# Don Capital Case Problem

**Topics:**Standard deviation, Normal distribution, Arithmetic mean

**Pages:**2 (436 words)

**Published:**October 4, 2012

FROM: THU NGUYEN

DATE: 02-21-12

SUBJECT: Case Problem #2

Problem Statement

Don Capita, the owner of Carpita Bottling Company in Lakeland, Wisconsin, received a phone call from the J.R. Summers grocery manager regarding the recently shipment. She requested Carpita Company to replace the entire of shipment because there were several bottles that weren’t qualified the fill level of 16-ounces. Therefore, Don decided to take a sample of 64 bottles of beer from the store to measure the average content in each bottle of beer. At the result, he found out that 15.993 ounces were the average content in each bottle of beers of the sample. However, those bottles were filled randomly among four bottling machines (one old machine and three new machines) from different warehouses. Therefore, Don needs help to determine if this filling error was more likely from the old machine or the new one so that he could reduce time and effort in checking all machines.

Statistical Results

Based on the information provided in this case, the sample size is 64-bottles of beers are large enough to assume the distribution for a probability is approximately normal

nπ=64(0.5)=32>5n1-π=641-0.5=32>5

Among the three new machines, the population mean was obtained at 16 ounces with a standard deviation of .16 ounces. By obtaining a Z-score at -.35 for the sampling distribution, it’s believed that the probability of the new machines producing a group of 64 bottles with a mean of 15.993 ounces will be 36.32%

In another hand, the population mean of the old machine was obtained at 16.05 ounces to compensate for the inconsistencies in the bottling process. Besides, its standard population was observed at .24 ounces. By obtaining a Z-score at -1.9 for the sampling distribution, it’s believed that the probability of the old machines producing a group of 64 bottles with a mean of 15.993 ounces will be 2.87%

The Excel formulas were used to calculate the...

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