Question

If the median is 240 so find the value of f. 0-100 15 100-200. 17 200-300. F 300-400. 12 400-500. 9 500-600. 5 600-700. 2

Answer 1

Value of f is 20.
I hope it will help you.

Answer 2

Given is the median for frequency distribution, Find the value of 'f'.

Explanation:

• Given below is the table showing the frequency distribution for each class and its corresponding cumulative frequency:               $Class\ \ \ \ \ \ \ \ \ \ frequency\ \ \ \ \ \ cf\\0-100\ \ \ \ \ \ \ \ \ \ \ 15\ \ \ \ \ \ \ \ \ \ \ \ \ \ 15 \\100-200\ \ \ \ \ \ \ \ 17\ \ \ \ \ \ \ \ \ \ \ \ \ \ 32\\200-300\ \ \ \ \ \ \ \ F\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 32+F\\300-400\ \ \ \ \ \ \ \ 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ 44+F\\400-500\ \ \ \ \ \ \ \ 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 53+F\\ 500-600\ \ \ \ \ \ \ \ 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 58+F\\600-700\ \ \ \ \ \ \ \ 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 60+F$  
• We have the class size $C=100$ and the total number of frequencies is N, then  $\frac{N}{2}=\frac{60+F}{2}\ \ \ \ -> \frac{N}{2}=30+\frac{F}{2}$
• Hence the median class is the one having cf greater than or equal to $\frac{N}{2}$ which is, $200-300$
• The median is given by , $M_d=l+(\frac{\frac{N}{2} -cf'}{f_m} )C$  
• Here, 'cf' ' is the cumulative frequency of class before the median class, 'l' is the lower boundary, 'fm' is the frequency of median class.
• Hence putting the values and median in the above formula we get ,
• $M_d=240\\->200+(\frac{30+\frac{F}{2}-32 }{F} )100=240\\->200F+50F-200=240F\\->F=20$   ------>ANSWER