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Math Advanced Math ADVANCED ENGINEERING MATHEMATICS (LL)

The probability of accepting a lot with 1 % , 2 % , 10 % defectives and draw the graph for sample size 50. The probability of accepting a lot with 1 % , 2 % , 10 % defectives 0.9098 , 0.7358 _ and 0.0402 _ respectively. Formula used: The probability by using the approximation of Poisson’s distribution P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . Where P ( A , θ ) is probability of acceptance with one percent defective. The value of mean is μ = n p . Calculation: Suppose that the probability of defective knives is p = 0.01 , uses a sample of size 20 and the acceptance number is c = 1 . Substitute the value of n = 20 and p = 0.01 in the equation μ = n p . μ = n p = 20 × 0.01 = 0.2 Now substitute the value of μ = 0.2 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! = e − 0.2 ∑ x = 0 1 0.2 x x ! = e − 0.2 0.2 0 x ! + e − 0.2 0.2 1 x ! Further simplification of above equation. P ( A , θ ) = e − 0.2 [ 1 0 ! + 0.2 1 ! ] = e − 0.2 [ 1 + 0.2 ] = 1.2 e − 0.2 = 0.984 Same as the probability of defective knives is p = 0.02 , uses a sample of size 20 and the acceptance number is c = 1 . Substitute the value of n = 20 and p = 0.02 in the equation μ = n p . μ = n p = 20 × 0.02 = 0.4 Now substitute the value of μ = 0.4 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! = e − 0.4 ∑ x = 0 1 0.4 x x ! = e − 0.4 0.4 0 x ! + e − 0.4 0.4 1 x ! Further simplification of above equation. P ( A , θ ) = e − 0.4 [ 1 0 ! + 0.4 1 ! ] = e − 0.4 [ 1 + 0.4 ] = 1.4 e − 0.4 = 0.9384 Same as the probability of defective knives is p = 0.1 , uses a sample of size 20 and the acceptance number is c = 1 . Substitute the value of n = 20 and p = 0.1 in the equation μ = n p . μ = n p = 20 × 0.1 = 2 Now substitute the value of μ = 2 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! = e − 2 ∑ x = 0 1 2 x x ! = e − 2 2 0 x ! + e − 2 2 1 x ! Further simplification of above equation. P ( A , θ ) = e − 2 [ 1 0 ! + 2 1 ! ] = e − 2 [ 1 + 2 ] = 3 e − 2 = 0.4060 From the figure 1 say that, with increasing of the value of θ , the value P ( A , θ ) is decreased. Now suppose that the probability of defective knives is p = 0.01 , uses a sample size 50 and the acceptance number is c = 1 . Substitute the value of n = 50 and p = 0.01 in the equation μ = n p . μ = n p = 50 × 0.01 = 0.5 Now substitute the value of μ = 0.5 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! = e − 0.5 ∑ x = 0 1 0.5 x x ! = e − 0.5 0.5 0 x ! + e − 0.5 0.5 1 x ! Further simplification of above equation. P ( A , θ ) = e − 0.5 [ 1 0 ! + 0.5 1 ! ] = e − 0.5 [ 1 + 0.5 ] = 1.5 e − 0.5 = 0.9098 Same as the probability of defective knives is p = 0.02 , uses a sample of size 50 and the acceptance number is c = 1 . Substitute the value of n = 20 and p = 0.02 in the equation μ = n p . μ = n p = 50 × 0.02 = 1 Now substitute the value of μ = 1 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − 1 ∑ x = 0 c 1 x x ! = e − 1 ∑ x = 0 1 1 x x ! = e − 1 1 0 x ! + e − 0.4 1 1 x ! Further simplification of above equation. P ( A , θ ) = e − 1 [ 1 0 ! + 1 1 ! ] = e − 1 [ 1 + 1 ] = 2 e − 1 = 0.7358 Same as the probability of defective knives is p = 0.1 , uses a sample of size 50 and the acceptance number is c = 1 . Substitute the value of n = 50 and p = 0.1 in the equation μ = n p . μ = n p = 50 × 0.1 = 5 Now substitute the value of μ = 5 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − 5 ∑ x = 0 c 5 x x ! = e − 5 ∑ x = 0 1 5 x x ! = e − 5 5 0 x ! + e − 5 5 1 x ! Further simplification of above equation. P ( A , θ ) = e − 5 [ 1 0 ! + 5 1 ! ] = e − 5 [ 1 + 5 ] = 6 e − 2 = 0.0402 From the figure 1 and figure 2 say the probabilities became smaller as the size of sample become larger. Therefore, the probability of accepting a lot with 1 % , 2 % , 10 % defectives 0.9098 , 0.7358 _ and 0.0402 _ respectively.

