SOLUTION: 1. Twenty-three percent of Ole Miss students are from out of state. In a random sample of 6 students, what is the probability that at least 2 of them are from out of state?(round
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Question 1087753: 1. Twenty-three percent of Ole Miss students are from out of state. In a random sample of 6 students, what is the probability that at least 2 of them are from out of state?(round to whole number)
a. 42% b. 7% c. 28% d. 64% e. none of these
2. For the problem above what is the expected value of x.
a. 4.2 b. 2.7 c. 1.4 d. .5 e. none of these
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
1. Twenty-three percent of Ole Miss students are from out of state. In a random sample of 6 students, what is the probability that at least 2 of them are from out of state?(round to whole number)
a. 42% b. 7% c. 28% d. 64% e. none of these
2. For the problem above what is the expected value of x.
a. 4.2 b. 2.7 c. 1.4 d. .5 e. none of these
Solution:
Since the probability is known and constant (large population), the binomial distribution can be used to model the situation.
p=0.23
n=6
(1) At least two student from out-of-state.
P(x)=C(n,x)p^(x)*(1-p)^(n-x)
=1-C(6,0)(0.23)^0*(0.77)^6-C(6,1)(0.23)^1*(0.77)^5
=1-0.2084-0.3735
=0.4180
(2) Expected value of x
Expected value of x (in a binomial distribution)
=np
=6*0.23
=1.38
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