The number of pages printed before replacing the cartridge in a laser printer is normally distributed to a mean of 11,500 pages
and a standard deviation of 800 pages. A new cartridge has just been installed. a. What is the probability that the printer produces more than 12,000 pages before this cartridge must be replaced? b. What is the probability that the printer produces fewer than 10,000 pages?
<span>a. What is the probability that the printer produces more than 12,000 pages before this cartridge must be replaced? solution to solve the equation we need to find the z-score: z=(x-</span>μ)/σ where: x=12000 μ=mean=11500 σ=standard deviation=800 thus z=(12000-11500)/800=0.625 thus P(x>12000)=1-P(x<12000)=1-0.7357=0.2643
<span>b. What is the probability that the printer produces fewer than 10,000 pages? z=(10000-11500)/800 z=-1.875 hence P(x<10000) =0.0301
