Question

Suppose that 20% of the subscribers of a cable television company watch the shopping channel at least once a week. The cable company is trying to decide whether to replace this channel with a new local station. A survey of 100 subscribers will be undertaken. The cable company has decided to keep the shopping channel if the sample proportion is greater than 0.27. What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only 0.20? (Round your answer to four decimal places.)

Answer 1

Answer:

4.01% probability that the cable company will keep the shopping channel.

Step-by-step explanation:

For each subscriber, there are only two possible outcomes. Either they watch the shopping channel at least once a week, or they do not. This means that we can solve this problem using the binomial probability distribution.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 100, p = 0.20$

So

$E(X) = 100*0.20 = 20$

$\sqrt{V(X)} = \sqrt{100*0.20*0.80} = 4$

The cable company has decided to keep the shopping channel if the sample proportion is greater than 0.27. What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only 0.20?

This probability is 1 subtracted by the pvalue of Z when $X = 0.27*100 = 27$. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{27 - 20}{4}$

$Z = 1.75$

$Z = 1.75$ has a pvalue of 0.9599.

So there is a 1-0.9599 = 0.0401 = 4.01% probability that the cable company will keep the shopping channel.