question_answer

  The mean of the following frequency table is 53. But the frequencies and in the classes 20-40 and 60-80 are missing. Find the missing frequencies.

Age (in years)	0-20	20-40	40-60	60-80	80-100	Total
Number of people	15		21		17	100

OR Find the median of the following frequency distribution :

Marks	Frequency
0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 800-900 900-100	2 5 9 12 17 20 15 9 7 4

 

Answer:

  Let assumed mean a = 50 and h = 20

C.I.		Mid-value
0-20 20-40 40-60 60-80 80-100	15 21 17	10 30 50 70 90	-2 -1 0 1 2	-30 0 34

                                                                    ?(1) Mean i.e., Adding (1) and (2)        ?(2) Putting the value of in   OR

Age	Frequency	Cumulative frequency
0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 800-900 900- 1000	2 5 9 12 17 20 15 9 7 4	2 7 16 28 45 65 80 89 96 100

Cumulative frequency just greater than  is 65 and corresponding class is 500-600.  median class is Hence, Median