AP EAMCET201822 Apr 2018Morning ShiftMathematicsStatisticsActual
The mean deviation from the median for the following distribution (corrected to two decimals) is array lllllllll x _ i & 6 & 9 & 3 & 12 & 15 & 13 & 21 & 22 f _ i & 4 & 5 & 3 & 2 & 5 & 4 & 4 & 3 array
Options
- A13.42
- B5.40
- C4.97
- D11.25
Correct answer
B. 5.40
Step-by-step solution
array c|c|c|c|c x _ i & f _ i & array c Cumulative frequency array & ( d _ i )= | x _ i - 1 5 | & f _ i | d _ i | 6 & 4 & 4 & 9 & 36 9 & 5 & 9 & 6 & 30 3 & 3 & 12 & 12 & 36 12 & 2 & 14 & 3 & 6 15 & 5 & 19 & 0 & 0 13 & 4 & 23 & 2 & 8 21 & 4 & 27 & 6 & 24 22 & 3 & 30 & 7 & 21 & array c N= f_i =30 array & & & array c f d_i =161 array array Clearly, aligned N & =30 N 2 & =15 aligned The cumulative frequency just greater than N 2 is 19 and corresponding value of x is 15 . Therefore, median =15 . Clearly, f_i |x_i-15 |=