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Question 7 (1 mark) Attempt 1 Each hour an operator inspects 120 pencils and determines if they are acceptable. Historically the proportion of unacceptable pencils has been 0.016 Find the probability that there are more than (greater than) 4 unacceptable pencils Use the Normal approximation to the Binomial distribution Your answer can be rounded to three decimal digit accuracy when entered Probability isSkipped

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Answer #1

= 120(0016) 1 .92 vnpg =:/120 (0.016)(1.-0.016) 1.375 = 1-7 T X-1.92 4-1.92 1.3745 1.3745 . -1-P(Zs1.51) = 1-0.9345 = 0.066 Thus the probability that more than 4 unacceptable pencils is 0.066.

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