anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
so, what is your process?
amistre64
  • amistre64
are we to assume a normal distribution?
anonymous
  • anonymous
I'm not sure how to approach it exactly. I believe it would be something like P(15) = ?

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amistre64
  • 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
  • anonymous
it's a binomial distribution problem; we have \(p=0.8\) and we're dealing with estimates \(\hat p=\bar x/n\)
anonymous
  • 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
  • anonymous
actually, it's not using estimates, just \(X\sim B(n=20,p=0.8)\)
anonymous
  • anonymous
X∼B(n=20,p=0.8) is the formula I need to use?
anonymous
  • 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
  • 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
  • amistre64
hmm, 20 = n ... had two competing thoughts in my head
anonymous
  • anonymous
so I just plug in the values and use that formula?
amistre64
  • amistre64
well yeah, if you want to sum up the discrete probabilities of the binomial cdf
amistre64
  • amistre64
16 to 20 is P(x>15) 14 to 18 is P(14 <= x <= 18) etc..
amistre64
  • 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
  • amistre64
k=14 to 18 .. not to 28 :)
anonymous
  • anonymous
Thank you amistre. I will take a look and see if I can plug these in and figure it out
anonymous
  • anonymous
@amistre64 how do you plug this into the calculator?
anonymous
  • anonymous
@work12345 tediously :) but your calculator might have a binomcdf function in its PROB menu
anonymous
  • anonymous
@oldrin.bataku what numbers do I plugin for that?
anonymous
  • anonymous
http://tibasicdev.wikidot.com/binomcdf
anonymous
  • anonymous
thank you
anonymous
  • anonymous
@oldrin.bataku is there another way to write out the formula besides the way amistre wrote it?
anonymous
  • 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
  • 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
  • anonymous
thanks so much. Do you know how to answer c? I need to find the mean and standard deviation of x
anonymous
  • 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
  • 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
  • anonymous
Thanks so much! I'm going to look it over to try and digest it

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