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 thất have 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) = 0.5314 Interpret this probability. 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. This is the probability that at least 3 out of 6 selected households have cable TV. 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). P(xS 5) = 0.46856

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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.)
%3D
(a) Calculate p(3) = P(x = 3).
%3D
p(3) = 0.5314
%3D
Interpret this probability.
This is the probability that at least 3 out of 10 selected households have cable TV.
This is the probability that exactly 3 out of 6 selected households have cable TV.
This is the probability that at least 3 out of 6 selected households have cable TV.
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 < 5).
P(x < 5) = 0.46856
Transcribed Image Text: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.) %3D (a) Calculate p(3) = P(x = 3). %3D p(3) = 0.5314 %3D Interpret this probability. This is the probability that at least 3 out of 10 selected households have cable TV. This is the probability that exactly 3 out of 6 selected households have cable TV. This is the probability that at least 3 out of 6 selected households have cable TV. 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 < 5). P(x < 5) = 0.46856
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