Now, select the blue pencil in the Compass Tool and create a perpendicular bisector of the diameter of your circle. Place the ce
nter of the compass on one of the points where the diameter intersects the circle. Adjust the compass so that the pencil tip extends beyond the center of the circle. Draw an arc through the circle. Repeat this on the other side without changing the radius of your compass tool. Using the straightedge tool, draw a line through the intersections of the two arcs. (8 points: 4 points for correctly constructing the arcs, 4 points for constructing the bisector)
1 answer:
You might be interested in
Given
This is a one-tailed test.
Since the p-value of the sample statistic (0.00383) is less that the significant level (0.025), we reject the null hypothesis.
This is a one-tailed test.
Since the p-value of the sample statistic (0.00383) is less that the significant level (0.025), we reject the null hypothesis.
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.