SOLUTION: The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,450. The distribution of pages printed per cartridge clo

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Question 1037032: The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,450. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 570 pages. The manufacturer wants to provide guidelines to potential customers as to how long they can expect a cartridge to last.

How many pages should the manufacturer advertise for each cartridge if it wants to be correct 90 percent of the time? (Round z value to 2 decimal places. Round your answer to the nearest whole number.)

Pages

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
mean = 12450
s = 570.

you want to be confident that a random cartridge sold will provide more than the number of pages quoted 90% of the time.

this means that only 10% of the cartridges sold will provide a number of pages less than that.

you would use a z-score calculator to determine the z-score that has the probability of being less than the indicated z-score only 10% of the time.

that z-score will be -1.29 with an accuracy to the nearest 2 decimal places.

i used a z-score calculator that says that the actual z-score is -1.2815515767.

that would round to -1.28, but i chose -1.29 for the reasons that will be explained further below.

if you look up the the z-score in the z-score table, it will tell you that .....

a z-score of -1.28 has .1003 proportion of the normal distribution curve to the left of it.

a z-score of -1.29 has .0985 proportion of the normal distribution curve to the left of it.

you could use the calculator, or you could interpolate, or you could take the conservative approach and use -1.29 as your z-factor.

i chose -1.29 because this will ensure you that, at least 90% of the cartridges selected will have a number of pages greater than that.

the actual number could be higher, but it is highly improbable that it will be lower.

given a z-score of -1.29, you would calculate the raw score as follows:

z = (x - m) / s

z is the z-score.
x is the raw score.
m is the mean.
s is the standard deviation.

the formula becomes -1.29 = (x - 12450) / 570.

solve for x to get x = -1.29 * 570 + 12450 = 11714.7.

you can predict that any cartridge that was sold will be able to print at least 11715 pages 90% of the time.

there is a graphing calculator that will show you this.

that calculator can be found at http://davidmlane.com/hyperstat/z_table.html

plug in the numbers and the calculator tells you that you can achieve a raw score greater than 11715 for approximately 90.14% of the time.

here's an output from the calculator.