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 17.49 ppm and a standard deviation of 4 ppm. Suppose that you draw a random sample of 14 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).

__________Probability (as a proportion)

Answer 1

Solution :

Given that ,

mean = = 17.49

standard deviation = = 4

= / n = 4 / 14 = 1.0690

P( > 17.99) = 1 - P( < 17.99)

= 1 - P[( - ) / < (17.99 - 17.49) /1.0690 ]

= 1 - P(z < 0.467727)

= 1 - 0.679999 = 0.320,Probability = 0.320