Question

The mean of the following frequency distribution is 62.8 and the sum of all frequency is 50. Compute the missing frequency f1 and f2 :​

Answer 1

Step-by-step explanation:

CLASS CLASS fi fixi

MARK (Xi).

0-20. 10. 5 50

20-40. 30. f1. 30f1

40-60. 50. 10 500

60-80. 70. f2. 70f2

80-100. 90. 7. 630

100-120 110 8. 880

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30+f1+f2. 2060+30 f1+70 f2

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submission of fi = 50 = 30+f1+f2

f1+f2 = 50-30

f1 + f2 = 20 ,,,,,,,,,,,,,,,,,,,,,,,,(1)

_

mean ( x ) = submission fixi

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submission fi

62.8 = 2060 +30f1+70f2

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50

62.8 × 50 = 2060 + 30 f1 +70 f2

3140.0 = 2060 +30 f1 + 70 f2

3140 - 2060 = 30 f1 + 70 f2

30f1 + 70 f2 = 1080

3 f1 + 7 f2 = 108. ,,,,,,,,,,,,,,,,,,,,,,,,,,,,(2)

express f1 in term of f2 from eq (1) ,We get

f1 = 20 - f2

put this value of f1 in eq (2) ,We get

3 ( 20 - f2) +7 f2 = 108

60 - 3 f2 +7 f2 = 108

4 f2 = 108 - 60

f2 = 48/4

f2 = 12

now put this value of f2 in eq (1) , We get

f1 = 20 - f2

f1 = 20 - 12

f1 = 8

SO f1 = 8 and f2 = 12

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