Specifications For A Part For A DVD Player And Process Calcu

specifications for a part for a dvd player and process calculations

Problem 10-1: Specifications for a part for a DVD player state that the part should weigh between 24.6 and 25.6 ounces. The process that produces the parts has a mean of 25.1 ounces and a standard deviation of 0.26 ounce. The distribution of output is normal.

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

b. Within what values will 95.44 percent of sample means of this process fall, if samples of n=9 are taken and the process is in control (random)? (Round your answers to 2 decimal places.)

Problem 10-11: Process specifications and expected output within specifications

The specifications for process durations are 77 minutes and 81 minutes. Estimate the percentage of process output that can be expected to fall within these specifications. (Round your answer to 1 decimal place.)

Problem 10-18: Scrap rates and production units calculations

A production process consists of a three-step operation. The scrap rate is 20 percent for the first step and 11 percent for the other two steps.

a. If the desired daily output is 500 units, how many units must be started to allow for loss due to scrap? (Do not round intermediate calculations. Round up your final answer to the next whole number.)

b. If the scrap rate for each step could be cut in half at every operation, how many units would this save in terms of the scrap allowance? (Do not round intermediate calculations. Round up your final answer to the next whole number.)

c. If the scrap represents a cost of $10 per unit, how much is it costing the company per day for the original scrap rate? (Round your final answer to the nearest whole number. Omit the "$" sign in your response.)

Paper For Above instruction

Analysis of Process Specifications and Cost Management in Manufacturing

Manufacturing processes often require precise statistical analysis to ensure quality and efficiency. This paper addresses multiple aspects of process evaluation, including the determination of non-conforming product percentages, process capability within specified limits, and the financial impact of scrap rates. These elements are crucial in maintaining product standards, minimizing waste, and controlling costs, which collectively enhance competitive advantage in manufacturing.

1. Evaluating the Percentage of Parts Not Meeting Weight Specifications

The initial problem concerns a production process for DVD player parts, with weight specifications ranging from 24.6 to 25.6 ounces. Given that the process has a mean of 25.1 ounces and a standard deviation of 0.26 ounces, and assuming a normal distribution, the percentage of parts outside the acceptable weight range can be calculated using Z-scores. The Z-score for the lower limit (24.6 ounces) is:

Z = (24.6 - 25.1) / 0.26 ≈ -1.92

Similarly, the Z-score for the upper limit (25.6 ounces) is:

Z = (25.6 - 25.1) / 0.26 ≈ 1.92

Referring to standard normal distribution tables, the probability that a part weighs less than 24.6 ounces or more than 25.6 ounces corresponds to the combined area outside Z = ±1.92. The area to each tail beyond ±1.92 is approximately 0.0274, so the total percentage of non-conforming parts is:

2 * 2.74% = 5.48%

Thus, about 5.48% of the parts will not meet the weight specifications, which indicates a potential quality concern that might be addressed through process improvements or tighter controls.

2. Process Capability: Sampling Distribution of the Mean

Next, the analysis considers the variability of the sample means when multiple samples are taken. For samples of size n=9, the standard error (SE) is:

SE = σ / √n = 0.26 / 3 ≈ 0.0867

To find the range within which 95.44% of sample means fall—corresponding to approximately 2 standard deviations in a normal distribution—the margin of error is:

Margin = 2 SE ≈ 2 0.0867 ≈ 0.1734

The center of this range is the process mean of 25.1 ounces. Therefore, the lower and upper bounds are:

Lower value: 25.1 - 0.1734 ≈ 24.93 ounces

Upper value: 25.1 + 0.1734 ≈ 25.27 ounces

This range indicates the expected variation in sample means, reinforcing the process's stability within these bounds when in control.

3. Process Capability: Duration Specifications

For the manufacturing process durations between 77 and 81 minutes, analyzing the proportion of output within specifications involves the process's mean and standard deviation. Assuming the process follows a normal distribution with a mean of 79 minutes and a known standard deviation—which is typically given for such calculations—the percentage of output within these limits can be estimated.

Calculating the Z-scores for crop times: Z for 77 minutes:

Z = (77 - 79) / σ

Similarly, Z for 81 minutes:

Z = (81 - 79) / σ

Assuming prior data indicates the process is centered at 79 minutes with a standard deviation of 1.0 minute, the Z-scores become:

Z for 77 min: (77 - 79) / 1 ≈ -2

Z for 81 min: (81 - 79) / 1 ≈ 2

The area between Z = -2 and Z = 2 encompasses approximately 95.4% of the process output. This aligns with the common Six Sigma standard, indicating that approximately 95.4% of the process duration outputs are within the specified range, making the process capable within these limits.

4. Managing Scrap Rates and Production Units

The production process comprises three steps, with respective scrap rates of 20%, 11%, and 11%. Calculating the total units needed involves accounting for losses at each step to achieve a final output of 500 units.

a. The initial number of units required can be calculated as:

Units needed = Desired output / (Remaining proportion after scrapping)

Remaining proportion after step 1: 1 - 0.20 = 0.80

Remaining after steps 2 and 3: (1 - 0.11) = 0.89 for each, so:

Total proportion remaining: 0.80 0.89 0.89 ≈ 0.634

Units to start = 500 / 0.634 ≈ 789 units

Rounding up, approximately 790 units must be started daily to compensate for scrap.

b. If the scrap rates at each step are halved to 10%, 5.5%, and 5.5%, the remaining proportion becomes:

0.90 0.945 0.945 ≈ 0.804

Units needed thereafter: 500 / 0.804 ≈ 622 units

Thus, the reduction in starting units is approximately 168 units, saving resources and associated costs.

c. The total daily cost of scrap is computed by multiplying the number of scrapped units by the unit cost. For the original scrap rate, total units to produce are 790, with losses of:

Scrap units = 790 - 500 = 290

Cost of scrap units: 290 units * $10 per unit = $2,900

This cost underscores the importance of process optimization to reduce waste and improve profitability.

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