On a Friday night, 26% of the students at a university go out for pizza, 10% see
ID: 3131301 • Letter: O
Question
On a Friday night, 26% of the students at a university go out for pizza, 10% see a movie, 26% go to a party, and 38% stay home to study. 24 students are randomly selected. Can the probability that exactly 7 out of these students stay home be computed using a binomial model?
a)No, because there are more than two possible outcomes
b)Yes, because this can be restated as a binomial model with 38% of the people staying home and the other 62% not
c)Yes, because the expected number of people who stay home is greater than 7
d)No, because less than 50% of the people stay home
Airlines, in some cases, overbook flights because past experience has shown that some passengers fail to show up. Let x be the discrete random variable that represent the number of passengers who can't board the plane because there are more passengers than seats.
What is the probability that atmost 2 individuals will not be able to board the plane?
a)0.052
b)0.978
c)0.022
d)0.926
What of the following pairs of probability values would complete the following probability distribution?
-3
-2
-1
0
1
2
0.15
0.24
p1
p2
0.12
0.24
p1 = 0.14, p2 = 0.39
p1 = –0.1, p2 = 0.35
p1 = 0.25, p2= 0.15
p1= 0.1, p2 = 0.15
X 0 1 2 3 4 P(X) 0.802 0.124 0.052 0.009 0.013Explanation / Answer
Can the probability that exactly 7 out of these students stay home be computed using a binomial model?
b)Yes, because this can be restated as a binomial model with 38% of the people staying home and the other 62% not
What is the probability that atmost 2 individuals will not be able to board the plane?
P( x <= 2 ) = 0.802 + 0.124 + 0.052 = 0.978
What of the following pairs of probability values would complete the following probability distribution?
0.15 + 0.24 + p1 + p2 + 0.12 + 0.24 = 1
p1 + p2 = 1 - 0.75 = 0.25
p1 = 0.1
p2 = 0.15
0.1 + 0.15 = 0.25
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