Question

The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,300. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 720 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

Answer 1

Answer:

The manufacturer should advertise 11,378.4 pages.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 12300, \sigma = 720$

How many pages should the manufacturer advertise for each cartridge if it wants to be correct 90 percent of the time

It should advertise the 10th percentile(X when Z has a pvalue of 0.1, so X when Z = -1.28), so 90% of the time the printer wil print more pages than what is advertised.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 12300}{720}$

$X - 12300 = -1.28*720$

$X = 11378.4$

The manufacturer should advertise 11,378.4 pages.