Certainly! Here's a structured answer to each question based on the data and instructions. --- ## **1. Calculate the mean and range for the above samples** ### **Steps:** - Find the mean for each sample (average of 4 values). Find the range for each sample (max value - min value). ### **Calculations:** | Sample Number | Data (IN OHMS) | Mean | Range | |----------------|----------------|-------|--------| | 1 | 1014, 1019, 976, 1018 | (1014+1019+976+1018)/4 = 1021.8 | 1019 - 976 = 43 | | 2 | 989, 1009, 995, 984 | (989+1009+995+984)/4 = 994.3 | 1009 - 984 = 25 | | 3 | 102, 998, 1012, 1003 | (102+998+1012+1003)/4 = 1008.8 | 102 - 998 = 22 | | 4 | 985, 1014, 1021, 1027 | (985+1014+1021+1027)/4 = 10011.8| 1027 - 985 = 42 | | 5 | 976, 984, 996, 1006 | (976+984+996+1006)/4 = 987.5 | 1006 - 976 = 30 | | 6 | 1021, 982, 1006, 1011 | (1021+982+1006+1011)/4 = 1005. | 1021 - 982 = 39 | | 7 | 983, 1017, 988, 978 | (983+1017+988+978)/4 = 991.3 | 1017 - 978 = 39 | | 8 | 1011, 971, 978, 1015 | (1011+971+978+1015)/4 = 987.5 | 1015 - 971 = 44 | | 9 | 990, 973, 1002, 103 | (990+973+1002+103)/4 = 998.8 | 103 - 973 = 57 | | 10 | 983, 1002, 1005, 1001| (983+1002+1005+1001)/4 = 994.3 | 1005 - 983 = 22 | | 11 | 101, 1001, 978, 900 | (101+1001+978+900)/4 = 972.3 | 101 - 900 = 110 | | 12 | 994, 974, 977, 990 | (994+974+977+990)/4 = 983.8 | 994 - 974 = 20 | | 13 | 999, 983, 974, 103 | (999+983+974+103)/4 = 995. | 103 - 974 = 56 | | 14 | 1002, 1008, 991, 980| (1002+1008+991+980)/4 = 995.3 | 1008 - 980 = 28 | | 15 | 986, 1023, 1006, 989| (986+1023+1006+989)/4 = 1001. | 1023 - 986 = 37 | --- ## **2. Determine \( \bar{X} \) and R** - **Grand Mean (\( \bar{X} \))**: Average of all sample means. \[ \bar{X} = \frac{\sum \text{Sample Means}}{15} = \frac{1021.8 + 994.3 + 1008.8 + 1011.8 + 987.5 + 1005. + 991.3 + 987.5 + 998.8 + 994.3 + 972.3 + 983.8 + 995. + 995.3 + 1001.}{15} \approx 995.6 \] - **Range (R)**: Max of all sample ranges, which is 110. --- ## **3. Determine UCL and LCL for an X-chart** Using standard formulas: \[ UCL = \bar{X} + 3 \times \frac{R_{avg}}{d_2} \] \[ LCL = \bar{X} - 3 \times \frac{R_{avg}}{d_2} \] Where \( R_{avg} = 43+25+22+42+30+39+39+44+57+22+110+20+56+28+37 \div 15 \approx 44.2 \) For 4 samples per subgroup, \( d_2 \approx 2.059 \). \[ \text{UCL} \approx 995.6 + 3 \times \frac{44.2}{2.059} \approx 995.6 + 3 \times 21.5 \approx 995.6 + 64.5 \approx 106.1 \] \[ \text{LCL} \approx 995.6 - 64.5 \approx 931.1 \] --- ## **4. Determine UCL and LCL for an R-chart** \[ UCL_R = R_{avg} \times D_4 \approx 44.2 \times 2.282 \approx 100.9 \] \[ LCL_R = R_{avg} \times D_3 \approx 44.2 \times \approx \] (Note: \( D_3 = \) for sample size 4) --- ## **5. Comments about the process** - **The process is in statistical control** if all sample points lie within the control limits and no patterns or trends are evident. - **The process appears stable** based on the calculations, with most points within the control limits and no visible patterns. --- ## **Final Summary** - The mean of the samples is approximately **995.6**. - The overall range of sample ranges is about **44.2**. - The control limits for the X-chart are approximately **UCL = 106.1** and **LCL = 931.1**. - The control limits for the R-chart are approximately **UCL = 100.9** and **LCL = **. - The process appears to be **in statistical control** based on the data and control limits. --- Let me know if you'd like me to fill in the actual table or provide further explanations!