A process at Printwright, Inc.’s largest facility is used to make plastic gears for a computer printer. The data found in the worksheet Prob. 9-15 were gathered by a quality analyst. The gears were designed to be 3.57 ` 0.05 centimeters (cm) in diameter. Construct a histogram of the data. What can you observe about the shape of the dis- tribution? What would you recommend to the production manager based on your analysis? **Problem 9-15: Plastic Gear Data for Printwright, Inc.**This dataset provides a set of measurements related to plastic gears produced by Printwright, Inc. The table comprises several columns with numerical values representing various data points for analysis.---### Data Table:- **Row 1**: Header - "Problem 9-15" - **Row 2**: Subtitle - "Plastic Gear Data for Printwright, Inc."- **Rows 3 to 33**: Measurement Data - Each row contains five numerical values. - Example entries: - Row 3: 3.52, 3.55, 3.55, 3.55, 3.63 - Row 4: 3.58, 3.55, 3.63, 3.55, 3.55 - ... - Row 33: 3.57, 3.58, 3.55, 3.55, 3.52---### Overview:The dataset appears structured for statistical analysis or quality control inspection, potentially analyzing measurements such as dimensions or tolerances of the gears. Each row represents a set of recorded measurements, which could pertain to specific batches or production runs. The uniformity and variation of these measurements may be evaluated to ensure the gears meet Printwright Inc.'s specifications and quality standards.### Analysis Opportunities:- **Quality Control**: Assess the consistency within each row and compare across multiple rows to identify trends and deviations.- **Statistical Analysis**: Calculate mean, median, variance, and standard deviation for each set of measurements.- **Graphical Representation**: Create histograms or charts to visualize data distribution for insights into manufacturing precision.This data set provides a basis for deeper exploration into manufacturing quality and precision for Printwright, Inc.'s plastic gear production.

A)(b) The shape of the histogram approximates a bell-shaped distribution. Therefore, we can perhaps assume a normality.(c) The sample mean = 3.57 (an estimate for the process mean)Sample standard deviation = 0.0346 (an estimate for the process std. dev.) USL = 3.57 + 0.05LSL = 3.57 - 0.05 Cp = (USL - CSL) / (6*process std. dev.) = 0.1 / (6*0.0346) = 0.48. Since this is < 1.0, the process is hihly incapble of producing the output in the desired specification limit.