Question 1125355: How would I compute this problem by hand; what is the formula?
A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 17 randomly selected pens yields no more than two defective pens.
(a) Find the probability that this shipment is accepted if 5% of the total shipment is defective. (Use 3 decimal
places.)
(b) Find the probability that this shipment is not accepted if 15% of the total shipment is defective. (Use 3 decimal
places.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! this looks like a binomial probability type of problem.
the sample size is 17.
the shipment is accepted if the number of defective pens is less than or equal to 2.
otherwise, it is not accepted.
if 5% of the shipment is defective, then the probability of randomly picking 1 defective pen out of the shipment is .05.
the binomial probability formula is p(x) = p^x * q^(n-x) * c(n,x)
p(x) is the probability of getting x defective.
p is the probability that a randomly selected pen will be defective.
q is equal to 1 - p which is the probability that a randomly selected pen is not defective.
c(n,x) is the number of ways to select a set of x elements out of a set of n elements where order is not important.
the formula for c(n,x) is c(n,x) = n! / (x! * (n-x)!).
you are looking for the probability that less than or equal to 2 are defective when p = .05.
when p = .05 and the sample size is 17, then:
p = .05, q = .95, n = 17, x = 0 to 17, c(n,x) = c(17,x).
the formula is, once again p(x) = p^x * q^(n-x) * c(n,x).
less than or equal to 2 defective is 0, 1, or 2 defective only.
p(0) = .05^0 * .95^17 * c(17,0) = 1 * .4181203352 * 1 = .4181203353
p(1) = .05^1 * .95^16 * c(17,1) = .05 * .4401266687 * 17 = .3741076684
p(2) = .05^2 * .95^15 * c(17,2) = .0025 * .4632912302 * 136 = .1575190183
the sum of p(0) + p(1) + p(3) = .9497470218 which rounds to .95.
when p = .15 and the sample size is 17, then:
p = .15, q = .85, n = 17, x = 0 to 17, c(n,x) = c(17,x).
the formula is, once again p(x) = p^x * q^(n-x) * c(n,x).
less than or equal to 2 defective is 0, 1, or 2 defective only.
p(0) = .15^0 * .85^17 * c(17,0) = 1 * .0631134233 * 1 = .0631134233
p(1) = .15^1 * .85^16 * c(17,1) = .15 * .0742510862 * 17 = .1893402699
p(2) = .15^2 * .85^15 * c(17,2) = .0225 * .0873542191 * 136 = .2673039104
the sum of p(0) + p(1) + p(3) = .5197576037 which rounds to .52.
that's the probability that less than or equal to 2 are defective.
the probability that more than 2 are defective is therefore 1 - .52 = .48.
your solution should be.
the probability that the shipment is accepted if 5% are defective would be .95.
the probability that the shipment is not accepted if 15% are defective would be .48.
that's what i get if i analyzed the problem correctly.
i used excel to calculate all the probabilities.
the sum of all probabilities is equal to 1 as it should be.
here's the table for 5% defective.
here's the table for 15% defective.
if you need to show your answer to 3 decimal places, then .950 and .480.
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