Question

1 The monthly sales of mufflers follow the normal distribution with a mean of 1250 and a standard deviation of 255. The manufacturer would like to establish inventory levels such that there is only a 2% chance of running out of stock. Refer to the table in Appendix B.1.  

Where should the manufacturer set the inventory levels?

part B

1 The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 11,300. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 680 pages. The manufacturer wants to provide guidelines to potential customers as to how long they can expect a cartridge to last. Refer to the table in Appendix B.1.  

How many pages should the manufacturer advertise for each cartridge if it wants to be correct 95 percent of the time? (Round z-value to 2 decimal places and the final answer to the nearest whole number.)

Pages           

Answer 1

Solution:-

1) Given that,

mean = = 1250

standard deviation = = 255

Using standard normal table,

P(Z < z) = 2%

= P(Z < z ) = 0.02

= P(Z < -2.05 ) = 0.02  

z = -2.05

Using z-score formula,

x = z * +

x = -2.05 * 255 + 1250

x = 727.25

= 727 the manufacturer set the inventory levels.

2) Given that,

mean = = 11300

standard deviation = = 680

Using standard normal table,

P(Z > z) = 95%

= 1 - P(Z < z) = 0.95  

= P(Z < z) = 1 - 0.95

= P(Z < z ) = 0.05

= P(Z < -1.65 ) = 0.05  

z = -1.65

Using z-score formula,

x = z * +

x = -1.65 * 680 + 11300

x = 10178 pages.