The agewise participation of students in the Annual Function of a school in the shown in the following distribution.
Age                                   number of students
5-7                                     x               
7-9                                     15              
9-11                                   18              
11-13                                   30             
13-15                                   50             
15-17                                    48            
17-19                                    x           
Find the missing frequencies when the sum of frequencies is 181.Also,find the mode of data.
In the process of finding mode of data can mode be less than the lower class limit of the modal class interval.

Asked by ranjitrk1 | 19th Jun, 2016, 02:49: PM

Expert Answer:

begin mathsize 12px style Given space that space straight x plus 15 plus 18 plus 30 plus 50 plus 48 plus straight x space equals space 181 rightwards double arrow 2 straight x plus 161 equals 181 rightwards double arrow 2 straight x equals 20 rightwards double arrow straight x equals 10 So comma space the space missing space frequency space is space 10. Here space the space maximum space class space frequency space is space 50 comma space and space the space class space corresponding space to space this space frequency space is space 13 minus 15. So comma space the space modal space class space is space 13 minus 15. Now comma space modal space class space equals space 13 space minus 15 lower space limit space left parenthesis straight l right parenthesis space of space modal space class space equals space 13 class space size space left parenthesis straight h right parenthesis space equals space 2 frequency space left parenthesis straight f subscript 1 right parenthesis space of space the space modal space class space equals space 50 frequency space left parenthesis straight f subscript 0 right parenthesis space of space class space preceding space the space modal space class space equals space 30 frequency space left parenthesis straight f subscript 2 right parenthesis space of space class space succeeding space the space modal space class space equals 48 space Now comma space let space us space substitute space these space values space in space the space formula colon Mode equals straight l plus open parentheses fraction numerator straight f subscript 1 minus straight f subscript 0 over denominator 2 straight f subscript 1 minus straight f subscript 0 minus straight f subscript 2 end fraction close parentheses cross times straight h space space space space space space space space space space equals 13 plus open parentheses fraction numerator 50 minus 30 over denominator 2 left parenthesis 50 right parenthesis minus 30 minus 48 end fraction close parentheses cross times 2 space space space space space space space space space space equals 13 plus open parentheses 20 over 22 close parentheses cross times 2 equals 13 plus 20 over 11 equals 14.82 Therefore comma space the space mode space of space the space data space is space 14.82. Modal space class space is space the space class space with space the space maximum space frequency. The space mode space is space straight a space value space inside space the space modal space class space and space hence space cannot space be space less space than the space lower space class space limit space of space the space modal space class space interval. end style

Answered by Rebecca Fernandes | 19th Jun, 2016, 08:04: PM

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