Question

According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answer the following questions.
a. Suppose a random sample of 150 couples is asked, "Did you meet online>" Describe the sampling distribution of p, the proportion of couples that met online.
b. What is the probability that in a random sample of 150 couples more than 25% met online?
c. What is the probability that in a random sample of 150 couples between 15% and 20% met online?

Answer 1

Using the normal distribution and the central limit theorem, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

• By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$, as long as $np \geq 10$ and $n(1 - p) \geq 10$.

In this problem:

• 22% of couples meet online, hence p = 0.22.

• A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

$\mu = p = 0.22$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338$

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is one subtracted by the p-value of Z when X = 0.25, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{0.25 - 0.22}{0.0338}$

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15, hence:

X = 0.2:

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

$Z = \frac{0.2 - 0.22}{0.0338}$

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

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

$Z = \frac{0.15 - 0.22}{0.0338}$

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the normal distribution and the central limit theorem, you can check brainly.com/question/24663213