In a sample of 900900 U.S. adults, 214214 dine out at a resaurant more than once
ID: 3358165 • Letter: I
Question
In a sample of
900900
U.S. adults,
214214
dine out at a resaurant more than once per week.
TwoTwo
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
nothing.
(RoundIn a sample of 900900 U.S. adults, 214214 dine out at a resaurant more than once per week. TwoTwo 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
nothing.
(Round to three decimal places as needed.)
to three decimal places as needed.)
Explanation / Answer
Solution:- given that sample = 900, x = 214
a) Probability one adult will dine out more than once a week is 214/900
Probability both selected will dine out more than once a week is (214/900)(213/899) = 0.056
2 method
a) Binomial distribution:-n= 2,p = 0.238
P(x = 2) = 2C2* 0.2382= 0.056
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