A drilling machine bores holes with a mean diameter of 0.5230 cm and S.D. of 0.0032cm. Calculate the2-sigma and 3-sigma upper and lower control limits for mean of sample of 4
Question
A drilling machine bores holes with a mean diameter of 0.5230 cm and S.D. of 0.0032cm. Calculate the2-sigma and 3-sigma upper and lower control limits for mean of sample of 4
Solution
To calculate the 2-sigma and 3-sigma upper and lower control limits, we first need to understand that the sigma (σ) represents the standard deviation, which in this case is 0.0032 cm. The mean (μ) is 0.5230 cm.
The control limits are calculated using the formula: Control Limit = μ ± zσ/√n
Where: μ = mean = 0.5230 cm σ = standard deviation = 0.0032 cm n = sample size = 4 z = z-score (which is 2 for 2-sigma limits and 3 for 3-sigma limits)
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For 2-sigma control limits: Upper Control Limit = μ + 2σ/√n Lower Control Limit = μ - 2σ/√n
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For 3-sigma control limits: Upper Control Limit = μ + 3σ/√n Lower Control Limit = μ - 3σ/√n
Now, let's calculate:
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2-sigma control limits: Upper Control Limit = 0.5230 + 2*(0.0032/√4) = 0.5230 + 0.0032 = 0.5262 cm Lower Control Limit = 0.5230 - 2*(0.0032/√4) = 0.5230 - 0.0032 = 0.5198 cm
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3-sigma control limits: Upper Control Limit = 0.5230 + 3*(0.0032/√4) = 0.5230 + 0.0048 = 0.5278 cm Lower Control Limit = 0.5230 - 3*(0.0032/√4) = 0.5230 - 0.0048 = 0.5182 cm
So, the 2-sigma control limits are 0.5262 cm (upper) and 0.5198 cm (lower), and the 3-sigma control limits are 0.5278 cm (upper) and 0.5182 cm (lower).
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