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:
On the graphing calculator, use the function normCdf, where
- lower bound = -9999
- upper bound = 210
- mean = 250
- standard deviation = 46
It will result in normCdf(-9999,210,250,46) ≈ 0.192269 or 19.2269%
The correct answer is y=2/3x+4 which is the last option :)
Answer:
Step-by-step explanation:
Given
Midpoint is 
Endpoints are
and 
Mid point of any two point is given by

and 
So,


Also



Therefore 