Question

The Cincinnati Enquirer reported that, in the United States, 66% of adults and 87% of youths ages 12 to 17 use the Internet. Use the reported numbers as the population proportions and assume that samples of 300 adults and 300 youths will be used to learn about attitudes toward Internet security. What is the probability that the sample proportion of adults using the Internet will be within or - 0.04 of the population proportion

Answer 1

Answer:

85.58% probability that the sample proportion of adults using the Internet will be within + or - 0.04 of the population proportion

Step-by-step explanation:

We use the binomial approximation to the normal to solve this problem.

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:

$n = 300, p = 0.66$

So

$\mu = E(X) = 300*0.66 = 198$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.66*0.34} = 8.2049$

What is the probability that the sample proportion of adults using the Internet will be within + or - 0.04 of the population proportion

This is the pvalue of Z when X = (0.66+0.04)*300 = 210 subtracted by the pvalue of Z when X =  (0.66-0.04)*300 = 186. So

X = 210

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

$Z = \frac{210 - 198}{8.2049}$

$Z = 1.46$

$Z = 1.46$ has a pvalue of 0.9279

X = 186

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

$Z = \frac{186 - 198}{8.2049}$

$Z = -1.46$

$Z = -1.46$ has a pvalue of 0.0721

0.9279 - 0.0721 = 0.8558

85.58% probability that the sample proportion of adults using the Internet will be within + or - 0.04 of the population proportion