The probability of accepting a lot with 1 % , 2 % , 10 % defectives and draw the graph for sample size 50. The probability of accepting a lot with 1 % , 2 % , 10 % defectives 0.9098 , 0.7358 _ and 0.0402 _ respectively. Formula used: The probability by using the approximation of Poisson’s distribution P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . Where P ( A , θ ) is probability of acceptance with one percent defective. The value of mean is μ = n p . Calculation: Suppose that the probability of defective knives is p = 0.01 , uses a sample of size 20 and the acceptance number is c = 1 . Substitute the value of n = 20 and p = 0.01 in the equation μ = n p . μ = n p = 20 × 0.01 = 0.2 Now substitute the value of μ = 0.2 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! = e − 0.2 ∑ x = 0 1 0.2 x x ! = e − 0.2 0.2 0 x ! + e − 0.2 0.2 1 x ! Further simplification of above equation. P ( A , θ ) = e − 0.2 [ 1 0 ! + 0.2 1 ! ] = e − 0.2 [ 1 + 0.2 ] = 1.2 e − 0.2 = 0.984 Same as the probability of defective knives is p = 0.02 , uses a sample of size 20 and the acceptance number is c = 1 . Substitute the value of n = 20 and p = 0.02 in the equation μ = n p . μ = n p = 20 × 0.02 = 0.4 Now substitute the value of μ = 0.4 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! = e − 0.4 ∑ x = 0 1 0.4 x x ! = e − 0.4 0.4 0 x ! + e − 0.4 0.4 1 x ! Further simplification of above equation. P ( A , θ ) = e − 0.4 [ 1 0 ! + 0.4 1 ! ] = e − 0.4 [ 1 + 0.4 ] = 1.4 e − 0.4 = 0.9384 Same as the probability of defective knives is p = 0.1 , uses a sample of size 20 and the acceptance number is c = 1 . Substitute the value of n = 20 and p = 0.1 in the equation μ = n p . μ = n p = 20 × 0.1 = 2 Now substitute the value of μ = 2 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! = e − 2 ∑ x = 0 1 2 x x ! = e − 2 2 0 x ! + e − 2 2 1 x ! Further simplification of above equation. P ( A , θ ) = e − 2 [ 1 0 ! + 2 1 ! ] = e − 2 [ 1 + 2 ] = 3 e − 2 = 0.4060 From the figure 1 say that, with increasing of the value of θ , the value P ( A , θ ) is decreased. Now suppose that the probability of defective knives is p = 0.01 , uses a sample size 50 and the acceptance number is c = 1 . Substitute the value of n = 50 and p = 0.01 in the equation μ = n p . μ = n p = 50 × 0.01 = 0.5 Now substitute the value of μ = 0.5 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! = e − 0.5 ∑ x = 0 1 0.5 x x ! = e − 0.5 0.5 0 x ! + e − 0.5 0.5 1 x ! Further simplification of above equation. P ( A , θ ) = e − 0.5 [ 1 0 ! + 0.5 1 ! ] = e − 0.5 [ 1 + 0.5 ] = 1.5 e − 0.5 = 0.9098 Same as the probability of defective knives is p = 0.02 , uses a sample of size 50 and the acceptance number is c = 1 . Substitute the value of n = 20 and p = 0.02 in the equation μ = n p . μ = n p = 50 × 0.02 = 1 Now substitute the value of μ = 1 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − 1 ∑ x = 0 c 1 x x ! = e − 1 ∑ x = 0 1 1 x x ! = e − 1 1 0 x ! + e − 0.4 1 1 x ! Further simplification of above equation. P ( A , θ ) = e − 1 [ 1 0 ! + 1 1 ! ] = e − 1 [ 1 + 1 ] = 2 e − 1 = 0.7358 Same as the probability of defective knives is p = 0.1 , uses a sample of size 50 and the acceptance number is c = 1 . Substitute the value of n = 50 and p = 0.1 in the equation μ = n p . μ = n p = 50 × 0.1 = 5 Now substitute the value of μ = 5 and c = 1 in the equation P ( A , θ ) = e − μ ∑ x = 0 c μ x x ! . P ( A , θ ) = e − 5 ∑ x = 0 c 5 x x ! = e − 5 ∑ x = 0 1 5 x x ! = e − 5 5 0 x ! + e − 5 5 1 x ! Further simplification of above equation. P ( A , θ ) = e − 5 [ 1 0 ! + 5 1 ! ] = e − 5 [ 1 + 5 ] = 6 e − 2 = 0.0402 From the figure 1 and figure 2 say the probabilities became smaller as the size of sample become larger. Therefore, the probability of accepting a lot with 1 % , 2 % , 10 % defectives 0.9098 , 0.7358 _ and 0.0402 _ respectively.

