean formula, while computing $\sum {fx} $, don’t take the sum of $f$ and $x$ separately and then multiply them. It will be difficult. Students should carefully make the frequency distribution table; there are high chances of making mistakes while copying and computing data.
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