This Question: 4 pts In a sample of 900 U.S. adults, 219 dine out at a resaurant

Question

This Question: 4 pts In a sample of 900 U.S. adults, 219 dine out at a resaurant more than once per week. Two U.S adults are solected at random from the population of ai U S, aduis without replscoment. A 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 adul dines out more than once per weok is Round to three decimal places as needed.) (c) Find the probability that at least one of the two adults dines out more than once per week The probablity 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. Soloct all hat apply O A. The event in part (a) is unusual because its probability is less than or equal to o.05. Click to select your answer(s)

Explanation / Answer

Solution:- Given data n = 900 , X = 219

Probability one adult will dine out more than once a week is 219/900

a) Probability both selected will dine out more than once a week is
= (219/900)*(218/899) = 0.059

b) Probability neither dines out is = (681/900)*(680/899) = 0.572

c) Probability at least one should be the complement of = 1 - 0.572 = 0.428

dd) The probability that both adutls dine out more than once per week is usual because the probability 0.0 which is greater than to 0.059

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