Question 35473: According to the bureau of Naval Standards. 75% of all people are inners (their belly buttons poke inward) Everyone else is an outer.
Let the random variable x be the number of outers among 7 randomly selected persons.
A.Find the mean and variance of x
B.Find the probability that exactly 2 of the 7 persons are outers.
Answer by narayaba(40)   (Show Source):
You can put this solution on YOUR website! probability of being a inner is 3/4 and probability of being an outer is 1/4
x is random variable represneting the number of outer among 7 randomly slected persons
P(X = x) = C(n,x)*p^x*q^(n-x), where C(n,x) = n!/(x!*(n-x)!)
p = 1/4 q =1 - p =3/4 and n=7
mean of x
1*p(x=1) + 2*p(x=2) + 3*p(x=3) + 4*p(x=4) + 5*p(x=5) + 6*p(x=6) + 7*p(x=7)
mean of x = 7*1/4= 7/4
variance of x is n*p(1-p) = 7*(1/4)*(3/4) = 21/16
P(x=2) = C(7,2)*( 1/4)^2*(3/4)^5
= 21*(1/16)*(243/1024)
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