Question

A part, produced in high volumes, is dimensioned as shown. The machining process making this part is known to be statistically in control based on sampling data. The sampling data shows that D1 follows a normal distribution with a mean of 20 mm and a standard deviation of 0.3 mm, while D2 follows a normal distribution with a mean of 35 mm and a standard deviation of 0.4 mm. An inspection of dimension C is carried out in a sufficiently large number of parts. To be considered under six-sigma process control, the upper limit of dimension C should be ............... mm. (Rounded off to one decimal place)

Hint:

A key rule in the algebra of random variables is that {variances add} for both the sum and the difference of independent variables. A common mistake is to subtract the variances for a difference. Remember: \(Var(X-Y) = Var(X) + Var(Y)\) if X and Y are independent.

Answer 1

Step 1: Understanding the Concept:
This problem involves the statistics of tolerances. When a dimension is the result of the sum or difference of other dimensions that are normally distributed, the resulting dimension is also normally distributed. We need to find the mean and standard deviation of dimension C. Then, we apply the concept of process control limits. "Process control" limits are typically set at \(\pm 3\sigma\) from the mean. A "six-sigma process" is one where the specification limits are at least \(\pm 6\sigma\) from the mean, but the control limits are still at \(\pm 3\sigma\). The question asks for the "upper limit... to be considered under six-sigma process control", which in the context of statistical process control (SPC) refers to the Upper Control Limit (UCL).
Step 2: Key Formula or Approach:
Let \(C = D2 - D1\), where D1 and D2 are independent normal random variables. 1. Mean of C: The mean of the difference is the difference of the means. \[ \mu_C = \mu_{D2} - \mu_{D1} \] 2. Variance of C: The variance of the difference of independent variables is the sum of the variances. \[ \sigma_C^2 = \sigma_{D1}^2 + \sigma_{D2}^2 \] The standard deviation is \(\sigma_C = \sqrt{\sigma_{D1}^2 + \sigma_{D2}^2}\). 3. Upper Control Limit (UCL): For a process in statistical control, the UCL is typically set at 3 standard deviations above the mean. \[ \text{UCL} = \mu_C + 3\sigma_C \] Step 3: Detailed Calculation:
1. Given Data:
- For D1: \(\mu_{D1} = 20\) mm, \(\sigma_{D1} = 0.3\) mm
- For D2: \(\mu_{D2} = 35\) mm, \(\sigma_{D2} = 0.4\) mm
2. Calculate Mean of C: \[ \mu_C = 35 - 20 = 15 \text{ mm} \] 3. Calculate Standard Deviation of C: \[ \sigma_C = \sqrt{\sigma_{D1}^2 + \sigma_{D2}^2} = \sqrt{(0.3)^2 + (0.4)^2} \] \[ \sigma_C = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 \text{ mm} \] 4. Calculate the Upper Control Limit (UCL): \[ \text{UCL} = \mu_C + 3\sigma_C = 15 + 3(0.5) = 15 + 1.5 = 16.5 \text{ mm} \] Step 4: Final Answer:
The upper limit of dimension C should be 16.5 mm.
Step 5: Why This is Correct:
The solution correctly calculates the mean and standard deviation for the dimension C, which is a difference of two independent normal variables. It then correctly interprets "upper limit... for process control" as the 3-sigma Upper Control Limit (UCL), which is standard practice in statistical quality control. The calculation is accurate and properly rounded.