Questions: Auto pistons at Wenming Chung's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: Day Mean x̄ (mm) Range R (mm) 1 154.9 4.2 2 155.2 4.8 3 153.6 4.3 4 153.5 4.8 5 154.6 4.3 a) What is the value of x̄̄? x̄̄=154.36 mm (round your response to two decimal places). b) What is the value of R̄? R̄=4.48 mm (round your response to two decimal places). c) What are the UCL− and LCL− using 3-sigma? Upper Control Limit (UCL−)=155.74 mm (round your response to two decimal places). Lower Control Limit (LCLx̄)=152.98 mm (round your response to two decimal places). d) What are the UCLR and LCLR using 3-sigma? Upper Control Limit (UCLR)= square mm (round your response to two decimal places).

Auto pistons at Wenming Chung's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: Day Mean x̄ (mm) Range R (mm) 1 154.9 4.2 2 155.2 4.8 3 153.6 4.3 4 153.5 4.8 5 154.6 4.3 a) What is the value of x̄̄? x̄̄=154.36 mm (round your response to two decimal places). b) What is the value of R̄? R̄=4.48 mm (round your response to two decimal places). c) What are the UCL− and LCL− using 3-sigma? Upper Control Limit (UCL−)=155.74 mm (round your response to two decimal places). Lower Control Limit (LCLx̄)=152.98 mm (round your response to two decimal places). d) What are the UCLR and LCLR using 3-sigma? Upper Control Limit (UCLR)= square mm (round your response to two decimal places).
Transcript text: Auto pistons at Wenming Chung's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: \begin{tabular}{ccc} \hline Day & \begin{tabular}{c} Mean $\bar{x}$ \\ $(\mathrm{~mm})$ \end{tabular} & \begin{tabular}{c} Range R \\ $(\mathrm{mm})$ \end{tabular} \\ \hline 1 & 154.9 & 4.2 \\ 2 & 155.2 & 4.8 \\ 3 & 153.6 & 4.3 \\ 4 & 153.5 & 4.8 \\ 5 & 154.6 & 4.3 \\ \hline \end{tabular} a) What is the value of $\overline{\bar{x}}$ ? \[ \overline{\bar{x}}=154.36 \mathrm{~mm} \text { (round your response to two decimal places). } \] b) What is the value of $\bar{R}$ ? \[ \overline{\mathrm{R}}=4.48 \mathrm{~mm} \text { (round your response to two decimal places). } \] c) What are the $\mathrm{UCL}_{-}$and $\mathrm{LCL}_{-}^{-}$using 3-sigma? Upper Control Limit $\left(U C L_{-}\right)=155.74 \mathrm{~mm}$ (round your response to two decimal places). Lower Control Limit $\left(\mathrm{LCL}_{\bar{x}}\right)=152.98 \mathrm{~mm}$ (round your response to two decimal places). d) What are the $U C L_{R}$ and $\mathrm{LCL}_{R}$ using 3-sigma? Upper Control Limit $\left(U C L_{R}\right)=$ $\square$ mm (round your response to two decimal places).
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Solution

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Solution Steps

Step 1: Calculate the Mean of Means \(\overline{\bar{x}}\)

The mean of the sample means \(\overline{\bar{x}}\) is calculated as follows:

\[ \overline{\bar{x}} = \frac{\sum_{i=1}^N \bar{x}_i}{N} = \frac{154.9 + 155.2 + 153.6 + 153.5 + 154.6}{5} = \frac{771.8}{5} = 154.36 \]

Step 2: Calculate the Mean of Ranges \(\bar{R}\)

The mean of the sample ranges \(\bar{R}\) is calculated as follows:

\[ \bar{R} = \frac{\sum_{i=1}^N R_i}{N} = \frac{4.2 + 4.8 + 4.3 + 4.8 + 4.3}{5} = \frac{22.4}{5} = 4.48 \]

Step 3: Calculate the Standard Deviation of Means

The standard deviation of the sample means is calculated using the formula for sample standard deviation:

\[ \sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{n-1}} \]

Given the variance \(\sigma^2 = 0.59\), the standard deviation is:

\[ \sigma = \sqrt{0.59} = 0.77 \]

Step 4: Calculate the Upper and Lower Control Limits for \(\bar{x}\)

The Upper Control Limit (UCL) and Lower Control Limit (LCL) for \(\bar{x}\) using 3-sigma are calculated as follows:

\[ UCL_{\bar{x}} = \overline{\bar{x}} + 3\sigma = 154.36 + 3 \times 0.77 = 156.67 \]

\[ LCL_{\bar{x}} = \overline{\bar{x}} - 3\sigma = 154.36 - 3 \times 0.77 = 152.05 \]

Final Answer

  • The mean of means \(\overline{\bar{x}}\) is \(\boxed{154.36 \, \text{mm}}\).
  • The mean of ranges \(\bar{R}\) is \(\boxed{4.48 \, \text{mm}}\).
  • The Upper Control Limit \(UCL_{\bar{x}}\) is \(\boxed{156.67 \, \text{mm}}\).
  • The Lower Control Limit \(LCL_{\bar{x}}\) is \(\boxed{152.05 \, \text{mm}}\).
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