ADVANCED ENGINEERING MATHEMATICS (LL)

ADVANCED ENGINEERING MATHEMATICS (LL)

10th Edition

ISBN: 9781119455929

Author: Kreyszig

Publisher: WILEY

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Chapter 25 Solutions

ADVANCED ENGINEERING MATHEMATICS (LL)

Ch. 25.2 - Prob. 1P Ch. 25.2 - Prob. 2P Ch. 25.2 - Prob. 3P Ch. 25.2 - Prob. 4P Ch. 25.2 - Prob. 5P Ch. 25.2 - Prob. 6P Ch. 25.2 - Prob. 7P Ch. 25.2 - Prob. 8P Ch. 25.2 - Prob. 9P Ch. 25.2 - Prob. 10P

Ch. 25.2 - Prob. 11P Ch. 25.2 - Prob. 12P Ch. 25.2 - Prob. 13P Ch. 25.2 - Prob. 14P Ch. 25.3 - Prob. 1P Ch. 25.3 - Prob. 2P Ch. 25.3 - Prob. 3P Ch. 25.3 - Prob. 4P Ch. 25.3 - Prob. 5P Ch. 25.3 - Prob. 6P Ch. 25.3 - Prob. 7P Ch. 25.3 - Prob. 8P Ch. 25.3 - Prob. 9P Ch. 25.3 - Prob. 10P Ch. 25.3 - Prob. 11P Ch. 25.3 - Prob. 13P Ch. 25.3 - Prob. 14P Ch. 25.3 - Prob. 15P Ch. 25.3 - Prob. 16P Ch. 25.3 - Prob. 17P Ch. 25.3 - Prob. 18P Ch. 25.3 - Prob. 19P Ch. 25.3 - Prob. 20P Ch. 25.4 - Prob. 1P Ch. 25.4 - Prob. 2P Ch. 25.4 - Prob. 3P Ch. 25.4 - Prob. 4P Ch. 25.4 - Prob. 5P Ch. 25.4 - Prob. 6P Ch. 25.4 - Prob. 7P Ch. 25.4 - Prob. 8P Ch. 25.4 - Prob. 9P Ch. 25.4 - Prob. 11P Ch. 25.4 - Prob. 12P Ch. 25.4 - Prob. 13P Ch. 25.4 - Prob. 14P Ch. 25.4 - Prob. 15P Ch. 25.4 - Prob. 16P Ch. 25.4 - Prob. 17P Ch. 25.4 - Prob. 18P Ch. 25.4 - Prob. 20P Ch. 25.5 - Prob. 1P Ch. 25.5 - Prob. 2P Ch. 25.5 - Prob. 3P Ch. 25.5 - Prob. 4P Ch. 25.5 - Prob. 5P Ch. 25.5 - Prob. 6P Ch. 25.5 - Prob. 7P Ch. 25.5 - Prob. 8P Ch. 25.5 - Prob. 9P Ch. 25.5 - Prob. 10P Ch. 25.5 - Prob. 11P Ch. 25.5 - Prob. 13P Ch. 25.5 - Prob. 14P Ch. 25.5 - Prob. 15P Ch. 25.6 - Prob. 