MATH SOLVE

4 months ago

Q:
# The line width 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. Round your answers to 3 decimal places (e.g. 98.765). What is the probability that a line width is greater than 0.62 micrometer?

Accepted Solution

A:

Answer:There is a 0.82% probability that a line width is greater than 0.62 micrometer.Step-by-step explanation:Problems of normally distributed samples can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by[tex]Z = \frac{X - \mu}{\sigma}[/tex]After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.In this problemThe line width 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, so [tex]\mu = 0.5, \sigma = 0.05[/tex]. What is the probability that a line width is greater than 0.62 micrometer?That is [tex]P(X > 0.62)[/tex]So[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{0.62 - 0.5}{0.05}[/tex][tex]Z = 2.4[/tex]Z = 2.4 has a pvalue of 0.99180.This means that P(X \leq 0.62) = 0.99180.We also have that[tex]P(X \leq 0.62) + P(X > 0.62) = 1[/tex][tex]P(X > 0.62) = 1 - 0.99180 = 0.0082[/tex]There is a 0.82% probability that a line width is greater than 0.62 micrometer.