The injection molding department of a company uses an average of 30 gallons of special lubricant a day. The supply of the lubricant is replenished when the amount on hand is 170 gallons. It takes four days for an order to be delivered. Safety stock is 50 gallons, which provides a stockout risk of 9 percent. What amount of safety stock would provide a stockout risk of 3 percent?

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

2) Solution to the problem

For this problem we know that the service level = 1-0.09= 0.91 or 91%. Now we can find the z score that accumulates 0.91 of the area below and 0.09 of the area above.

And this value is Z=1.34. And we can find it with the following excel code"1.341"

And then we can find the Safety stock given by this formula:

Safety Stock = z * Standard deviation of lead time

50= 1.341 * sd

sd=37.286

For the other case when the service level is 1-0.03 = 0.97 or 97% the corresponding z score is z=1.881

And the safety stock would be given by:

Safety Stock = 1.881 *37.286=70.134

topeadeniran2 topeadeniran2 · Answer 2 · 2022-04-05T16:35:54+02:00

The probability computed shows that the safety stock that would provide a stockout risk of 3 percent is 70 units.

How to calculate the probability?

Firstly, we will determine the value of the standard deviation which will be:

50 = 1.34 × x × ✓4

x = 50/(1.34 × 2)

x = 18.66

Therefore, the safety stock that would provide a stockout risk of 3 percent will be:

= 1.88 × 18.66 × 2

=70 units.

In conclusion, the correct option is 70 units.

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