Question

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?

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

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 problem

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, so $\mu = 0.5, \sigma = 0.05$.

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

That is $P(X > 0.62)$

So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.62 - 0.5}{0.05}$

$Z = 2.4$

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

$P(X \leq 0.62) + P(X > 0.62) = 1$

$P(X > 0.62) = 1 - 0.99180 = 0.0082$

There is a 0.82% probability that a line width is greater than 0.62 micrometer.