Find Missing Frequency

You are given that the Median value is 46.

   (a)       Using the Median formula fill up the missing frequencies.

   (b)       Calculate the arithmetic mean of the completed table,

Solution: (a)       Let the missing frequency be f1 for class-interval 30—40 and f2 for class-interval 50—60.

Computation of Median

Variable Frequency (f) Cumulative Frequency (c.f.)

10—20 12 1220—30 30 4230—40 f1 42+f1

40—50 65 107+f1

50—60 f2 107+f1+f2

60—70 25 132+f1+f2

70—80 18 150+f1+f2

 ∑f or N= 150 +f1+f2     = 229 (given)

 Middle item is N/2 =229/2 =114.5Median value is 46 (given) and it lies in the class-interval 40—50. Using the Median formula .

      M = l1+(l2-l1)/f1  (m-c) we get

    46 = 40+(50-40)/65 [114.5-(42+f1)]

    46 =40+(10/65×72.5-f1 )

   46 = 40+(725-10f_1)/65  or 6 = (725-10f_1)/65

   390 = 725-10f1 or -335 = -10f1f1 = 33.5 or 34     since the frequency cannot be in fraction

Now ∑f = 150+f1+f2 (Given)  or  229 = 150+34+f2     or f2 = 45

(b)     Now mean can be calculated by completing the series by putting the value of f1 and f2. The calculation of mean would be done by the method explained earlier.The value of the mean in the question would come to 45.83.

Solved Problem Of Mode

Example: The median and the mode of the following distribution are known to be $ 335 and $ 340 respectively. Three frequency values from the table are, however, missing. Find the missing values :                      

Wages in Frequency0—100 4

100—200 16200—300 60300—400 ?400—500 ?500—600 ?600—700 4

  230

Solution: Let the missing frequencies be x1 x2 and x3 for the class-intervals 300—400, 400—500 and 500—600 respectively. Now 230 = 84+x1+x2+x3 or x3=230—84—x1—x2 or x3 = 146— (x3+x2)

Now we will recast the table as follows :  

Wages. Frequency. Cumulative frequency.        ($)               

0—100 4 4100—200 16 20200-300 60 80

300—400 x1 80+x1

400-500 X2 80+x1+x2

500—600 x3 80+x1+x2+x3

600—700 4 84+x1+x2+x3

Median and Mode, as given, both lie in 300- 400 class interval.

Median            = 335  =l1+(l2-l1)/f1  (m-c)  = 300 +100/x1  (115—80) or

                335 = 300 +3500/x1   or 35 = 3500/x1  or x1 = 100    Mode            = 340 = l1+(f1-f0)/(2f1-f0-f2     )     (l2-l1)

                        = 300 + (x1-60)/(2x1-60-x2 × 100

                    340 = 300 + (100-60)/(200-60-x2 ) × 100

 or                  40 = 4000/(140-x2 ) or 40 (140—x2) = 4000 or x2 = 40

                      X3 = 146— x1—x2 = 146— 100—40 = 6

 Thus the missing values are 100, 40 and 6 respectively.

Median in Cumulative Series- Solved Examples

Example: Calculate the median from the following data:

Value Frequency     Value     Frequency Less than    10       4 Less than   50   96“       “       20     16 “     60 112“       “       30     40 “     70 120“       “       40     76 “     80 125

                                            (C.A. Nov., 1977)

Solution: Computation of Median                       Value Frequency Cumulative

Frequency0—10 4 410—20 12 1620—30 24 4030—40 36 7640—50 20 9650—60 16 11260—70 8 12070—80 5 125

Middle item is 125/2 or 62.5 which lies in 30—40 group                 

                            

Cumulative frequency table can be of 'more than' type also. In such cases also the data have to be converted into a simple continuous series and median calculated according to the rule explained in example. The  following illustration would clarify the point.

or                    f2 = 100-(56+f1)

or                    f2 = 44—f1

         

  Calculation of Median

Expenditure No. of families Cumulative frequency

0-20 14 1420-40 f1 14+f1

40-60 27 41+f1

60-80 44-f1 85+f1—f1 or 8580-100 15 100

     Median is given in this problem as 50.

Middle item of the series is also 100/2 or 50

which means it lies in the class-interval 40—60. Now,

               

                270 =720-20f1         or     -450=-20f1

                              or             f1 = 450/20¬ =22.5  

Since the frequency in this problem cannot be in fraction so f1 would be taken as 23.

f2 = 44—f1 or 44—23 or 21. Thus the missing values in the question are 23 and 21.

Calculate Median and Quartile

Example: From the following data, calculate the median and the first and third quartile wages:

Weekly

wages ($)

No. of workers

Weekly wages ($)

No. of workers

30—32 2 40—42 6232—34 9 42—44 3934—36 25 44—46 2036—38 30 46—48 1138—40 49 48—50 3

Solution: Computation of Median and Quartiles

  Weekly

wages (Rs.)No. of

workers (f)Cumulativefrequency

30—32 2 232—34 9 1134—36 25 3636—38 30 6638—40 49 11540—42 62 17742—44 39 21644—46 20 23646—48 11 24748—50 3 250

  N—250   

Median is the value of (250/2) th or 125th item which lies in the 40—42 group. Its value would be M = l2 + (l2-l1)/f1  (m-c)

   = 40+ 2/62 (125—115) =40.32

Lower Quartile is the value of (250/4)  th or 62.5th  item which lies in 36—38 group. Its value would be Q1 = l1+ (l2-l1)/f1   (q1—c) where q1 is the quartile number or 62.5         = 36 + (38-36)/ 30 (62.5-36)  =-37.75Upper Quartile is the value of (3(N))/4th or 187.5th  item which lies in 42—44 group. Its value would be,

Continuous Series-Exclusive

Example: (Continuous series-exclusive).

 

      Find the median of the following distribution. 

Class intervals

$

Frequencies Class-intervals     $ 

Frequencies

1—3 6 11—13 163—5 53 13—15 45—7 85 15—17 47—9 569—11 21

      245 Solution: Calculation of median

Class-intervals Frequency Cumulative frequency1-3 6 63 - 5 53 595- 7 85 1447- 9 56 2009-11 21 22111-13 16 23713-15 4 24115-17 4 245

Median=the value of N/2 i.e., 122.5th item,    which lies in 5—7 group

Applying the formula of interpolation.                    M = l1 + l2 - l1 / f1 (m - c)

 we have,   M = 5 + (7-5)/85(122.5—59) = 6.5

In the above example median is the value of 122.5th items which lies in 5—7 group. In this group the number of items is 85. On the presumption that these items are uniformly distributed in this class-interval, we can calculate median by direct arithmetical process also. 59th  item has the value of 5 and the next 85 items up to 144, are spread over two values from 5 to 7. From Wm to 122.5 there are 63.5 items. The value of 63.5 item after 59th  (or the value of 122.5 item) would exceed 5 (the value of 59th item) by 2/85X 63.5 or by 1.5. Thus the value of the 122.5th  item would be 5+1.5 or 6.5.

If the above problem is solved by the alternate formula given above the same result would be obtained. This will be clear from the following calculation.  Median                = L + N/2 - c.f. / f × i 

                             = 5 + 122.5 - 59 / 85 × 2    = 6.5