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

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- amistre64

so, what is your process?

- amistre64

are we to assume a normal distribution?

- anonymous

I'm not sure how to approach it exactly. I believe it would be something like P(15) = ?

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## More answers

- amistre64

more 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

it's a binomial distribution problem; we have \(p=0.8\) and we're dealing with estimates \(\hat p=\bar x/n\)

- anonymous

the sample size \(n=20\) is pretty small to use a normal approximation to the binomial distribution, but maybe they might allow it

- anonymous

actually, it's not using estimates, just \(X\sim B(n=20,p=0.8)\)

- anonymous

X∼B(n=20,p=0.8) is the formula I need to use?

- anonymous

I can see that n is the sample size and p = .8 is the probability, but not sure where to go from there

- amistre64

if we are using a binomial CDF
\[\sum_{k=a}^{b}\binom{20}{k}p^{k}q^{n-k}\]
seems fair to me

- amistre64

hmm, 20 = n ... had two competing thoughts in my head

- anonymous

so I just plug in the values and use that formula?

- amistre64

well yeah, if you want to sum up the discrete probabilities of the binomial cdf

- amistre64

16 to 20 is P(x>15)
14 to 18 is P(14 <= x <= 18)
etc..

- amistre64

\[P(x>15)=\sum_{k=16}^{20}\binom{20}{k}.8^{k}~.2^{20-k}\]
\[P(14\le x\le 18)=\sum_{k=14}^{28}\binom{20}{k}.8^{k}~.2^{20-k}\]

- amistre64

k=14 to 18 .. not to 28 :)

- anonymous

Thank you amistre. I will take a look and see if I can plug these in and figure it out

- anonymous

@amistre64 how do you plug this into the calculator?

- anonymous

@work12345 tediously :) but your calculator might have a binomcdf function in its PROB menu

- anonymous

@oldrin.bataku what numbers do I plugin for that?

- anonymous

http://tibasicdev.wikidot.com/binomcdf

- anonymous

thank you

- anonymous

@oldrin.bataku is there another way to write out the formula besides the way amistre wrote it?

- anonymous

binomcdf(n,p,x) computes \(P(X\le x)\), so \(P(X>15)=1-P(X\le 15)\) so 1 - binomcdf(20,0.8,15)

- anonymous

then \(P(14\le X\le 18)=P(X\le 18)-P(X\le 14)\) so binomcdf(20,0.8,18) - binomcdf(20,0.8,14)

- anonymous

thanks so much. Do you know how to answer c? I need to find the mean and standard deviation of x

- anonymous

well 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

the standard deviation follows from \(\sigma_{\hat p}=\sqrt{p(1-p)}\) so \(\sigma_X=\sigma_{n\hat p}=\sqrt{n}\sigma_{\hat p}=\sqrt{n}\sqrt{p(1-p)}=\sqrt{np(1-p)}\)

- anonymous

Thanks so much! I'm going to look it over to try and digest it

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