## Step-by-Step Solution --- ### **Step 1: Find the sample means for each day** Calculate the average weight for each day. #### **Monday:** \[ \text{Mean} = \frac{23 + 22 + 19 + 20}{4} = \frac{84}{4} = 21 \] #### **Tuesday:** \[ \text{Mean} = \frac{23 + 21 + 19 + 23}{4} = \frac{86}{4} = 21.5 \] #### **Wednesday:** \[ \text{Mean} = \frac{20 + 19 + 20 + 21}{4} = \frac{80}{4} = 20 \] #### **Thursday:** \[ \text{Mean} = \frac{18 + 19 + 20 + 19}{4} = \frac{76}{4} = 19 \] #### **Friday:** \[ \text{Mean} = \frac{18 + 20 + 22 + 20}{4} = \frac{80}{4} = 20 \] --- ### **Step 2: List the sample means** | Day | Mean Weight | |----------|-------------| | Monday | 21 | | Tuesday | 21.5 | | Wednesday| 20 | | Thursday | 19 | | Friday | 20 | --- ### **Step 3: Answer part (a) – What value of \( z \) is used?** The control limits are set as: - UCL (Upper Control Limit) = 21 - LCL (Lower Control Limit) = 19 The formula for control limits is: \[ \text{UCL, LCL} = \mu \pm z \left(\frac{\sigma}{\sqrt{n}}\right) \] Given: - \(\mu = 20\) (mean) - \(\sigma = 2\) (std dev) - \(n = 4\) (sample size) Calculate \(\frac{\sigma}{\sqrt{n}}\): \[ \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{4}} = \frac{2}{2} = 1 \] So, \[ \text{UCL} = 20 + z \cdot 1 \] \[ \text{LCL} = 20 - z \cdot 1 \] Set UCL = 21 and LCL = 19: \[ 21 = 20 + z \implies z = 1 \] \[ 19 = 20 - z \implies z = 1 \] #### **Answer to (a):** The control chart is based on a \( z \) value of **1**. --- ### **Step 4: Answer part (b) – Is the process in control?** Check if all sample means are between 19 (LCL) and 21 (UCL): - Monday: 21 (on UCL) - Tuesday: 21.5 (**above UCL**) - Wednesday: 20 (within limits) - Thursday: 19 (on LCL) - Friday: 20 (within limits) #### **Conclusion:** - **Tuesday's mean (21.5) is above the UCL (21).** - At least one point outside control limits → **Process is not in control**. --- ## **Final Summary** 1. **The control chart uses a \( z \) value of 1** for the control limits. 2. **The process is not in control**, since the sample mean for Tuesday (21.5) is above the upper control limit (21). --- **If you have follow-up questions or need more details, let me know!**

## Step-by-Step Solution Summary --- ### **Step 1: Calculate Daily Sample Means** Calculated the average weight for each day from the provided data. The means were found to be 21, 21.5, 20, 19, and 20 for Monday through Friday, respectively. --- ### **Step 2: List Sample Means** Created a table to summarize the daily means of the package weights. The table provides a clear view of the average weights for each day. --- ### **Step 3: Determine \( z \) Value for Control Limits** Used the control limit formulas to find the \( z \) value when the upper control limit (UCL) is 21 and the lower control limit (LCL) is 19. Concluded that the control chart is based on a \( z \) value of 1. --- ### **Step 4: Assess Process Control** Evaluated the daily sample means against the control limits. Concluded that the process is not in control due to Tuesday's mean exceeding the UCL. --- ## **Final Summary** 1. A \( z \) value of 1 was determined for the control limits of the sample mean chart. 2. The process is not in control because one sample mean (Tuesday) exceeded the upper control limit.