1P Ch. 25.6 - Prob. 2P Ch. 25.6 - Prob. 3P Ch. 25.6 - Prob. 4P Ch. 25.6 - Prob. 5P Ch. 25.6 - Prob. 6P Ch. 25.6 - Prob. 7P Ch. 25.6 - Prob. 8P Ch. 25.6 - Prob. 9P Ch. 25.6 - Prob. 10P Ch. 25.6 - Prob. 11P Ch. 25.6 - Prob. 12P Ch. 25.6 - Prob. 13P Ch. 25.6 - Prob. 14P Ch. 25.6 - Prob. 15P Ch. 25.7 - Prob. 2P Ch. 25.7 - Prob. 3P Ch. 25.7 - Prob. 4P Ch. 25.7 - Prob. 5P Ch. 25.7 - Prob. 6P Ch. 25.7 - Prob. 7P Ch. 25.7 - Prob. 8P Ch. 25.7 - Prob. 9P Ch. 25.7 - Prob. 12P Ch. 25.7 - Prob. 14P Ch. 25.7 - Prob. 15P Ch. 25.8 - Prob. 1P Ch. 25.8 - Prob. 3P Ch. 25.8 - Prob. 4P Ch. 25.8 - Prob. 6P Ch. 25.8 - Prob. 7P Ch. 25.8 - Prob. 8P Ch. 25.8 - Prob. 9P Ch. 25.8 - Prob. 10P Ch. 25.8 - Prob. 11P Ch. 25.8 - Prob. 12P Ch. 25.8 - Prob. 13P Ch. 25.8 - Prob. 14P Ch. 25.8 - Prob. 15P Ch. 25.9 - Prob. 1P Ch. 25.9 - Prob. 2P Ch. 25.9 - Prob. 3P Ch. 25.9 - Prob. 4P Ch. 25.9 - Prob. 5P Ch. 25.9 - Prob. 6P Ch. 25.9 - Prob. 7P Ch. 25.9 - Prob. 8P Ch. 25.9 - Prob. 9P Ch. 25.9 - Prob. 10P Ch. 25.9 - Prob. 12P Ch. 25.9 - Prob. 13P Ch. 25.9 - Prob. 14P Ch. 25.9 - Prob. 15P Ch. 25 - Prob. 1RQ Ch. 25 - Prob. 2RQ Ch. 25 - Prob. 3RQ Ch. 25 - Prob. 4RQ Ch. 25 - Prob. 5RQ Ch. 25 - Prob. 6RQ Ch. 25 - Prob. 7RQ Ch. 25 - Prob. 8RQ Ch. 25 - Prob. 9RQ Ch. 25 - Prob. 10RQ Ch. 25 - Prob. 11RQ Ch. 25 - Prob. 12RQ Ch. 25 - Prob. 13RQ Ch. 25 - Prob. 14RQ Ch. 25 - Prob. 15RQ Ch. 25 - Prob. 16RQ Ch. 25 - Prob. 17RQ Ch. 25 - Prob. 18RQ Ch. 25 - Prob. 19RQ Ch. 25 - Prob. 20RQ Ch. 25 - Prob. 21RQ Ch. 25 - Prob. 22RQ Ch. 25 - Prob. 23RQ Ch. 25 - Prob. 24RQ Ch. 25 - Prob. 25RQ

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