Question

The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of ten syringes taken from the batch. Suppose the batch contains 5% defective syringes. Find μ (in terms of the number of syringes). (Enter a number. Enter your answer to two decimal places.) What is the expected number of defective syringes the inspector will find? (Enter a number. Enter your answer to two decimal places.) What is the probability that the batch will be accepted? (Enter a number. Round your answer to three decimal places.) Find σ (in terms of the number of syringes). (Enter a number. Round your answer to three decimal places.)

Answer #1

This is binomial distribution with n = 10 , p = 0.05

μ = np = 10 * 0.05 = 0.5

Expected number of defective defective syringes = 10 * 0.05 = 0.5

P( batch will be accepted) = p( x < 2 ) = p( x<=1)

= p( x = 0) + p( x = 1)

For binomial distribution,

p(x) = ^{n}C_{x} p^{x}
(1-p)^{n-x}

Therefore

p(x=0) + p(x=1) = ^{10}C_{0} 0.05^{0}
0.95^{10} +^{10}C_{1} 0.05^{1}
0.95^{9}

= 0.914

Therefore probability that batch will be accepted is 0.914

Standard deviation = Sqrt(np(1-p))

= sqrt(10*0.05*0.95)

= 0.689

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