Suppose that a quality characteristic is normally distribute

Suppose that a quality characteristic is normally distributed with specification limits (1.64, 1.84). The process standard deviation is 0.1. Suppose that the process mean is 1.71 Determine the natural tolerance limits. Calculate the fraction defective. Calculate the process capability ratio.

Solution

We are given that a quallity characteristicis normally distributed with specification limits (1.64, 1.84).

process standard deviation (sd) = 0.1

process mean (mean) = 1.71

a) natural tolerance limits :

Natural tolerances are the control limits placed at three times the standard deviation from the process average. These limits are some times refered to as Three Sigma Limits.

lower limit = mean - 3*sd = 1.41

upper limit = mean + 3*sd = 2.01

The natural tolerance limits are (1.41, 2.01).

Fraction defective :

Specification limits are defined by your customer and represent the desired performance of your process.

fraction defevtive (p) = D/n

where D is binomial random variable with unknown parameter p.

Process capability ratio, Cp, is the tolerance width divided by 6 standard deviations (process variability).

Cp = (Upper specification - Lower specification) / 6*sd

Cp = (2.01-1.41) / 6*0.1

Cp = 1