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.)
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) = nCx px (1-p)n-x
Therefore
p(x=0) + p(x=1) = 10C0 0.050 0.9510 +10C1 0.051 0.959
= 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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