Question

n 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 17.39 ppm and a standard deviation of 4.25 ppm. Suppose that you draw a random sample of 11 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 17.99 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:

0.321 is the probability that their mean printing speed of the sample is greater than 17.99 ppm.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 17.39 ppm

Standard Deviation, σ = 4.25 ppm

Sample size = 11

We are given that the distribution of printing speed is a bell shaped distribution that is a normal distribution.

Formula:  

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$  

P(printing speed of the sample is greater than 17.99 ppm.)  

P(x > 17.99)  

$P( x > 17.99) = P( z > \displaystyle\frac{17.99-17.39}{\frac{4.25}{\sqrt{11}}}) = P(z > 0.468229)\\\\P( z > 0.468229) = 1 - P(z < 0.468229)$

Calculating the value from the standard normal table we have,

$1 - 0.679 = 0321= 32.1\%\\P( x > 17.99) = 32.1\%$

Thus, 0.321 is the probability that their mean printing speed of the sample is greater than 17.99 ppm.

Answer 2

Using the normal distribution and the central limit theorem, it is found that there is a 0.319 = 31.9% probability that the mean printing speed of the sample is greater than 17.99 ppm.

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 samples of size n, the standard deviation is $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• Mean of 17.39 ppm, thus $\mu = 17.39$
• Standard deviation of 4.25 ppm, thus $\sigma = 4.25$
• Sample of 11 printers, thus $n = 11, s = \frac{4.25}{\sqrt{11}}$.

The probability is 1 subtracted by the p-value of Z when X = 17.99, thus:

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

By the Central Limit Theorem

$Z = \frac{17.99 - 17.39}{\frac{4.25}{\sqrt{11}}$

$Z = 0.47$

$Z = 0.47$ has a p-value of 0.681.

1 - 0.681 = 0.319

0.319 = 31.9% probability that the mean printing speed of the sample is greater than 17.99 ppm.

A similar problem is given at https://brainly.com/question/17153877