• Q: Find the missing frequency in the following frequency distribution table,
    if N = 75 and median is 41 ,

    C 8-18 18-28 28-38 38-48 48-58 58-68 68-78  T
    f   22 6 f1 8 14 f2 5 75

    where:-
    C → class , f → frequency. T → Total. Share to Whatsapp



  • Given that data :
    Let f1 and f2 be the frequencies of class intervals 28-38 and 58-68 respectively. then,
    22 + 6 + f1 + 8 + 14 + f2 + 5 = 75
    f1 + f2 = 75 - 55

    f1 + f2 = 20
    f1 = 20 - f2 .................(1)

  • Median is 41 , which lies in 38-48. so, the median class is 38-48,
    since, L = 38, h = 10, f = 8, N = 75 and
      cf = 22 + 6 + f1 = 28 + f1

    Now,median = L + [h{(N/2)-cf}/f]
    =>41 = 38 + [ 10{(75/2)-(28 + f1)}/8 ]

    =>41 = 38 + [ 10{(37.5)-(28 + f1)}/8 ]
    =>41 = [38*8 + 10{(37.5)-(28 + f1)}]/(8)

    =>41 * 8 = [38*8 + 10{ 9.5 - f1}]
    => 328 = [304 + 95 - 10 *f1]

    => 328 = [399 - 10 *f1]
    => 10 *f1 = [399 - 328]

  • => 10 *f1 = 71
    => f1 = 7.1
    => f1 = 7
    From 1 , we get,
    f1 = 20 - f2
    => 7 = 20 - f2
    => f2 = 20 - 7
    => f2 = 13

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