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

a. 0.82%

b. 71.11%

c. 0.564 micrometer

Step-by-step explanation:

What we must do is calculate the z value for each value and thus find what percentage each represents and the subtraction would be the percentage between those two values.

We have that z is equal to:

z = (x - m) / (sd)

x is the value to evaluate, m the mean, sd the standard deviation

a.

So for 0.62 copies we have:

z = (0.62 - 0.5) / (0.05)

z = 2.4

and this value represents 0.9918

p (x > 0.62) = 1 - 0.9918

p (x > 0.62) =  0.0082 = 0.82%

b.

So for 0.47 copies we have:

z = (0.47 - 0.5) / (0.05)

z = -0.6

and this value represents 0.2742

So for 0.63 copies we have:

z = (0.63 - 0.5) / (0.05)

z = -2.6

and this value represents 0.9953

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

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

c.

So for x copies we have:

p = 0.9, It represents z =1.28

1.28 = (x - 0.5) / (0.05)

x = 1.28*0.05 + 0.5

x = 0.564 micrometer

asdsaddsa

step-by-step explanation: