Question

A consumer advocate claims that 65 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 8 cable subscribers is selected. Assuming independence, use an appropriate formula to compute the probability that 7 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 place. Round other final answers to 4 decimal places.)

(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 2 decimal place. Round other final answers to 4 decimal places.)

1. The probability that 14 or fewer subscribers in the sample are not satisfied with their service.

2. The probability that more than 15 subscribers in the sample are not satisfied with their service.

3. The probability that between 15 and 18 (inclusive) subscribers in the sample are not satisfied with their service.

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

(c) Suppose that when we survey 20 randomly selected cable television subscribers, we find that 14 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.)

Answer 1

a) n = 8, p = 0.65

Probability of 7 or more, P(X ≥ 7) =    1 - P(X ≤ 6)

Using excel binomial function:     

= 1 - BINOM.DIST(6, 8, 0.65, 1)     

= 0.1691  

b) n = 20, p = 0.65

1)

Probability of 14 or fewer, P(X ≤ 14) =

Using excel binomial function:

= BINOM.DIST(14, 20, 0.65, 1)

= 0.7546

2)

Probability of more than 15, P(X > 15) =    1 - P(X ≤ 15)

Using excel binomial function:     

= 1 - BINOM.DIST(15, 20, 0.65, 1)     

= 0.1182  

3)

Probability between 15 and 18, P(15 ≤ X ≤ 18) = P(X ≤ 18) - P(X < 15) =

= P(X ≤ 18) - P(X ≤ 14)

= BINOM.DIST(18, 20, 0.65, 1) - BINOM.DIST(14, 20, 0.65, 1)

= 0.2433

4)

Probability of exactly 18, P(X = 18) =

Using excel binomial function:

= BINOM.DIST(18, 20, 0.65, 0)

= 0.0100

c) Yes; if the claim is true, the probability that 14 or fewer are satisfied is only 0.7546.