Question

Calculate Arithmetic Mean from the following table
Wages in Rupees
No. of Workers(f)
Less than 48
Less than 56
48-64
64 and above
72-80
80 and above
88 and above
5
15
29
31
8
19
5

Answer 1

16

Arthimetic mean $=$$\frac{5+15+29+31+8+19+5}{7} =16$

Answer 2

Find the mean of the given data.

Explanation:

• Given the frequencies f1, f2, ...., fi of individual class intervals than the mean of the grouped data is given by,                                                                        $\bar{x}= \frac{\sum f_i m_i}{\sum f} \\mid\ point\ of\ each\ interval\ ->m_i=\frac{upper\ limit + lower\ limit}{2}$  
• We have the data of wages and no. of workers as given below,                               $Wages\ in\ Rupees\ \ \ \ \ \ \ \ \ \ No.\ of\ Workers(f)\\ Less\ than\ 48\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5\\Less\ than\ 56\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 15\\48-64\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 29\\64\ and\ above\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 31\\72-80\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8\\80\ and\ above\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 19\\88\ and\ above\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5$    
• above data is not properly arranged. Hence, arranging it in proper intervals we get,                                                                                                   $Wages\ in\ Rupees\ \ \ \ \ \ \ \ \ \ No.\ of\ Workers(f)\\ Less\ than\ 48\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5\\48-56\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 10\\56-64\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 19\\64-72\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\\72-80\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8\\80-88\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 12\\88\ and\ above\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5$                                                    
• we get the sum of individual frequencies of each class interval as,             $\sum f_im_i= 5(\frac{48+0}{2})+10(\frac{48+56}{2})+19(\frac{56+64}{2})+6(\frac{64+72}{2})+8(\frac{72+80}{2})+12(\frac{88+80}{2})+5(\frac{88}{2}) \\->\sum f_im_i= 4024$
• Now, we get the sum of all frequencies  as,                                                        $\sum f_i = 5+10+19+6+8+12+5\ \ \ \ \ \ \ ->\sum f_i=65$  
• Hence, we get the arithmetic mean of the data as,                                       $\bar{x}=\frac{4024}{65}\ \ \ \ \ \ \ \ \ \ ->\bar{x}= 61.91(approx.)$