Question

The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:
No. of calls (x):
0
1
2
3
4
5
6
No. of intervals (f):
15
24
29
49
54
43
39
Compute the mean number of calls per intervals.

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 = 135 ,Σfi = 250 , A = 3

ARITHMETIC MEAN =  A +  Σfidi / Σfi

ARITHMETIC MEAN =  3 + (135/250)

= 3 + 27/50

= 3 + 0.54  

ARITHMETIC MEAN = 3.54  

Hence, the mean number of calls per interval is 3. 54 .  

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer :

The mean number of calls per intervals is 3.54.

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 3.

Frequency Distribution Table -

$\begin{tabular}{| c | c | c | c |}\cline{1-4}x_i & f_i & d_i=x_{i}-A & f_{i}d_{i} \\ \cline{1-4}0 & 15 & -3 & -45 \\ \cline{1-4}1 & 24 & -2 & -48 \\ \cline{1-4}2 & 29 & -1 & -29 \\ \cline{1-4}3 & 46 & 0 & 0 \\ \cline{1-4}4 & 54 & 1 & 54 \\ \cline{1-4}5 & 43 & 2 & 86 \\ \cline{1-4}6 & 39 & 3 & 117 \\ \cline{1-4} & \Sigma f_{i}=250 & & \Sigma f_{i}d_{i}=135\\ \cline{1-4}\end{tabular}$

Since, Arithmetic mean -

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

$\implies 3+\frac{135}{250}$

$\implies 3+\frac{27}{50}$

$\implies 3+0.54$

$\implies 3.54$