In a sample of 800 U.S. adults, 202 dine out at a resaurant more than once per w
ID: 3294867 • Letter: I
Question
In a sample of 800 U.S. adults, 202 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 probability 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. None of these events are unusualNone of these events are unusual.
B. The event in part (c) is unusual because its probability is less than or equal to 0.05.
C. The event in part (b) is unusual because its probability is less than or equal to 0.05.
D. The event in part (a) is unusual because its probability is less than or equal to 0.05
Explanation / Answer
a) P( Both adult dine out ) = (202/800)*(201/799) = 0.064
B) P( neither dine out ) = (598/800)*(597/799) = 0.559
C) P(at least one dine out ) = 1-0.559 = 0.441
D) Option A is correct because no one event has probability less than 0.05
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