The following is the frequency distribution of marks obtained by 10th class students of a school in secondary school examination in mathematics. 

Marks            No of students
30-40                         8    
40-50                         7
50-60                         10
60-70                         15
70-80                          5
80-90                          8
90-100                       7

By using the above information, calculate mean, median, and mode.

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Dear Student,

Please find below the solution to the asked query:

We form table from given information , As  :

Marks	Mid point ( x )	Number of students ( f )	Cumulative frequency Cf
30 - 40	35	8	8
40 - 50	45	7	15
50 - 60	55	10	25
60 - 70	65	15	40
70 - 80	75	5	45
80 - 90	85	8	53
90 - 100	95	7	60
		$\sum _{}$f = 60

fx  = 280 , 315 , 550 , 975 , 375 , 680 , 665 $\sum _{}$fx = 3840
So,
Mean  = $\frac{{\sum }_{}^{}fx}{{\displaystyle \sum _{}}f}$ = $\frac{3840}{{\displaystyle 60}} = \mathbf{64}$                                                         ( Ans )
And
Median group is  60 - 70  , As we have mean 64

And
we know formula for median  = $L +\frac{{\displaystyle \frac{n}{2}} - C{f}_b}{f_m}\times w$
Here

L is the lower class boundary of the group containing the median =  60
n is the total number of data =  60
cf_b is the cumulative frequency of the groups before the median group =  25
f_m is the frequency of the median group = 15
w is the group width =  10
So,
Median  = $60 +\frac{{\displaystyle \frac{60}{2} - 25}}{15}\times 10 = 60 + \frac{30 - 25}{15}\times 10 = 60 + \frac{5}{15}\times 10= 60 + \frac{10}{3} = \frac{180 +10}{3} = \frac{190}{3} = \mathbf{63}\mathbf{.}\mathbf{33}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$
And
Modal group( class with highest frequency ) = 60 - 70
we know formula for mode  = $L +\frac{{\displaystyle f_m - f_{m-1}}}{\left(f_m - f_{m-1}\right) + \left(f_m - f_{m+1}\right)}\times w$
Here

L is the lower class boundary of the modal group =  60
f_{m - 1} is the frequency of the group before the modal group = 10
f_m  is the frequency of the modal group = 15
f_{m + 1} is the frequency of the group after the modal group = 5
w is the group width =  10
So,
Mode = $60 +\frac{{\displaystyle 15 - 10}}{\left(15 - 10\right) + \left(15 -5\right)}\times 10 =60 +\frac{{\displaystyle 5}}{\left(5\right) + \left(10\right)}\times 10 \mathbf{ }\mathbf{=}\mathbf{ }60 + \frac{10}{3} = \frac{180 +10}{3} = \frac{190}{3} =\mathbf{ }\mathbf{ }\mathbf{63}\mathbf{.}\mathbf{33}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$

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