(The decimal spaces mentioned are super important) Suppose that in a certain met
ID: 3351469 • Letter: #
Question
(The decimal spaces mentioned are super important) Suppose that in a certain metropolitan area, nine out of 10 households have cable TV. Let x denote the number among six randomly selected households that have cable TV, so x is a binomial random variable with n = 6 and p = 0.9. (Round your answers to four decimal places.) (a) Calculate p(3)-P(x-3). p(3)- Interpret this probability. O This is the probability that at least 3 out of 6 selected households have cable TV. O This is the probability that at least 3 out of 10 selected households have cable TV. O This is the probability that exactly 3 out of 6 selected households have cable TV O This is the probability that exactly 3 out of 10 selected households have cable TV. (b) Calculate p(6), the probability that all six selected households have cable TV p(6)- (c) Determine P(x s 5)
Explanation / Answer
a)
n=6
p=0.9
q=1-p
=1-0.9
=0.1
P=nCr p^r q^(n-r)
=6C3 (0.9)^3 (0.1)^(6-3)
=20*0.729*0.001
=0.0146
This is the probability that exactly 3 out og 6 selected households have cable tv
b) P=nCr p^r q^(n-r)
= 6C6 (0.9)^6 (0.1)^(6-6)
=1*0.531441*1
=0.5314
c) P(x5) = 1-P(x>5)
= 1-P(x=6)
=1-0.5314
=0.4686
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