The line width of a tool used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micromete

Question

3 months ago

15

The line width of a tool used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micromete

r and a standard deviation of 0.05 micrometer.
What is the probability that a line width is greater than 0.62 micrometer?
What is the probability that a line width is between 0.47 and 0.63 micrometer?
The line width of 90% of samples is below what value?

Mathematics

1 answer:

zzz [12.3K] · Answer 1 · 2026-02-20T16:07:07+03:00

Answer:

a. 0.82%

b. 71.11%

c. 0.564 micrometer

Step-by-step explanation:

To find the z value for each measurement, we must calculate and determine the percentage they correspond to, and the difference will give us the percentage between those two statistics.

The equation for z is:

z = (x - m) / (sd)

Here, x is the value being evaluated, m denotes the mean, and sd represents the standard deviation.

a.

For 0.62 copies, we calculate:

z = (0.62 - 0.5) / (0.05)

z = 2.4

This translates to 0.9918.

Therefore, p (x > 0.62) = 1 - 0.9918

p (x > 0.62) = 0.0082 = 0.82%

b.

For 0.47 copies, we have:

z = (0.47 - 0.5) / (0.05)

z = -0.6, which equates to 0.2742.

For 0.63 copies:

z = (0.63 - 0.5) / (0.05)

z = -2.6, which yields 0.9953.

Hence, p (0.47 > x > 0.63) = 0.9953 - 0.2742

p (0.47 > x > 0.63) = 0.7211 = 72.11 %

c.

For x copies, we find:

p = 0.9 corresponds to z = 1.28.

Thus, 1.28 = (x - 0.5) / (0.05)

From which we derive:

x = 1.28*0.05 + 0.5

x = 0.564 micrometer