Question

A consumer advocate claims that 70 percent of cable television subscribers are not satisfied with their cable service. In an attempt to justify this claim, a randomly selected sample of cable subscribers will be polled on this issue.

(a)

Suppose that the advocate's claim is true, and suppose that a random sample of 4 cable subscribers is selected. Assuming independence, use an appropriate formula to compute the probability that 3 or more subscribers in the sample are not satisfied with their service. (Do not round intermediate calculations.Round final answer to p in 2 decimal places. Round other final answers to 4 decimal places.)

Binomial, n = , p =

Probability =

(b)	Suppose that the advocate's claim is true, and suppose that a random sample of 20 cable subscribers is selected. Assuming independence, find: (Do not round intermediate calculations. Round final answer to p in 1 decimal place. Round other final answers to 4 decimal places.)

Binomial, n = , p =

1.	The probability that 12 or fewer subscribers in the sample are not satisfied with their service.

Probability

2.	The probability that more than 14 subscribers in the sample are not satisfied with their service.

Probability

3.	The probability that between 14 and 18 (inclusive) subscribers in the sample are not satisfied with their service.

Probability

4.	The probability that exactly 18 subscribers in the sample are not satisfied with their service.

Probability

(c)

Suppose that when we survey 20 randomly selected cable television subscribers, we find that 12 are actually not satisfied with their service. Using a probability you found in this exercise as the basis for your answer, do you believe the consumer advocate's claim? Explain. (Round your answer to 4 decimal places.)

(Click to select)YesNo ; if the claim is true, the probability that 12 or fewer (Click to select)are notare satisfied is   only .

Answer 1

a)Binomial, n =4 , p = 0.70

Probability : =P(X>=3) =P(X=3)+P(X=4) = $\binom{4}{3}(0.7)^{3}(0.3)^{1}+\binom{4}{4}(0.7)^{4}(0.3)^{0}$

=0.4116+0.2401 =0.6517

b)

Binomial, n = 20, p = 0.7

1)probability that 12 or fewer subscribers in the sample are not satisfied with their service =P(X<=12)

= $\sum_{x=0}^{12}\binom{20}{x}(0.7)^{x}(0.3)^{20-x}$ =0.2277

2) P(X>14) = $\sum_{x=15}^{20}\binom{20}{x}(0.7)^{x}(0.3)^{20-x}$ =0.4164

3)P(14<=X<=18)=0.6004

4)probability that exactly 18 subscribers in the sample are not satisfied with their service.=P(X=18)=0.0278

d)

Yes ; if the claim is true, the probability that 12 or fewer  are not  satisfied is 0.2277