Respuesta :
Answer:
There is approximately 4% chance that the cable company will keep the shopping channel, even though the true proportion who watch it is only 0.20.
Step-by-step explanation:
The random variable X can be defined as the number of subscribers of a cable television company who watch the shopping channel at least once a week.
X follows a Binomial distribution with parameters n = 100 and p = 0.20.
But the sample selected is too large and the probability of success is close to 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:
1. np ≥ 10
2. n(1 - p) ≥ 10
Check the conditions as follows:
 [tex]np=100\times 0.20=20>10\\\\n(1-p)=100\times 0.80=80>10[/tex]
Thus, a Normal approximation to binomial can be applied.
[tex]\hat p\sim N(p,\frac{p(1-p)}{n})[/tex]
Compute the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only 0.20 as follows:
[tex]P(\hat p>0.27)=P(\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}>\frac{0.27-0.20}{\sqrt{\frac{0.20(1-0.20}{100}}})[/tex]
          [tex]=P(Z>1.75)\\=1-P(Z<1.75)\\=1-0.95994\\=0.04006\\\approx 0.04[/tex]
Thus, there is approximately 4% chance that the cable company will keep the shopping channel, even though the true proportion who watch it is only 0.20.
