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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 10—20 20—30 30—40 40—50 50—60 60—70 70—80 ∑f or N= 150 +f1+f2

Frequency (f) 12 30 f1 65 f2 25 18 = 229 (given)

Cumulative Frequency (c.f.) 12 42 42+f1 107+f1 107+f1+f2 132+f1+f2 150+f1+f2

Middle item is N/2 =229/2 =114.5 Median 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 = -10f1 f1 = 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 Frequency 0—100 4 100—200 16 200—300 60 300—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 4 100—200 16 20 200-300 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 Less than “ “ “ “ “ “

10 20 30 40

Frequency 4 16 40 76

Value Less than 50 “ 60 “ 70 “ 80

Frequency 96 112 120 125

(C.A. Nov., 1977) Solution: Computation of Median Value Frequency

Cumulative Frequency 0—10 4 4 10—20 12 16 20—30 24 40 30—40 36 76 40—50 20 96 50—60 16 112 60—70 8 120 70—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

0-20 20-40 40-60 60-80 80-100

14 f1 27 44-f1 15

Cumulative frequency 14 14+f1 41+f1 85+f1—f1 or 85 100

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

or

-450=-20f1

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 ($) 30—32 32—34 34—36 36—38 38—40

No. of workers

Weekly wages ($)

No. of workers

2 9 25 30 49

40—42 42—44 44—46 46—48 48—50

62 39 20 11 3

Solution: Computation of Median and Quartiles

Weekly wages (Rs.) 30—32 32—34 34—36 36—38 38—40 40—42 42—44 44—46 46—48 48—50

No. of workers (f) 2 9 25 30 49 62 39 20 11 3 N—250

Cumulative frequency 2 11 36 66 115 177 216 236 247 250

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) =-37.75 Upper 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 3—5 5—7 7—9 9—11

6 53 85 56 21

11—13 13—15 15—17

16 4 4

245 Solution: Calculation of median Class-intervals Frequency Cumulative frequency 1-3 6 6 3-5 53 59 5- 7 85 144 7- 9 56 200 9-11 21 221 11-13 16 237 13-15 4 241 15-17 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

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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 10—20 20—30 30—40 40—50 50—60 60—70 70—80 ∑f or N= 150 +f1+f2

Frequency (f) 12 30 f1 65 f2 25 18 = 229 (given)

Cumulative Frequency (c.f.) 12 42 42+f1 107+f1 107+f1+f2 132+f1+f2 150+f1+f2

Middle item is N/2 =229/2 =114.5 Median 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 = -10f1 f1 = 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 Frequency 0—100 4 100—200 16 200—300 60 300—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 4 100—200 16 20 200-300 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 Less than “ “ “ “ “ “

10 20 30 40

Frequency 4 16 40 76

Value Less than 50 “ 60 “ 70 “ 80

Frequency 96 112 120 125

(C.A. Nov., 1977) Solution: Computation of Median Value Frequency

Cumulative Frequency 0—10 4 4 10—20 12 16 20—30 24 40 30—40 36 76 40—50 20 96 50—60 16 112 60—70 8 120 70—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

0-20 20-40 40-60 60-80 80-100

14 f1 27 44-f1 15

Cumulative frequency 14 14+f1 41+f1 85+f1—f1 or 85 100

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

or

-450=-20f1

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 ($) 30—32 32—34 34—36 36—38 38—40

No. of workers

Weekly wages ($)

No. of workers

2 9 25 30 49

40—42 42—44 44—46 46—48 48—50

62 39 20 11 3

Solution: Computation of Median and Quartiles

Weekly wages (Rs.) 30—32 32—34 34—36 36—38 38—40 40—42 42—44 44—46 46—48 48—50

No. of workers (f) 2 9 25 30 49 62 39 20 11 3 N—250

Cumulative frequency 2 11 36 66 115 177 216 236 247 250

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) =-37.75 Upper 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 3—5 5—7 7—9 9—11

6 53 85 56 21

11—13 13—15 15—17

16 4 4

245 Solution: Calculation of median Class-intervals Frequency Cumulative frequency 1-3 6 6 3-5 53 59 5- 7 85 144 7- 9 56 200 9-11 21 221 11-13 16 237 13-15 4 241 15-17 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

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