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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

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  1. amistre64
    • one year ago
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    so, what is your process?

  2. amistre64
    • one year ago
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    are we to assume a normal distribution?

  3. anonymous
    • one year ago
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    I'm not sure how to approach it exactly. I believe it would be something like P(15) = ?

  4. amistre64
    • one year ago
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    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

  5. anonymous
    • one year ago
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    it's a binomial distribution problem; we have \(p=0.8\) and we're dealing with estimates \(\hat p=\bar x/n\)

  6. anonymous
    • one year ago
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    the sample size \(n=20\) is pretty small to use a normal approximation to the binomial distribution, but maybe they might allow it

  7. anonymous
    • one year ago
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    actually, it's not using estimates, just \(X\sim B(n=20,p=0.8)\)

  8. anonymous
    • one year ago
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    X∼B(n=20,p=0.8) is the formula I need to use?

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

  10. amistre64
    • one year ago
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    if we are using a binomial CDF \[\sum_{k=a}^{b}\binom{20}{k}p^{k}q^{n-k}\] seems fair to me

  11. amistre64
    • one year ago
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    hmm, 20 = n ... had two competing thoughts in my head

  12. anonymous
    • one year ago
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    so I just plug in the values and use that formula?

  13. amistre64
    • one year ago
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    well yeah, if you want to sum up the discrete probabilities of the binomial cdf

  14. amistre64
    • one year ago
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    16 to 20 is P(x>15) 14 to 18 is P(14 <= x <= 18) etc..

  15. amistre64
    • one year ago
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    \[P(x>15)=\sum_{k=16}^{20}\binom{20}{k}.8^{k}~.2^{20-k}\] \[P(14\le x\le 18)=\sum_{k=14}^{28}\binom{20}{k}.8^{k}~.2^{20-k}\]

  16. amistre64
    • one year ago
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    k=14 to 18 .. not to 28 :)

  17. anonymous
    • one year ago
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    Thank you amistre. I will take a look and see if I can plug these in and figure it out

  18. anonymous
    • one year ago
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    @amistre64 how do you plug this into the calculator?

  19. anonymous
    • one year ago
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    @work12345 tediously :) but your calculator might have a binomcdf function in its PROB menu

  20. anonymous
    • one year ago
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    @oldrin.bataku what numbers do I plugin for that?

  21. anonymous
    • one year ago
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    http://tibasicdev.wikidot.com/binomcdf

  22. anonymous
    • one year ago
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    thank you

  23. anonymous
    • one year ago
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    @oldrin.bataku is there another way to write out the formula besides the way amistre wrote it?

  24. anonymous
    • one year ago
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    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)

  25. anonymous
    • one year ago
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    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)

  26. anonymous
    • one year ago
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    thanks so much. Do you know how to answer c? I need to find the mean and standard deviation of x

  27. anonymous
    • one year ago
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    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 \(\mu=np=20(0.8)=16\)

  28. anonymous
    • one year ago
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    the standard deviation follows from \(\sigma_{\hat p}=\sqrt{p(1-p)}\) so \(\sigma_X=\sigma_{n\hat p}=\sqrt{n}\sigma_{\hat p}=\sqrt{n}\sqrt{p(1-p)}=\sqrt{np(1-p)}\)

  29. anonymous
    • one year ago
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    Thanks so much! I'm going to look it over to try and digest it

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