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 micrometer. a. What is the probability that a line width is greater than 0.62 micrometer? b. What is the probability that a line width is between 0.4 and 0.63 micrometer? c. The line width of 90% of samples is below what value?

a) 0.0081975

b) 0.97259

Step-by-step explanation:

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 micrometer

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

a. What is the probability that a line width is greater than 0.62 micrometer?

z = 0.62 - 0.5/0.05

z =2.4

Probability value from Z-Table:

P(x<0.62) = 0.9918

P(x>0.62) = 1 - P(x<0.62)

= 0.0081975

b. What is the probability that a line width is between 0.4 and 0.63 micrometer?

For 0.4

z = 0.4 - 0.5/0.05

= -2

Probability value from Z-Table:

P(x = 0.4) = 0.02275

For 0.63

z = 0.63 - 0.5/0.05

= 2.6

Probability value from Z-Table:

P(x = 0.63) = 0.99534

P(x = 0.63) - P(x = 0.4)

= 0.99534 - 0.02275

= 0.97259

c. The line width of 90% of samples is below what value?

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation: