Question

Suppose an article reported that 16% of unmarried couples in the United States are mixed racially or ethnically. Consider the population consisting of all unmarried couples in the United States.

When n = 100, what is the probability that the proportion of couples in the sample who are racially or ethnically mixed will be greater than 0.17? (Round the answer to four decimal places. Type NONE if it is not appropriate to use the normal distribution.)

Answer 1

Answer:

39.36% probability that the proportion of couples in the sample who are racially or ethnically mixed will be greater than 0.17.

Step-by-step explanation:

For each couple, there are only two possible outcomes. Either they are mixed racially or ethnically, or they are not. This means that the binomial probability distribution is used to solve this problem.

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:

$n = 100, p = 0.16$

So

$E(X) = 100*0.16 = 16$

$\sqrt{Var(X)} = \sqrt{100*0.16*0.84} = 3.67$

When n = 100, what is the probability that the proportion of couples in the sample who are racially or ethnically mixed will be greater than 0.17?

This is 1 subtracted by the pvalue of Z when $X = 100*0.17 = 17$. So:

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

$Z = \frac{17 - 16}{3.67}$

$Z = 0.27$

$Z = 0.27$ has a pvalue of 0.6064.

So there is a 1-0.6064 = 0.3936 = 39.36% probability that the proportion of couples in the sample who are racially or ethnically mixed will be greater than 0.17.