An HP laser printer is advertised to print text documents at

An HP laser printer is advertised to print text documents at a speed of 22 ppm (pages per minute). The manufacturer tells you that the printing speed is actually a Normal random variable with a mean of 21.2 ppm and a standard deviation of 3.66 ppm. Suppose that you draw a random sample of 41 printers.

Use normal approximation to find the probability that more than 44% of the sampled printers operate at the advertised speed, i.e. the printing speed is equal to or greater than 22 ppm. (Carry answers to at least six decimal places in intermediate steps. Give your final answer to the nearest three decimal places)

If you just simply do Z score calculation, it comes about 0.081 but that\'s not the answer. I don\'t know what to do with the inforation about 44%.

Solution

Mean ( u ) =21.2
Standard Deviation ( sd )=3.66
Number ( n ) = 41
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
P(X < 22) = (22-21.2)/3.66/ Sqrt ( 41 )
= 0.8/0.5716= 1.3996
= P ( Z <1.3996) From Standard NOrmal Table
= 0.9192                  
P(X > 22) = (22-21.2)/3.66/ Sqrt ( 41 )
= 0.8/0.572= 1.3996
= P ( Z >1.3996) From Standard Normal Table
= 0.081