Reference no: EM13947462

Case Study: discusses American Oil, a company that uses advanced equipment to detect oil under the earth's surface. Chad Williams is a field geologist who has recently learned that a design-engineering group in the company has come up with an enhancement to their current equipment that will greatly improve their ability to detect and find oil. However, this enhancement requires the use of 800 electrical capacitors. These capacitors must operate within 0.5 microns from the standard of 12 microns. This causes a problem for Chad, because he must order these capacitors, and the supplier can only provide capacitors that operate at a normal distribution, with a mean of 12 microns, and a standard deviation of one micron. Since these capacitors are very expensive, American Oil wants to take a 98% chance that they will receive 800 capacitors that fit their needs for the enhancement, and Chad must figure out how many of them to order.

To do this, you can use z score to find out the area of the normally distributed bell curve that will fit American Oil's needs. The top of the bell curve is centered on the mean, which is 12 microns. Knowing this, you will use the standardized normal z value formula. This is your value you are finding minus the mean and divided by the standard deviation. You will do this for both plus and minus 0.5 microns. Doing this will give you a z-value of 0.5, and -0.5. You will then take these z-values and look them up in the normal distribution table. You will find out that a z score of 0.5 will produce .1915. Since both sides of the bell curve have the same z score, then they both have the area of .1915. Adding these two together gives you .383, or the total area of capacitors that the producer can give that will be useful to American Oil.

Since American Oil needs their capacitors to be within 0.5 microns from the standard mean of 12 microns and the producer can only provide capacitors with a standard deviation of one micron from the standard mean, American Oil must order more than they need. They want to choose a value high enough so that 38.3 percent of what is ordered is equal to 800. To find out how many capacitors are needed, you will take the amount needed (800) and divide it by 38.3%. Doing this will yield 2088.77. Since it is impossible to have 0.77 of a capacitor, you will round up to 2089 capacitors. This amount guarantees that American Oil will receive 800 capacitors that work to their specifications. However, since the capacitors are very expensive, American Oil is willing to take a 98% chance that they will receive their 800 capacitors. Figuring this number out is as easy as multiplying the total needed (2089) by 98%. This will give you an answer of 2047.22. Again, since you cannot have 0.22 of a capacitor, you would round up to 2048.

American Oil is a company that needs capacitors that operate within 0.5 microns from the standard mean of 12 microns. The producers can only provide American Oil with capacitors that operate with a mean of 12 microns and a standard deviation of one. American Oil needs 800 capacitors that fit their needs. They are willing to take a 98% chance that they will receive enough capacitors to fulfill their needs. This is because of the high price for the capacitors. Using the basic statistical technique of finding the z-score, we found that American Oil needs to order at least 2048 capacitors to fit all of the criteria.

Review Case "American Oil Company" (Groebner p.254). Make the necessary computations and write a short report discussing your answers and findings.