Question

The line width for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometers what is the probability that a line width is greater than 0.62 micrometer?

Answer 1

To solve the question we first calculate the z-score:
z=(x-μ)/σ
where:
μ-mean
σ-standard deviation
thus from the information given we shall have:
z=(0.62-0.5)/0.05
z=2.4
Thus
P(x>0.62)=1-P(x<0.62)
=1-P(z=2.4)=1-0.9918
=0.0082