Question

Suppose that material hardness is normally distributed with a mean of 52 and a standard deviation of 1. Specification limits for hardness are from 45 to 55. What is the fraction defective? What value for the process mean will minimize the fraction defective? When the fraction defective is 0.0027 this corresponds to what PPM?

Answer 1

part i)

A material is defective if it's hardness lies outside the specification limit. Hence the estimate of fraction defective is,

Fraction defective P(X < LSL)P(X > USL)

Since material harness is normally distributed, the required probability is obtained by applying central limit theorem,

USL- LSL- +P Z> PZ < P(X < LSL)+P(X> USL)

USL- LSL- 1- PZ< PZ < P(X < LSL)+P(X> USL)

45 52 55 52 P(X <LSL)+P(X> USL) PZ< 1 PZ 1

P(X<LSL) P(X> USL) P (Z-7) (1- P (Z < 3))

Using the standard normal distribution table for z<-7 and z<3, the probabilities are

PX<LSL) P(X> USL) 0.00000+(1- 0.99865) 0.00135

part ii)

The process mean will minimize the fraction defective if it lies at the center of specification such that process mean = (45+55)/2=50

part iii)

Fraction defective 0.0027 0.0027 x 100000 27OPPM