Consider the following frequency distribution table Class interal 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 180 f 1 34 180 136 f 2 50 If the total frequency is 685 and median is 42 .6, then the values of f1 and f2 are

82, 23 Explanation: (d) Class Frequency Cumulative Frequency 10 – 20 180 180 20 – 30 f₁ 180 + f₁ 30 – 40 34 214 + f₁ 40 – 50 180 394 + f₁ 50 – 60 136 530 + f₁ 60 – 70 f₂ 530 + f₁ + f₂ 70 – 80 50 580 + f₁ + f₂ Total frequency = 685 580 + f₁ + f₂ = 685 f₁ + f₂ = 105 Median = 42.6 (given) lies in (40 – 50) interval Class = 40 – 50 Median = ℓ + – C · f f = h ℓ = 40, h = 10 C · f = 214 + f₁ 42.6 = 40 + – (214 + f₁) 2.6 × 18 = 342.5 – 214 – f₁ f₁ = 81.7 ≈ 82 f₂ = 105 – 82 = 23

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NIMCET 2023 — Mathematics PYQ

NIMCET | Mathematics | 2023

Consider the following frequency distribution table

Class interal	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	180	f₁	34	180	136	f₂	50

If the total frequency is 685 and median is 42 .6, then the values of f1 and f2 are

NIMCET 2023 — Mathematics PYQ

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