10.8 - 10.8 = 0.10 = 0.10 Pa = 0.02 Pt = 0.10 The required...

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10.8 = 0.10, β = 0.10 P a = 0.02, P t = 0.10 The required sample size n is, 2 1 1 1 1 2 1 1 1 1 ) 02 . 0 ( ) 10 . 0 ( ) 10 . 0 ( ) 90 . 0 ( ) ( ) ( ) ( ) 1 ( Φ Φ Φ Φ = Φ Φ Φ Φ = a t P P n α β 11 77 . 10 78 . 0 56 . 2 ) 06 . 2 ( 28 . 1 ) 28 . 1 ( 28 . 1 2 2 = = = The corresponding tolerable fraction defective M is
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