With one method of a procedure called acceptance sampling, a sample of items is

Question

With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay. Among 810 airport baggage scales, 102 are defective (based on data from the New York City Department of Consumer Affairs). If four of the scales are randomly selected and tested, what is the probability that the entire batch will be accepted? Is this scheme likely to detect the large number of With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay. Among 810 airport baggage scales, 102 are defective (based on data from the New York City Department of Consumer Affairs). If four of the scales are randomly selected and tested, what is the probability that the entire batch will be accepted? Is this scheme likely to detect the large number of With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay. Among 810 airport baggage scales, 102 are defective (based on data from the New York City Department of Consumer Affairs). If four of the scales are randomly selected and tested, what is the probability that the entire batch will be accepted? Is this scheme likely to detect the large number of

Explanation / Answer

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
p = 102/810 = 0.12593
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 4 * 0.12593
= 0.5037
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 4 * 0.12593 * 0.87407
= 0.4403
III.
standard deviation = sqrt( variance ) = sqrt(0.4403)
=0.6635
P( X = 0 ) = ( 4 0 ) * ( 0.12593^0) * ( 1 - 0.12593 )^4
= 0.5837

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