Homework SourseThe manufacturer of a laser printer reports the mean number
The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,275. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 745 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 99 percent of the time? (Round z value to 2 decimal places. Round your answer to the nearest whole number.)
| The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,275. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 745 pages. The manufacturer wants to provide guidelines to potential customers as to how long they can expect a cartridge to last. |
Solution
Normal Distribution
Mean ( u ) =12275
Standard Deviation ( sd )=745
Normal Distribution = Z= X- u / sd ~ N(0,1)
P ( Z < x ) = 0.99
Value of z to the cumulative probability of 0.99 from normal table is 2.326
P( x-u/s.d < x - 12275/745 ) = 0.99
That is, ( x - 12275/745 ) = 2.33
--> x = 2.33 * 745 + 12275 = 14007.87
