Question

A quality-control technician has been monitoring the output of a milling machine. Each day, the technician selects a random sample of 20 parts to measure and plot on the control chart. Over 10 days, the average diameter was 1.327 millimeters with a standard deviation of 0.0525 millimeters. What is the lower control limit (LCL) for an X-bar chart of this data? ________ millimeters

Answer 1

Answer: 1.292 millimeters

Step-by-step explanation:

Given the following ;

Number of Random sample (n) = 20

Average diameter(mean) = 1.327 millimeters

Standard deviation(std) = 0.0525 millimeters

To calculate the Lower Control Limit (LCL) :

Recall:

LCL = mean - [3(std÷sqrt(n))]

LCL = 1.327 - [3(0.0525÷sqrt(20))]

LCL = 1.327 - [3(0.0525 ÷ 4.472135955)]

LCL = 1.327 - [3 × 0.0117393568818]

LCL = 1.327 - 0.0352180706456216

LCL = 1.2917819293543784

LCL = 1.292 millimeters