Question

The following table gives the number of branches and number of plants in the garden of a school.
No of branches (x): 2 3 4 5 6
No of plants (f): 49 43 57 38 13
Calculate the average number of branches per plant

Answer 1

ASSUMED MEAN METHOD :  

In this method, first of all, one among xi 's is chosen as the assumed mean denoted  by ‘A’. After that the difference ‘di’ between ‘A’ and each of the xi's i.e di = xi - A is calculated .  

ARITHMETIC MEAN =  A +  Σfidi / Σfi

[‘Σ’ Sigma means ‘summation’ ]

★★ We may take Assumed mean 'A’ to be that xi which lies in the middle of x1 ,x2 …..xn.

FREQUENCY DISTRIBUTION TABLE IS IN THE ATTACHMENT  

From the table : Σfidi = - 77 ,Σfi = 200 , A = 4

ARITHMETIC MEAN =  A +  Σfidi / Σfi

ARITHMETIC MEAN =  4 + (- 77/200)

= 4 -  77/200

= 4 -  0.385

ARITHMETIC MEAN = 3.615 ≈ 3.62 (approximate)  

Hence, the average number of branches per plant is ≈ 3.62(approximate).  

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer :

The mean number of branches per plant is 3.62 .

Step-by-step explanation :

Assumed mean method -

In assumed mean method, at first one of the $x_{i}$ is chosen as assumed mean which is denoted by "A".

Then we calculate the difference, $d_{i}$ by using the given formula i.e.,

$d_{i}=x_{i}-A$  

$Arithmetic\:mean=A+\frac{\Sigma f_{i}d_{i}}{\Sigma f_{i}}$

where $\Sigma$ means summation.

Let us take the assumed mean to be 4.

Frequency Distribution Table -

$\begin{tabular}{| c | c | c | c |}\cline{1-4}x_i & f_i & d_i=x_{i}-A & f_{i}d_{i} \\ \cline{1-4}2 & 49 & -2 & -98 \\ \cline{1-4}3 & 43 & -1 & -43 \\ \cline{1-4}4 & 57 & 0 & 0 \\ \cline{1-4}5 & 38 & 1 & 38 \\ \cline{1-4}6 & 13 & 2 & 26 \\ \cline{1-4} & \Sigma f_{i}=200 & & \Sigma f_{i}d_{i}=-77\\ \cline{1-4}\end{tabular}$

Since, Arithmetic mean -

$\implies A+\frac{\Sigma f_{i}d_{i}}{\Sigma f_{i}}$

$\implies 4+\frac{-77}{200}$

$\implies 4-\frac{77}{200}$

$\implies 4-0.385$

$\implies 3.615$

$\implies \approx 3.62$