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.