Question

What is the probability that the proportion of people who will order coffee with their meal is between 2% of the population mean? If in a restaurant, the proportion of people who order coffee with their dinner is .9. A simple random sample of 144 patrons of the restaurant is taken.

Answer 1

Answer:

57.62% probability that the proportion of people who will order coffee with their meal is within 2% of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, we have that $\mu = p, s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

$p = 0.9, n = 144$

So

$\mu = 0.9, s = \sqrt{\frac{0.9*0.1}{144}} = 0.025$

What is the probability that the proportion of people who will order coffee with their meal is within 2% of the population mean?

This is the pvalue of Z when X = 0.9 + 0.02 = 0.92 subtracted by the pvalue of Z when X = 0.9 - 0.02 = 0.88. So

X = 0.92

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

By the Central Limit Theorem

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

$Z = \frac{0.92 - 0.9}{0.025}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881

X = 0.88

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

$Z = \frac{0.88 - 0.9}{0.025}$

$Z = -0.8$

$Z = -0.8$ has a pvalue of 0.2119

0.7881 - 0.2119 = 0.5762

57.62% probability that the proportion of people who will order coffee with their meal is within 2% of the population mean.