Question

A television station plans to ask a random sample of 400 city residents if they can name the news anchor on the evening news at their station. They plan to fire the news anchor if fewer than 10% of the residents in the sample can do so. Suppose that in fact 12% of city residents could name the anchor if asked. What is the approximate probability that the anchor will be fired?

Answer 1

Answer:

There is an 8.38% probability that the anchor will be fired.

Step-by-step explanation:

For each resident, there is only two possible outcomes. Either they can name the news anchor on the evening news at their station, or they cannot.

This means that the binomial probability distribution will be used in our solution.

However, we are working with samples that are considerably big. So i am going to aaproximate 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:

400 city residents are going to be asked. So $n = 400$.

Suppose that in fact 12% of city residents could name the anchor if asked. This means that $p = 0.12$.

So,

$\mu = E(X) = 400*0.12 = 48$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{400*0.12*0.88} = 6.5$.

They plan to fire the news anchor if fewer than 10% of the residents in the sample can do so.

What is the approximate probability that the anchor will be fired?

10% of the residents is 0.10*400 = 40 residents.

Fewer than 10% is 39 residents. So the probability that the anchor will be fired is the pvalue of Z when $X = 39$.

So:

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

$Z = \frac{39 - 48}{6.5}$

$Z = -1.38$

$Z = -1.38$ has a pvalue of 0.0838. This means that there is an 8.38% probability that the anchor will be fired.