Mathematics, 27.03.2020 03:25, aurelio1121

Suppose that 50% 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 .58. What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only .50

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answered: hannahelisabeth19

5.48% probability that the cable company will keep the shopping channel

Step-by-step explanation:

I will use the binomial approximation to the normal to solve this question.

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 = 0.5

$\mu = E(X) = np = 100*0.5 = 50$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.5*0.5} = 5$

What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only .50

Only if the sample proportion is greater than 0.58, which means that this probability is 1 subtracted by the pvalue of Z when X = 0.58*100 = 58.

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

$Z = \frac{58 - 50}{5}$

Z = 1.6

Z = 1.6 has a pvalue of 0.9452

1 - 0.9452 = 0.0548

5.48% probability that the cable company will keep the shopping channel

answered: Guest

it’s the first one your welcome

answered: Guest

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

⇒ option b is correct.

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Suppose that 50% of the subscribers of a cable television company watch the shopping channel at leas...