Question

A coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standard deviation of 0.2 ounces. If a sample of 9 cups is selected, find the probability that the mean of the sample will be less than 12.1 ounces.

Answer 1

Answer:

0.9332 = 93.32% probability that the mean of the sample will be less than 12.1 ounces.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, 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.

Mean of 12 ounces and a standard deviation of 0.2 ounces.

This means that $\mu = 12, \sigma = 0.2$

Sample of 9:

This means that $n = 9, s = \frac{0.2}{\sqrt{9}}$

Find the probability that the mean of the sample will be less than 12.1 ounces.

This is the pvalue of Z when X = 12.1. So

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

By the Central Limit Theorem

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

$Z = \frac{12.1 - 12}{\frac{0.2}{\sqrt{9}}}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9332 = 93.32% probability that the mean of the sample will be less than 12.1 ounces.