Ex.14.1 Q1 Statistics Solution - NCERT Maths Class 10
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?
Text Solution
What is known?
The number of plants in \(20\) houses in a locality.
What is unknown?
The mean number of plants per house and the method used for finding the mean.
Reasoning:
We can solve this question by any method of finding mean but here we will use direct method to solve this question because the data given is small.
The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations.
We know that if \(\ x_1, x_2, \ldots, x_n \) are observations with respective frequencies \(f_{1}, f_{2}, \ldots, f_{n}\) then this means observation \(\ x_1 \text { occurs } f_1 \text { times, } x_2 \text { occurs } f_2\) times, and so on.
\(\ x\) is the class mark for each interval, you can find the value of \(\ x\) by using
class mark
\[\begin{align}\left(x_i\right) =\frac{\text { upper limit + lower limit }}{2}\end{align}\]
Now, the sum of the values of all the observations \(=\ f_1 \ x_1+\ f_2 x_2+\ldots+\ f_n \ x_n\) and the number of observations \(=f_1+f_2+\ldots+f_n\).
So, the mean of the data is given by
\[\begin{align}\ \overline x\!=\!\frac{f_{1} x_{1}\!+\!f_{2} x_{2}\!+\!\ \ldots \ldots\!+\!f_{n} x_{n}}{f_{1}\!+\!f_{2}\!+\!\ldots \ldots\!+\!f_{n}}\end{align}\]
\(\begin{align}\overline x=\frac{\sum f_{i} x_{i}}{\Sigma f_{i}}\end{align}\) where \(i\) varies from \(1\) to \(n\)
Steps:
Number of plants | Number of houses\(\ (f_i)\) | \(\ x_i\) | \(\ f_ix_i\) | |
---|---|---|---|---|
\(0 - 2\) | \(1\) | \(1\) | \(1\) | |
\(2 - 4\) | \(2\) | \(3\) | \(6\) | |
\(4 - 6\) | \(1\) | \(5\) | \(5\) | |
\(6 - 8\) | \(5\) | \(7\) | \(35\) | |
\(8 - 10\) | \(6\) | \(9\) | \(54\) | |
\(10 - 12\) | \(2\) | \(11\) | \(22\) | |
\(12 - 14\) | \(3\) | \(13\) | \(39\) | |
\(\Sigma f_i = 20\) | \(\Sigma f_ix_i = 162\) |
From the table it can be observed that,
\[\begin{align}\Sigma f_i=20, \\ \Sigma f_ix_i =162\end{align}\]
\[\begin{align} \text { Mean } \overline x &=\frac{\Sigma f_i x_i}{\Sigma f_i} \\ &=\frac{162}{20} \\ &=8.1 \end{align}\]
Thus, the mean number of plants each house has \(8.1.\)
Here, we have used the direct method because the value of \(x_i\) and \(f_i\) are small.