Sampling under normal distribution

Question

The concentration of impurities in a semiconducter in the production of microprocessors for computer is a normally distributed random variable with mean 127 parts per million and standard deviation 22 parts per million. A semiconducter is acceptable only if its concentration of impurities is below 150 parts per million.

What is the proportion of semiconducters that are acceptable for use?(the area under normal curve for the value of $Z=1.5$ is $0.668$).

Answer 1

Note that the acceptance value (150 ppm) is higher than the mean - which it had better be, if the supplier expects most of his materials to be accepted.

Note also that in the case were the impurities are < 127 ppm - or the lower half of the distribution - all the materials will pass. That lower half of the distribution - 50% of the materials - will always pass.

Now what about those materials on the upper half of the distribution - with impurities > 127 ppm? Only the fraction between 127 and 150 will pass. How many standard deviations is that from the mean?

$z = \frac{150-127}{22} = 1.0455$

What fraction of the cumulative normal distribution is that?

$P(0 < Z \leq 1.0455) = \Phi (1.0455) - \Phi (0) = 0.8508-0.5 = 0.3508$

So what is the total fraction that will pass?

This is the sum of the area where $P(Z \leq 0) = 0.5$ and the area where $P(0 < Z \leq 1.0455)=0.3508$. So then:

$P(Z \leq 0) + P(0 < Z \leq 1.0455) = 0.5 + 0.3588 = 0.8588$

This can be done in one step: $P(Z \leq 1.0455) = \Phi (1.0455) = 0.8588$

So why did I go to all the trouble of first solving it in two parts?

1. To give you better intuition into the answer by showing which parts of the area under the standard Normal curve that used.
2. Many times there are both a lower acceptance limit and and upper acceptance limit. Showing the answer in parts should help you solve the problem when there are an upper and lower limit.