The mean age of a group of 41 students is 17.45 years . Find the missing frequencies.
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  • Age (in years ).                 15.        16.     17.     18     19.    20 
  • Number of students            3.          ?        9.      11.    ?        3

Asked by abhi611 | 7th Jul, 2017, 07:38: PM

Expert Answer:

begin mathsize 16px style As space you space know space that space summation space of space frequencies space is space equal space to space the space number space of space students.
Let space frequency space of space 16 space be space straight x space and space frequency space of space 19 space be space straight y.
3 plus straight x plus 9 plus 11 plus straight y plus 3 equals 41
straight x plus straight y plus 26 equals 41
straight x plus straight y equals 15.............. left parenthesis straight i right parenthesis
Mean equals 17.45 space space space space space left parenthesis Given right parenthesis
fraction numerator 15 cross times 3 plus 16 straight x plus 17 cross times 9 plus 18 cross times 11 plus 19 straight y plus 20 cross times 3 over denominator 41 end fraction equals 17.45
45 plus 16 straight x plus 153 plus 198 plus 19 straight y plus 60 equals 17.45 cross times 41
456 plus 16 straight x plus 19 straight y equals 715.45
16 straight x plus 19 straight y equals 259.45................... left parenthesis ii right parenthesis
Solve space both space equations space simultaneously space straight o space get space the space values space of space straight x space and space straight y.

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Answered by Sneha shidid | 7th Jul, 2017, 09:12: PM