Follow the normal distribution with a mean of 1200 and a standard deviation of 225 and establish inventory levels such that there is only a 5% chance of running out of stock

Question

~~The monthly sales of mufflers in the Richmond, VA area follow the normal distribution with a mean of 1200 and a standard deviation of 225. The manufacturer would like to establish inventory levels such that there is only a 5% chance of running out of stock. Where should the manufacturer set the inventory levels?

Answer 1

The z-score is given by Z=(X-µ)/s, where s=standard deviation and µ=mean so X=sZ+µ=225Z+1200. What we need to do now is it find out the value of Z from the normal distribution table that gives us 100-5=95% probability that stocks of mufflers will not run out so as to meet the sales demand. The table gives a value of about 1.645 for Z. Therefore X=225*1.645+1200=1570. If the stock level (inventory level) each month is maintained at 1,570 mufflers, there's a 95% chance that supply will meet demand.