Question

10. Contacts. Assume that 30% of students at a university

Answer 1

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

a) A normal model with mean 0.3 and standard deviation of 0.0458 should be used.

b) There is a 0.2327 = 23.27% probability that more than one third of this sample wear contacts.

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:

• 30% of students at a university wear contact lenses, hence p = 0.3.

• We randomly pick 100 students, hence n = 100.

Item a:

$np = 100(0.3) = 30 \geq 10$

$n(1 - p) = 100(0.7) = 70 \geq 10$

Hence a normal model is appropriated.

The mean and the standard deviation are given as follows:

$\mu = p = 0.3$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.3(0.7)}{100}} = 0.0458$

Item b:

The probability is 1 subtracted by the p-value of Z when X = 1/3 = 0.3333, hence:

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

By the Central Limit Theorem

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

$Z = \frac{0.3333 - 0.3}{0.0458}$

$Z = 0.73$

$Z = 0.73$ has a p-value of 0.7673.

1 - 0.7673 = 0.2327.

0.2327 = 23.27% probability that more than one third of this sample wear contacts.

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