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anonymous
 one year ago
A consumer advocate claims that 80 percent of cable television subscribers are not satisfied with their cable service, and suppose that a random sample of 20 cable subscribers is selected. Let x be the number of cable television subscibers who are not satisfied with their cable service. Find the probability that
a. More than 15 subscribers are not satisfied with their service
b. 14 to 18 subscribers (inclusive) are not satisfied with their service
c. Find the mean and standard deviation of x
anonymous
 one year ago
A consumer advocate claims that 80 percent of cable television subscribers are not satisfied with their cable service, and suppose that a random sample of 20 cable subscribers is selected. Let x be the number of cable television subscibers who are not satisfied with their cable service. Find the probability that a. More than 15 subscribers are not satisfied with their service b. 14 to 18 subscribers (inclusive) are not satisfied with their service c. Find the mean and standard deviation of x

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.2so, what is your process?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2are we to assume a normal distribution?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure how to approach it exactly. I believe it would be something like P(15) = ?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2more than 15, so P(x>15) = ___ if we can assume a normal distribution, we can use the basic z score formula with a slight modification

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's a binomial distribution problem; we have \(p=0.8\) and we're dealing with estimates \(\hat p=\bar x/n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the sample size \(n=20\) is pretty small to use a normal approximation to the binomial distribution, but maybe they might allow it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually, it's not using estimates, just \(X\sim B(n=20,p=0.8)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0X∼B(n=20,p=0.8) is the formula I need to use?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can see that n is the sample size and p = .8 is the probability, but not sure where to go from there

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2if we are using a binomial CDF \[\sum_{k=a}^{b}\binom{20}{k}p^{k}q^{nk}\] seems fair to me

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2hmm, 20 = n ... had two competing thoughts in my head

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so I just plug in the values and use that formula?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2well yeah, if you want to sum up the discrete probabilities of the binomial cdf

amistre64
 one year ago
Best ResponseYou've already chosen the best response.216 to 20 is P(x>15) 14 to 18 is P(14 <= x <= 18) etc..

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2\[P(x>15)=\sum_{k=16}^{20}\binom{20}{k}.8^{k}~.2^{20k}\] \[P(14\le x\le 18)=\sum_{k=14}^{28}\binom{20}{k}.8^{k}~.2^{20k}\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2k=14 to 18 .. not to 28 :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you amistre. I will take a look and see if I can plug these in and figure it out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@amistre64 how do you plug this into the calculator?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@work12345 tediously :) but your calculator might have a binomcdf function in its PROB menu

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku what numbers do I plugin for that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku is there another way to write out the formula besides the way amistre wrote it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0binomcdf(n,p,x) computes \(P(X\le x)\), so \(P(X>15)=1P(X\le 15)\) so 1  binomcdf(20,0.8,15)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then \(P(14\le X\le 18)=P(X\le 18)P(X\le 14)\) so binomcdf(20,0.8,18)  binomcdf(20,0.8,14)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks so much. Do you know how to answer c? I need to find the mean and standard deviation of x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well the mean of the proportion estimate \(\hat p=X/n\) is just the true population proportion \(p=0.8\), so the mean of \(X=n\hat p\) is just \(\mu=np=20(0.8)=16\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the standard deviation follows from \(\sigma_{\hat p}=\sqrt{p(1p)}\) so \(\sigma_X=\sigma_{n\hat p}=\sqrt{n}\sigma_{\hat p}=\sqrt{n}\sqrt{p(1p)}=\sqrt{np(1p)}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much! I'm going to look it over to try and digest it
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