Mathematics High School

Assume the following for this question. Lower and Upper specification limits for a service time are 3 minutes and 5 minutes, respectively with the nominal expected service time at 4 minutes. The observed mean service time is 4 minutes with a standard deviation of 0.2 minutes. The current control limits are set at 3.1 and 4.9 minutes respectively.

Answers

Answer 1

The observed mean service time falls within the current control limits. We can conclude that the process is stable, the service time is in control, and it meets the required specifications.


1. Calculate the process capability index (Cpk) using the formula: Cpk = min((USL - mean)/3σ, (mean - LSL)/3σ), where USL is the upper specification limit, LSL is the lower specification limit, mean is the observed mean service time, and σ is the standard deviation.
2. Plug in the values: USL = 5 minutes, LSL = 3 minutes, mean = 4 minutes, σ = 0.2 minutes.
3. Calculate Cpk: Cpk = min((5-4)/(3*0.2), (4-3)/(3*0.2)) = min(0.556, 0.556) = 0.556.
4. Since the calculated Cpk is greater than 1, the process is considered capable and the service time is in control.
5. The current control limits (3.1 and 4.9 minutes) are wider than the specification limits (3 and 5 minutes) and the observed mean (4 minutes) falls within these control limits.
6. Therefore, the process is stable and meets the specifications.

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Related Questions

Suppose that in a particular sample, the mean is 50 and the standard deviation is 10. What is the z score associated with a raw score of 68?

Answers

The z-score associated with a raw score of 68 is 1.8.

Given mean = 50 and standard deviation = 10.

Z-score is also known as standard score gives us an idea of how far a data point is from the mean. It indicates how many standard deviations an element is from the mean. Hence, Z-Score is measured in terms of standard deviation from the mean.

The formula for calculating the z-score is given as

z = (X - μ) / σ

where X is the raw score, μ is the mean and σ is the standard deviation.

In this case, the raw score is X = 68.

Substituting the given values in the formula, we get

z = (68 - 50) / 10

z = 18 / 10

z = 1.8

Therefore, the z-score associated with a raw score of 68 is 1.8.

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