Question

epolled on this issue, A consumer advocate claims that 85 percent of cable television subscribers are not satisfied vwith their cable service. In n attempt to justify this claim, a randomly selected sample of cable subscribers will nthe sample are not satisfied formula to compute the probability that 6 or more subscribers h i end Do not round intormediat ealuations Rund Fnal anewor to in 2 decimal place Bound other final angere to d decimal pics 85 Binomial, n. 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 2 decimal place. Round other final answers to 4 decimal places.) 20.p | Binomial, n 85 The probability that 10 or fewer subscribers in the sample are not satisfied with their service. Probability The probability that more than 12 subscribers the sample are not satisfied with their service. Probability The probability that between 2 and 5 (inclusive) subscribers in the sample are not satisfied with their service. Probability The probability that exactly 15 subscribers in the sample not satisfied with their service. Probability 10 are actually not satisfied with their service. Using a probability vou found in this exercise as the basis for vour answer,. subscribers, we you believe the consumer ardwocate's claim? Explain, (Round vour answer to 4 decimal places.) (Click to select) if the claim is true, the probability that 10 or fewer (Click to select)satisfied is only

Answer 1

a)

Binomial probability is given by

P(X=x) = C(n,x)*px*(1-p)(n-x)

where  
Sample size , n =    7
Probability of an event of interest, p =   0.85

P(X≥6) = P(X=6)+ P(X=7)

P ( X =    6   ) = C(   7   ,   6   )*   0.85   ^   6   *   0.150   ^   1   =   0.3960

P ( X =    7   ) = C(   7   ,   7   )*   0.85   ^   7   *   0.150   ^   0   =   0.3206

P(X≥6) = P(X=6)+ P(X=7) = 0.3960 + 0.3206 = 0.7166(answer)

b)

1)

P(X≤10) = ΣC(n,x)*p^x*(1-p)^(n-x) where x goes from 0 to 10

so, P(10 or fewer) = 0.0002

2)

P(X>12) = 1 - P(X≤11) = 0.9941

3)

P(12≤x≤15) = P(X=12) + P(X=13) + P(X=14) + P(X=15) = 0.1688

4)

P ( X =    15   ) = C(   20   ,   15   )*   0.85   ^   15   *   0.150   ^   5   =   0.1028

c)

P(10 or fewer) = 0.0002

yes, if claim is true, the probability that 10 or fewer are satisfied is only 0.0002