Question

In: Operations Management

Q1: Golden Gopher Airline issues thousands of aircraft boarding passes to passengers each day. In some...

Q1: Golden Gopher Airline issues thousands of aircraft boarding passes to passengers each day. In some cases a boarding pass is spoiled for various reasons and discarded by the airline agent before the final boarding pass is issued to a customer. To control the process for issuing boarding passes, the airline has sampled the process for 100 days and determined the average proportion of defective passes is .006 (6 in every 1000 passes are spoiled and discarded). In the future, the airline plans to take a sample of 500 passes that are issued each day and calculate the proportion of spoiled passes in that sample for control chart purposes.

  • A. What is the sample size (n) for this problem? Is it 100, 500, or 1000? Explain the significance of the 100 days used to determine the average proportion defective.
  • B. Calculate the CL, UCL, and LCL, using three standard deviations for control purposes.

Solutions

Expert Solution

Solution:-

The sample size for the future plan is 500 spend every day.

The imperfection in the given case is deterioration of the passes, is estimated on the discrete scale (considered fortunate or unfortunate), in this manner such sort of quality measurement is called attribute types measurement.

At the point when the measurement is attribute type, the fitting quality control chart to utilize is p–chart for proportional defectives in the sample.

To figure normal proportional defectives per sample, enormous number of samples is taken, since defectives per sample may be less.

In this manner, 100 days sample is taken to decide suitable proportional defectives per sample.

For given procedure, the proportional defectives, pbar = 0.006 per sample

Sample size, n = 500 spends every day

Focal line of p-chart = CL = 0.006

Standard deviation of p-chart = Sp = √[(pbar)(1 - pbar)/n] = √[(0.006(1-0.006)/500]

Sp = 0.00345

Upper control limit = UCL = pbar + 3Sp = 0.006 + 3*0.00345 = 0.006 + 0.0104 = 0.0164

Lower control limit = LCL = pbar -3Sp = 0.006 - 3*0.00345 = 0.006 - 0.0104 = - 0.0044 ~0

Control limit = 0.006,
Upper control limit = 0.0164,
Lower control limit = 0


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