Question

Assume that the mean of the engineered part is 79.95 and standard deviation of 0.017291, with a lower specification limit of 79.9 and upper specification limit of 80.

a. Determine the probability that a part is defective.

b. Suppose that 1,000,000 parts are produced, what is the expected number of defective units? This is equivalent to finding the PPM (parts per million).

Answer 1

Following are the given data :

Upper specification Limit ( USL ) = 80

Lower specification Limit ( LSL ) = 79.9

Mean of engineering part = m = 79.95

Standard deviation of engineering part = Sd = 0.017291

Let z value corresponding to probability that engineering part dimension is maximum upto USL i.e. 80 = Z1

And

Z value corresponding to probability that engineering part dimension is maximum upto LSL i.e. 79.9 = Z2

Therefore ,

M + Z1 x Sd = 80

Or, 79.95 + 0.017291.Z1 = 80

Or Z1 = 0.05/0.017291

Or, Z1 = 2.89

Corresponding probability for Z1 = 2.89 as derived from standard normal distribution table 0.99807

Similarly ,

M + 0.017291.Z2 = 79.90

Or, 79.95 + 0.017291.Z2 = 79.90

Or, Z2 = - 0.05/0.017291

Or, Z2 = - 2.89

Corresponding probability for Z2 of – 2.89     will be 0.00193

Therefore , Probability that a part is not defective

= Probability that dimension of a part is within LSL- USL range

= Probability that dimension is maximum upto USL – Probability that dimension is maximum upto LSL

= 0.99807 – 0.00193

= 0.99614

Therefore , that the part is defective = 1 – probability that the part is not defective = 1 – 0.99614 = 0.00386

PROBABILITY THAT A PART IS DEFECTIVE = 0.00386

Expected number of defective units in 1 million units

= 1,000,000 x 0.00386

= 3860

EXPECTED NUMBER OF DEFECTIVE UNITS = 3860