Specifications For A Part For A DVD Player State That The PA

Specifications For A Part For A Dvd Player State That The Part Should

Specifications for a part for a DVD player state that the part should weigh between 25.5 and 26.5 ounces. The process that produces the parts has a mean of 26.0 ounces and a standard deviation of 0.22 ounce. The distribution of output is normal. Use Table-A.

a. What percentage of parts will not meet the weight specifications? (Round your "z" value and final answer to 2 decimal places.) Omit the "%" sign in your response.

b. Within what values will 99.74 percent of sample means of this process fall, if samples of n = 12 are taken and the process is in control (random)? (Round your answers to 2 decimal places.)

Paper For Above instruction

The precise determination of the percentage of parts that do not meet weight specifications and the interval within which the sample means will fall are fundamental in quality control processes. Both calculations rely on understanding the properties of the normal distribution, given the known mean, standard deviation, and sample size. This paper provides detailed calculations and interpretations of those figures for the specified DVD part manufacturing process.

Part A: Percentage of Parts Not Meeting the Weight Specifications

Given data indicates that the process producing DVD parts is normally distributed with a mean (μ) of 26.0 ounces and a standard deviation (σ) of 0.22 ounces. The weight specifications are between 25.5 and 26.5 ounces. To determine the percentage of parts outside these limits, we calculate the corresponding z-scores for both boundaries.

Calculations for the lower boundary (25.5 ounces):

z = (X - μ) / σ = (25.5 - 26.0) / 0.22 = -0.5 / 0.22 ≈ -2.27

Calculations for the upper boundary (26.5 ounces):

z = (26.5 - 26.0) / 0.22 = 0.5 / 0.22 ≈ 2.27

Using Table-A, which provides standard normal distribution values, the cumulative probability corresponding to z = -2.27 is approximately 0.0116, and for z = 2.27, it is approximately 0.9884.

The probability that a part falls outside the specifications is twice the probability of falling below the lower limit (or above the upper limit):

P(out of spec) = 2 (1 - P(z ≤ 2.27)) = 2 (1 - 0.9884) = 2 * 0.0116 = 0.0232

Therefore, approximately 2.32% of the parts will not meet the weight specifications. Rounded to two decimal places and omitting the percent sign, the answer is 2.32.

Part B: 99.74% Confidence Interval for Sample Means of Size 12

The second part involves determining the range within which 99.74% of the sample means will fall when samples of size n=12 are taken, assuming the process remains in control and random sampling is applicable. Given that the population standard deviation is 0.22 ounces and the sample size is 12, the standard error (SE) of the mean is calculated as:

SE = σ / √n = 0.22 / √12 ≈ 0.22 / 3.464 ≈ 0.0635

For a confidence level capturing 99.74% of the distribution, note that this corresponds approximately to z = ±3 (for a normal distribution, 99.7% is generally associated with 3 standard deviations). Hence, the margin of error (ME) is:

ME = z SE = 3 0.0635 ≈ 0.1905

Calculating the interval:

• Lower bound = μ - ME = 26.0 - 0.1905 ≈ 25.81
• Upper bound = μ + ME = 26.0 + 0.1905 ≈ 26.19

Therefore, 99.74% of sample means will fall between approximately 25.81 and 26.19 ounces when the process is in control. Rounded to two decimal places, the interval is 25.81, 26.19.

Conclusion

Quality control measures depend critically on accurate calculations of process variability and constraints. The findings indicate that about 2.32% of parts will be outside the specified weight range, highlighting areas where process improvements may be necessary to enhance output consistency. Additionally, the interval within which 99.74% of sample means fall demonstrates the expected variability for the process when performing statistical sampling, which is essential for ongoing process monitoring and control (Montgomery, 2012; Grant & Leavenworth, 2011).

References

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