Question

A process filling small bottles with baby formula has a target of 3 ounces +/- 0. 150 ounce. Two hundred bottles from the process were sampled. The results showed the average amount of formula placed in the bottles to be 3. 042ounces. The standard deviation of the amounts was 0. 034ounce. Determine the value of Cpk. Roughly what proportion of bottles meet the​specifications?

Part 2 The process capability index ​(​Cpk) is ​(round your response to three decimal​ places).

Part 3 Slightly more than (answer choice of ​99. 99, 95. 45, 90, 99. 73, or 80%) of the bottles meet the specifications

Answer 1

To determine the value of Cpk and the proportion of bottles that meet the specifications, we can use the following formulas:

Cpk = min((USL - μ) / (3σ), (μ - LSL) / (3σ))

Proportion within specifications = Normal Distribution Area between LSL and USL

Given the information provided:

Target value (μ) = 3 ounces
Standard deviation (σ) = 0.034 ounce
Upper Specification Limit (USL) = 3 + 0.150 = 3.150 ounces
Lower Specification Limit (LSL) = 3 - 0.150 = 2.850 ounces

Let's calculate each part:

Cpk = min((3.150 - 3.042) / (3 * 0.034), (3.042 - 2.850) / (3 * 0.034))

Cpk ≈ min(0.366, 2.824) ≈ 0.366

The process capability index (Cpk) is approximately 0.366.

To find the proportion of bottles that meet the specifications, we need to determine the area under the normal distribution curve between the LSL and USL. This can be done using statistical tables or software.

Based on the provided answer choices, the closest option is "Slightly more than 99.73%." However, to provide a precise proportion, we would need to calculate it using the z-score corresponding to the LSL and USL values and then find the area under the normal distribution curve between those z-scores.

Please note that I am unable to perform real-time statistical calculations, so I cannot provide the exact proportion. However, you can use the given information to calculate it using statistical tables or software.