Aspen Plastics produces plastic bottles to customer order. The quality inspector randomly selects four bottles from the bottle machine and measures the outside diameter of the bottle neck, a critical quality dimension that determines whether the bottle cap will fit properly. The dimensions (in.) from the last six samples are Bottle Sample123410.5920.6120.5980.62320.5840.5720.5840.57330.5960.6220.6180.58640.6060.6140.5850.59850.5820.5860.6080.61160.5830.5710.6050.625 LOADING...Click the icon to view the table of factors for calculating three-sigma limits for thex-chartand R-chart.Part 2Suppose that the specification for the bottle neck diameter is0.600±0.050in. and the population standard deviation is0.013in. a. What is the process capability index? TheCpkisenter your response here.(Enteryour response rounded to two decimalplaces.)

The given data is Bottle Sample 1 2 3 4 1 0.592 0.612 0.598 0.623 2 0.584 0.572 0.584 0.573 3 0.596 0.622 0.618 0.586 4 0.606 0.614 0.585 0.598 5 0.582 0.586 0.608 0.611 6 0.583 0.571 0.605 0.625a) The process capability index is Cpk=minUSL-μ3σ,μ-LSL3σ, where USL is the upper specification limit and LSL is the lower specification limit. Given that the specification for the bottle neck diameter is 0.600±0.050 in. So, LSL=0.600-0.050=0.550 and USL=0.600+0.050=0.650 The population standard deviation is 0.013 in. So, σ=0.013. Note that here μ is not given. So, we have to estimate it using the given data. The estimate of μ is x¯¯=∑i=1m∑j=1nxijmn, where xij is the diameter of the jth bottle neck in the ith sample,j=1,2,...,n and i=1,2,...,m. m is the total number of samples and n is the size of each sample.For the given data, m=6 and n=4. Bottle Sample 1 2 3 4 Sum 1 0.592 0.612 0.598 0.623 2.425 2 0.584 0.572 0.584 0.573 2.313 3 0.596 0.622 0.618 0.586 2.422 4 0.606 0.614 0.585 0.598 2.403 5 0.582 0.586 0.608 0.611 2.387 6 0.583 0.571 0.605 0.625 2.384 So, ∑i=1m∑j=1nxij=2.425+2.313+2.422+2.403+2.387+2.384=14.334 Now, x¯¯=∑i=1m∑j=1nxijmn=14.3346×4=0.59725 The process capability index is Cpk=minUSL-μ3σ,μ-LSL3σ=minUSL-x¯¯3σ,x¯¯-LSL3σ=min0.650-0.597253×0.013,0.59725-0.5503×0.013≈min(1.35,1.21)≈1.21 The Cpk is 1.21.