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:
Step-by-step explanation:
Forecast for period 1 is 5
Demand For Period 1 is 7
Demand for Period 2 is 9
Forecast can be given by

where





Forecast for Period 3


The answer is -7
The arrows both point toward the left side, and sine there are seven spaces the arrows are going through, you subtract 7 spaces and get -7
Hope that helps.
Answer:

Step-by-step explanation:





Remark
It looks like all you want is question 6. If that is the case, there are two ways to do it.
Algebra
<u>First answer</u>
abs(b - 22) = 5 Equate to + 5
b - 22 = 5 Add 22 to both sides.
b = 5 + 22
b = 27
<u>Second Answer</u>
Equate to - 5
b - 22 = -5
b = 22 - 5
b = 17
Method Two
<u>Graph the question</u>
The graph y = abs(b - 22) is shown below in red.
The values of y when y =5 are shown in blue.