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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 and standard deviation
, the zscore of a measure X is given by
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 .
What is the probability that a line width is greater than 0.62 micrometer?
That is
So
Z = 2.4 has a pvalue of 0.99180.
This means that P(X \leq 0.62) = 0.99180.
We also have that
There is a 0.82% probability that a line width is greater than 0.62 micrometer.
Answer:
The value of g(a) is:
- g(a) = 3 - 2a
Step-by-step explanation:
Given
As the function g(x) is given by
To determine
g(a) = ?
In order to determine g(a),
substitute x = a in the function g(x) = 3 - 2x
g(x) = 3 - 2x
g(a) = 3 - 2(a)
g(a) = 3 - 2a
Therefore, the value of g(a) is:
- g(a) = 3 - 2a