Question

An HP laser printer is advertised to print text documents at a speed of 18 ppm (pages per minute). The manufacturer tells you that the printing speed is actually a Normal random variable with a mean of 17.42 ppm and a standard deviation of 3.25 ppm. Suppose that you draw a random sample of 12 printers. Part i) Using the information about the distribution of the printing speeds given by the manufacturer, find the probability that the mean printing speed of the sample is greater than 18.12 ppm. (Please carry answers to at least six decimal places in intermediate steps. Give your final answer to the nearest three decimal places).

Answer 1

Answer:

0.227 = 22.7% probability that the mean printing speed of the sample is greater than 18.12 ppm.

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 17.42 ppm and a standard deviation of 3.25 ppm.

This means that $\mu = 17.42, \sigma = 3.25$

Sample of 12:

This means that $n = 12, s = \frac{3.25}{\sqrt{12}}$

Find the probability that the mean printing speed of the sample is greater than 18.12 ppm.

This is 1 subtracted by the p-value of Z when X = 18.12.

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

By the Central Limit Theorem

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

$Z = \frac{18.12 - 17.42}{\frac{3.25}{\sqrt{12}}}$

$Z = 0.75$

$Z = 0.75$ has a pvalue of 0.773.

1 - 0.773 = 0.227

0.227 = 22.7% probability that the mean printing speed of the sample is greater than 18.12 ppm.