In a sample of 1100 US adults, 218 dine out at a restaurant more than once per w
ID: 3270540 • Letter: I
Question
In a sample of 1100 US adults, 218 dine out at a restaurant more than once per week. Two US adults are selected at random from the population of all US adults without replacement. Assuming the sample is (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. The event in part (c) is unusual because its probability is less than or equal to 0.05. B. The event in part (b) is unusual because its probability is less than or equal to 0.05. C. None of these events are unusual. D. The event in part (a) is unusual because its probability is less than or equal to 0.05.Explanation / Answer
Solution:-
X = 218, N = 1100
p = 0.218
a) The probability that both adults dine out more than once per week is 0.04752.
p = 0.218, n = 2
By applying binomial distribution:-
P(x, n, p) = nCx*p x *(1 - p)(n - x)
P(x = 2) = 0.04752
b) The probability that neither adults dine out more than once per week is 0.6115.
p = 0.218, n = 2
By applying binomial distribution:-
P(x, n, p) = nCx*p x *(1 - p)(n - x)
P(x = 0) = 0.6115
c) The probability that atleast one adults dine out more than once per week is 0.3885.
p = 0.218, n = 2
By applying binomial distribution:-
P(x, n, p) = nCx*p x *(1 - p)(n - x)
P(x > 1) = P(x = 1) + P(x = 2)
P(x > 1) = 0.3885
d) The event in part (a) is unusual beacuse it's probability is less than or equal to 0.05.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.