In a sample of 1000 U.S. adults, 188 dine out at a resaurant more thar parts (a)
ID: 3074261 • Letter: I
Question
In a sample of 1000 U.S. adults, 188 dine out at a resaurant more thar parts (a) through (d). (Round to three decimal places as needed.) (b) Find the probability that neither (Round to three decimal places as needed.) Round to three decimal places as needed.) (d) Which of the events can be considered unusual? Ex A. The event in part (b) is unusual because its probability is less than or equal to 0.05. B. The event in part (a) is unusual because its probability is less than or equal to 0.05. D. None of these events are unusual (5th)Explanation / Answer
Total 1000 US adults. 188 dine out and the remaining 1000 - 188 = 812, do not dine out.
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(a) Probability that both dine out = (188/1000) * (187/999) = 0.035
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(b) Probability that neither dine out = (812/1000) * (811/999) = 0.659
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(c) At least 1 of them Dines out = P(1st dines out and second does not) + P(1st does not dine out, but the second does) + P(That both dine out)
= [(188/1000) * (812/999)] + [(812/1000) * (188/999)] + [(188/1000) * (187/999)]
= (152656 + 152656 + 35156)/(1000*999) = 0.341
(We could do this in another way. The sum of probabilities is = 1. In this case therefore P(Both Dine out) + P(1st dines out and second does not) + P(1st does not dine out, but the second does) + P(That both dont dine out) = 1
From (b) P(Both dont dine out) = 0.659
Therefore the P( at least 1 dines out) = P(1st dines out and second does not) + P(1st does not dine out, but the second does) + P(That both dine out) = 1 - P(Both dont dine out) = 1 - 0.659 = 0.341
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(d) Typically a situation or an event having a probability of less than 5% or 0.05 chance of occurring is unusual. We see that the probability in (a) is 0.035 and hence Option B.
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