Introduction satistics

Question

The diameter of a shaft in an optical storage drive is normally distributed with a mean of 0.2508 inch and a standard deviation of 0.0005 inch. The specifications on the shaft are 0.2500± 0.0015 inch. What proportions of the shafts conforms to the specification?

Answer 1

Let X be the random variable denoting the diameter of a shaft selected at random. Then

P(0.2500-0.0015 ≤ X ≤ 0.2500+0.0015)

= P(0.2485 ≤ X ≤ 0.2515)

From here, most calculators and software will let you compute this directly. On a TI-84, use 2nd > VARS > normalcdf and enter normalcdf(lower bound = 0.2485, upper bound = 0.2515, μ = 0.2508, σ = 0.0005)≈0.92.

In Excel, use =NORM.DIST(0.2515,0.2508,0.0005,TRUE)-NORM.DIST(0.2485,0.2508,0.0005,TRUE)≈0.92.

The proportion of shafts within spec is about 92/100.