Question

In a recent year, about 22% of Americans 18 years and older are single. What is the probability that in a random sample of 200 Americans 18 or older more than 30 are single? (Hint: use binomial to normal approximation techniques) Round your z-score to 2 decimal places before referencing table E. Then provide answer out to 4 decimal places.

Answer 1

Using the normal approximation to the binomial, it is found that there is a 0.0107 = 1.07% probability that more than 30 are single.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$.

In this problem, the proportion and the sample size are, respectively, p = 0.22 and n = 200, hence:

$\mu = np = 200(0.22) = 44$

$\sigma = \sqrt{np(1 - p)} = \sqrt{200(0.22)(0.78)} = 5.8583$

The probability that more than 30 are single, using continuity correction, is P(X > 30.5), which is 1 subtracted by the p-value of Z when X = 30.5, hence:

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

$Z = \frac{30.5 - 44}{5.8583}$

Z = -2.3

Z = -2.3 has a p-value of 0.0107.

0.0107 = 1.07% probability that more than 30 are single.

More can be learned about the normal distribution at brainly.com/question/24663213