Question

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

Answer 1

so, what is your process?

Answer 2

are we to assume a normal distribution?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

if we are using a binomial CDF

\[\sum_{k=a}^{b}\binom{20}{k}p^{k}q^{n-k}\]
seems fair to me

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

16 to 20 is P(x>15)

14 to 18 is P(14 <= x <= 18)

etc..

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

@amistre64 how do you plug this into the calculator?

Answer 19

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

Answer 20

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

Answer 21

http://tibasicdev.wikidot.com/binomcdf

Answer 22

thank you

Answer 23

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

Answer 24

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)

Answer 25

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)

Answer 26

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

Answer 27

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 \(\mu=np=20(0.8)=16\)

Answer 28

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

Answer 29

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