In a sample of 800 U.S. adults, 171 dine out at a resaurant more than once per w
ID: 3306264 • Letter: I
Question
In a sample of 800 U.S. adults, 171 dine out at a resaurant more than once per week. Two U.S. adults are selected at random from the population of all U.S. adults without replacement. Assuming the sample is representative of all U.S. adults complete parts (a) through (d) (a) Find the probability that both adults dine out more than once per week The probability that both adults dine out more than once per week is Round to three decimal places as needed.) (b) Find the probability that neither adult dines out more than once per week The probability that neither adult dines out more than once per week is Round to three decimal places as needed.) (c) Find the probabiity that at least one of the two adults dines out more than once per week. The probability that at least one of the two adults dines out more than once per week is Round to three decimal places as needed.) (d) Which of the events can be considered unusual? Explain. Select all that apply. A. The event in part (c) is unusual because its probability is less than or equal to 0.05. B. c. D. None of these events are unusual. The event in part(a) is unusual because its probability is less than or equal to 0.05. The event in part (b) is unusual because its probability is less than or equal to 0.05.Explanation / Answer
(a) Let n be the sample size and p be the number of favourable cases i.e. the number of people who dine out more than once a week.
Then if a sample of size 2 is drawn without replacement then the required probability is given by: 171/800x170/799 = 0.0455.
(b) The required probability is given by: (800-171)/800x(799-171)/799 = 0.6180
(c) The required prabability is given by: 1-P(none of the adults dine out once a week) = 1-0.6180 = 0.382
(d) The event in part (a) is unusual because the probability that two randomly selected adults go out to dine in a restaurant is expected to be moderately high. (option C)
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