The quality-control inspector of a production plant will reject a batch of syrin

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 (a) Make a histogram showing the probabilities of r = 0, 1, 2, 3, , 9 and 10 defective syringes in a random sample of ten syringes 0.6 0.5 0.3 04 Pi) 0.2 P() 0.3 0.2 0.1 0.1 0 123 45 678 9 10 0 1 23 45678 9 10 0.3 0.35 0.3 0.25 0.2 0.15 0.1 0.25 0.2 Pir) 0.15 Pir) 0.1 0.05 0.05 0 1 2 3 4567 8 9 10 01 2 3 4 567 8 9 10

Explanation / Answer

b) this is a binomial distribution and therefore E(X) = n*p = 10 * 5% = 0.05 =

Also, he will find 0.05 defective syringes.

c) P(X<2) = 1 - P(X=0) - P(X=1)
P(X=0) = 10C0 *(0.05)^0 * (0.95)^10 = 0.95^10 = 0.599
P(X=1) = 10C1 *(0.05)^1 * (0.95)^9 = 10 * 0.05 * (0.95)^9 = 0.315
1 - P(X=0) - P(X=1) = 1 - 0.599 - 0.315 = 0.086

d) I think that's supposed to be a sigma and therefore the standard deviation: for binomial distribution variance = n*p*(1-p) = 10 * 0.05 * 0.95 = 0.475 Therefore the standard deviation is sqrt(0.475) = 0.689

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